Answer:
The probability of the stick's weight being 2.33 oz or greater is 0.41%.
Step-by-step explanation:
Test statistic (z) = (weight - mean)/sd
weight of stick = 2.33 oz
mean = 1.75 oz
sd = 0.22 oz
z = (2.33 - 1.75)/0.22 = 2.64
The cumulative area of the test statistic is the probability that the weight is 2.33 oz or less. The cumulative area is 0.9959.
The probability the weight is 2.33 or greater = 1 - 0.9959 = 0.0041 = 0.41%
The total number of parking spaces in a parking garage is calculated by adding the area of the lower level given by 22x2, the area of the upper level given by 20x2, and a compact car section given by 12x for a total of 414 parking spaces. Which equation could be used to solve for the number of compact car parking spaces?
A) 22x2 − 20 x2 − 12x = 414
B) 22x2 − 20 x2 + 12x = 414
C) 22x2 + 20 x2 − 12x = 414
D) 22x2 + 20 x2 + 12x = 414
Answer:
D
Step-by-step explanation:
Determine whether each pair of triangles is similar. If yes, state the similarity property that supports it, if not, explain why.
Answer:
Step-by-step explanation:
If two triangles are equal, it means that the ratio of the length of each side of one triangle to the length of the corresponding side of the other triangle is constant. Also, corresponding angles are congruent.
1) Triangle TUV is similar to triangle SQR because
Angle Q is congruent to angle U
TU/SQ = UV/QR = 2
2) Triangle ABC is not similar to triangle DEF because the ratio of AB to DF is not constant.
Consider the probability that no less than 92 out of 159 students will pass their college placement exams. Assume the probability that a given student will pass their college placement exam is 55%. Approximate the probability using the normal distribution. Round your answer to four decimal places.
Answer:
0.2843 = 28.43% probability that no less than 92 out of 159 students will pass their college placement exams.
Step-by-step explanation:
I am going to use the binomial approximation to the normal to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]p = 0.55, n = 159[/tex]. So
[tex]\mu = E(X) = 159*0.55 = 87.45[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{159*0.55*0.45} = 6.27[/tex]
Probability that no less than 92 out of 159 students will pass their college placement exams.
No less than 92 is more than 91, which is 1 subtracted by the pvalue of Z when X = 91. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{91 - 87.45}{6.27}[/tex]
[tex]Z = 0.57[/tex]
[tex]Z = 0.57[/tex] has a pvalue of 0.7157
1 - 0.7157 = 0.2843
0.2843 = 28.43% probability that no less than 92 out of 159 students will pass their college placement exams.
To approximate the probability that at least 92 out of 159 students will pass their college placement exams using the normal distribution, we will follow these steps:
Step 1: Determine the parameters of the binomial distribution.
In a binomial distribution the parameters are the number of trials (n) and the probability of success in a single trial (p). For this particular case:
n = 159 (the number of students)
p = 0.55 (the probability that a given student will pass)
Step 2: Calculate the mean (μ) and standard deviation (σ) of the binomial distribution.
The mean (μ) of a binomial distribution is given by:
μ = n * p
The standard deviation (σ) is given by:
σ = sqrt(n * p * (1 - p))
For our case:
μ = 159 * 0.55 ≈ 87.45
σ = sqrt(159 * 0.55 * (1 - 0.55)) ≈ sqrt(159 * 0.55 * 0.45) ≈ sqrt(71.775) ≈ 8.472 (rounded to three decimal places for intermediate calculation)
Step 3: Apply the continuity correction.
Because we're approximating a discrete distribution with a continuous one, we use a continuity correction. To find the probability of at least 92 students passing, we'll look for the probability of X > 91.5 (since X is a discrete random variable, X ≥ 92 is equivalent to X > 91.5 when using a continuous approximation).
Step 4: Calculate the z-score for the corrected value.
The z-score is calculated by taking the value of interest, applying the continuity correction, subtracting the mean, and then dividing by the standard deviation:
z = (X - μ) / σ
Using the corrected value:
z = (91.5 - 87.45) / 8.472 ≈ 0.4774 (rounded to four decimal places for the calculation)
Step 5: Find the corresponding probability.
The z-score tells us how many standard deviations away from the mean our value of interest is. To find the probability that at least 92 students pass (i.e., P(X ≥ 92), we need to find 1 - P(Z < z) because the normal distribution table or a calculator gives us P(Z < z), which is the probability of being less than a certain z value.
We are looking for the probability that z is greater than our calculated value, which is the upper tail of the distribution.
Step 6: Consult the standard normal distribution table or use a calculator.
The z-score of approximately 0.4774 corresponds to a percentage in the standard normal distribution. Since we want P(Z > z), we need to subtract P(Z < z) from 1. If we look up the probability for z=0.4774 in standard normal distribution tables or use a calculator, we find that P(Z < z) is approximately 0.6832.
Therefore, P(Z > z) = 1 - P(Z < z) = 1 - 0.6832 = 0.3168.
Step 7: Round the answer.
Rounding P(Z > z) = 0.3168 to four decimal places, the answer remains 0.3168.
So, the approximate probability that at least 92 out of 159 students will pass the exam, using the normal distribution approximation, is 0.3168 when rounded to four decimal places.
A river is 500 meters wide and has a current of 1 kilometer per hour. if tom can swim at a rate of 2 kilometers per hour at what angle to the shore should he swim if he wishes to cross the river to a point directly opposite bank
Answer:
[tex]63.44^{0}[/tex]
Step-by-step explanation:
Tom's resultant speed is calculated as [tex]\sqrt{1^{2}+2^{2} }[/tex]
= [tex]\sqrt{5}[/tex]
The distance tom swim is the hypotenuse of the right angle triangle.
sin = opposite/hypotenuse = [tex]\frac{1}{\sqrt{5} }[/tex] = 1/5 = 0.2
arcsin (0.2) = [tex]26.56^{0}[/tex]
upstream angle =
[tex]90^{0} - 26.56^{0}[/tex] =
[tex]63.44^{0}[/tex]
A website manager has noticed that during the evening hours, about 5 people per minute check out from their shopping cart and make an online purchase. She believes that each purchase is independent of the others and wants to model the number of purchases per minute. a) What model might you suggest to model the number of purchases per minute? b) What is the probability that in any one minute at least one purchase is made? c) What is the probability that seven people make a purchase in the next four minutes?
Answer:
a) Poisson distribution
b) 99.33% probability that in any one minute at least one purchase is made
c) 0.05% probability that seven people make a purchase in the next four minutes
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given time interval.
5 people per minute check out from their shopping cart and make an online purchase.
This means that [tex]\mu = 5[/tex]
a) What model might you suggest to model the number of purchases per minute?
The only information that we have is the mean number of an event(purchases) in a time interval. Each event is also independent fro each other. So you should suggest the Poisson distribution to model the number of purchases per minute.
b) What is the probability that in any one minute at least one purchase is made?
Either no purchases are made, or at least one is. The sum of the probabilities of these events is 1. So
[tex]P(X = 0) + P(X \geq 1) = 1[/tex]
We want to find [tex]P(X \geq 1)[/tex]
So
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
In which
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-5}*(5)^{0}}{(0)!} = 0.0067[/tex]
1 - 0.0067 = 0.9933.
99.33% probability that in any one minute at least one purchase is made
c) What is the probability that seven people make a purchase in the next four minutes?
The mean is 5 purchases in a minute. So, for 4 minutes
[tex]\mu = 4*5 = 20[/tex]
We have to find P(X = 7).
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-20}*(20)^{7}}{(7)!} = 0.0005[/tex]
0.05% probability that seven people make a purchase in the next four minutes
The Poisson distribution model is used when the data consist of counts of occurrences.
a) The Poisson distribution model is used when the data consist of counts of occurrences.
b) Given that: λ (mean number of occurrence) = 5 people per minute, hence:
[tex]P(X\ge 1)=1-P(X=0)=1-\frac{e^{-\lambda }\lambda^x}{x!}= 1-\frac{e^{-5 }5^0}{5!}=0.9999[/tex]
The probability that in any one minute at least one purchase is made is 0.9999.
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Evaluation of Proofs See the instructionsfor Exercise (19) on page 100 from Section 3.1. (a) Proposition. If m is an odd integer, then .mC6/ is an odd integer. Proof. For m C 6 to be an odd integer, there must exist an integer n such that mC6 D 2nC1: By subtracting 6 from both sides of this equation, we obtain m D 2n6C1 D 2.n3/C1: By the closure properties of the integers, .n3/ is an integer, and hence, the last equation implies that m is an odd integer. This proves that if m is an odd integer, then mC6 is an odd integer
Answer:
(A) Assume m is an odd integer.
Therefore m is of the m=2n-1 type where n is any number.
We have m + 6 = 2n-1 + 6 = 2n+5=2n+4 + 1 = 2(n+2) + 1, in which n + 2 is an integer, adding 6 to both ends.
Because m+6 is of the 2x + 1 type, where x = n + 2; then m + 6 is an odd integer too.
(B) Provided that mn is an integer also. For some integer y and m, n are integers, this means mn is of the form mn = 2y.
And y = mn/2.
Therefore either m is divided by 2 or n is divided by 2 since y, m, n are all integers. To put it another way, either m is a multiple of 2 or n is a multiple of 2, which means that m is even or n is even.
Final answer:
The proposition regarding odd integers is proven by correcting the initial flawed proof, demonstrating that adding 6 to an odd integer results in another odd integer by using the correct form of an odd integer in the calculation.
Explanation:
The given proposition states that adding 6 to an odd integer results in another odd integer. The proof starts with the assumption that there exists an integer n such that m + 6 = 2n + 1.
This equation is supposed to define m + 6 as an odd integer. By isolating m, the proof continues to show that m itself is expressed in the form of 2(n - 3) + 1, indicating that m is also an odd integer.
To correct the proof, we should start with an odd integer m such that m = 2k + 1, where k is an integer. Adding 6 to m gives m + 6 = 2k + 7, which can be written as 2(k + 3) + 1, clearly showing that m + 6 is an odd integer.
Therefore, this proves the original proposition correctly.
A classic counting problem is to determine the number of different ways that the letters of "possession" can be arranged. Find that number. The number of different ways that the letters of "possession" can be arranged is nothing.
Number of possible ways to arrange letters of word "possession" = 75600.
Step-by-step explanation:
A classic counting problem is to determine the number of different ways that the letters of "possession" can be arranged. We have, word "POSSESSION" which have 10 letters! as P,O,S,S,E,S,S,I,O,N . We have to find the number of ways letters of word "possession" can be arranged , in order to do that we will use simple logic as:
At first, we have a count of 10 letters and 10 places vacant to fill letters.
For first place we have a choice of 10 letters , after putting some letter from all 10 letters in first place now we are left with 9 places and 9 letters having 9 choices , similarly we'll be having 8 places , 8 letters and 8 choices and so on...... Therefore, Number of possible ways to arrange letters of word "possession" = [tex]10.9.8.7.6.5.4.3.2.1[/tex] = [tex]10![/tex] ( 10 factorial ) , but there are 4 letters "s" are repeated and 2 letters "o" are repeated so we need to eliminate the similar combinations ∴ Number of possible ways = [tex]\frac{10!}{4!(2!)}[/tex] = 75600.
∴Number of possible ways to arrange letters of word "possession" = 75600
The time that it takes a randomly selected employee to perform a certain task is approximately normally distributed with a mean value of 120 seconds and a standard deviation of 20 seconds. The slowest 10% (that is, the 10% with the longest times) are to be given remedial training. What times (the lowest value) qualify for the remedial training?
Answer:
145.6 seconds
Step-by-step explanation:
Mean time (μ) = 120 seconds
Standard deviation (σ) = 20 seconds
In a normal distribution, the z-score for any time, X, is given by:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
The slowest 10% correspond to the 90th percentile of a normal distribution, which has a corresponding z-score of roughly 1.28. The lowest time that requires remedial training is:
[tex]1.28=\frac{X-120}{20}\\X=145.6\ seconds[/tex]
Times of 145.6 seconds and over qualify for remedial training.
The data file wages contains monthly values of the average hourly wages (in dollars) for workers in the U.S. apparel and textile products industry for July 1981 through June 1987.
a. Display and interpret the time series plot for these data.
b. Use least squares to fit a linear time trend to this time series. Interpret the regression output. Save the standardized residuals from the fit for further analysis.
c. Construct and interpret the time series plot of the standardized residuals from part (b).
d. Use least squares to fit a quadratic time trend to the wages time series. (i.e y(t)=βo+β1t+β2t^2+et). Interpret the regression output. Save the standardized residuals from the fit for further analysis.
e. Construct and interpret the time series plot of the standardized residuals from part (d).
Answer:
a. data(wages)
plot(wages, type='o', ylab='wages per hour')
Step-by-step explanation:
a. Display and interpret the time series plot for these data.
#take data samples from wages
data(wages)
plot(wages, type='o', ylab='wages per hour')
see others below
b. Use least squares to fit a linear time trend to this time series. Interpret the regression output. Save the standardized residuals from the fit for further analysis.
#linear model
wages.lm = lm(wages~time(wages))
summary(wages.lm) #r square is correct
##
## Call:
## lm(formula = wages ~ time(wages))
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.23828 -0.04981 0.01942 0.05845 0.13136
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -5.490e+02 1.115e+01 -49.24 <2e-16 ***
## time(wages) 2.811e-01 5.618e-03 50.03 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.08257 on 70 degrees of freedom
## Multiple R-squared: 0.9728, Adjusted R-squared: 0.9724
## F-statistic: 2503 on 1 and 70 DF, p-value: < 2.2e-16
c. plot(y=rstandard(wages.lm), x=as.vector(time(wages)), type = 'o')
d. #we find Quadratic model trend
wages.qm = lm(wages ~ time(wages) + I(time(wages)^2))
summary(wages.qm)
##
## Call:
## lm(formula = wages ~ time(wages) + I(time(wages)^2))
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.148318 -0.041440 0.001563 0.050089 0.139839
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -8.495e+04 1.019e+04 -8.336 4.87e-12 ***
## time(wages) 8.534e+01 1.027e+01 8.309 5.44e-12 ***
## I(time(wages)^2) -2.143e-02 2.588e-03 -8.282 6.10e-12 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.05889 on 69 degrees of freedom
## Multiple R-squared: 0.9864, Adjusted R-squared: 0.986
## F-statistic: 2494 on 2 and 69 DF, p-value: < 2.2e-16
#time series plot of the standardized residuals
plot(y=rstandard(wages.qm), x=as.vector(time(wages)), type = 'o')
wages.qm = lm(wages ~ time(wages) + I(time(wages)^2))
summary(wages.qm)
##
## Call:
## lm(formula = wages ~ time(wages) + I(time(wages)^2))
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.148318 -0.041440 0.001563 0.050089 0.139839
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -8.495e+04 1.019e+04 -8.336 4.87e-12 ***
## time(wages) 8.534e+01 1.027e+01 8.309 5.44e-12 ***
## I(time(wages)^2) -2.143e-02 2.588e-03 -8.282 6.10e-12 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.05889 on 69 degrees of freedom
## Multiple R-squared: 0.9864, Adjusted R-squared: 0.986
## F-statistic: 2494 on 2 and 69 DF, p-value: < 2.2e-16
e. #time series plot of the standardized residuals
plot(y=rstandard(wages.qm), x=as.vector(time(wages)), type = 'o')
A local board of education conducted a survey of residents in the community concerning a property tax levy on the coming local ballot. They randomly selected 850 residents in the community and contacted them by telephone. Of the 850 residents surveyed, 410 supported the property tax levy. Let p represent the proportion of residents in the community that support the property tax levy.
A 90% confidence interval for p is (Use decimal notation. Give value to four decimal places and "z" value to three decimal places.)
A. 0.4489 to 0.5159.
B. 0.4542 to 0.5105.
C. 0.4487 to 0.5161.
D. 0.4463 to 0.5185.
Answer:
B. 0.4542 to 0.5105
Step-by-step explanation:
A 90% confidence interval for p is calculated as:
[tex]p-z_{\alpha /2}\sqrt{\frac{p(1-p)}{n} }\leq p\leq p+z_{\alpha /2}\sqrt{\frac{p(1-p)}{n} }[/tex]
This apply if n*p≥5 and n*(1-p)≥5
Where p is the proportion of sample, n is the size of the sample and [tex]z_{\alpha /2}[/tex] is equal to 1.645 for a 90% confidence.
Then, in this case p, n*p and n*(1-p) are calculated as:
[tex]p=\frac{410}{850} =0.4824[/tex]
n*p = (850)(0.4824) = 410
n*(1-p) = (850)(1-0.4824) = 440
So, replacing values we get:
[tex]0.4824-1.645\sqrt{\frac{0.4824(1-0.4824)}{850} }\leq p\leq 0.4824+1.645\sqrt{\frac{0.4824(1-0.4824)}{850} }[/tex]
[tex]0.4824-0.0282\leq p\leq 0.4824+0.0282[/tex]
[tex]0.4542\leq p\leq 0.5105[/tex]
It means that a 90% confidence interval for p is 0.4542 to 0.5105
Answer:
The correct answer in the option is;
B. 0.4542 to 0.5105.
Step-by-step explanation:
To solve the question, we note that
Total number of residents, n = 850
Number supporting property tax levy = 410
Proportion supporting tax levy, p = [tex]\frac{410}{850}[/tex] = 0.48235
The formula for confidence interval is
[tex]p +/-z*\sqrt{\frac{p(1-p)}{n} }[/tex]
Where
z = z value
The z value from the tables at 90 % = 1.64
Therefore we have
The confidence interval given as
[tex]0.48235 +/-1.64*\sqrt{\frac{0.48235(1-0.48235)}{850} }[/tex] = 0.48235 ± 2.811 × 10⁻²
= 0.4542 to 0.5105
The confidence interval is 0.4542 to 0.5105.
A guidance counselor at a university career center is interested in studying the earning potential of certain college majors. He claims that the proportion of graduates with degrees in engineering who earn more than $75,000 in their first year of work is not 15%. If the guidance counselor chooses a 5% significance level, what is/are the critical value(s) for the hypothesis test? 2010 20.05 20.025 1.960 20.01 2.326 20.005 2.576 1.282 1.645 Use the curve below to show your answer. Select the appropriate test by dragging the blue point to a right, left- or two tailed diagram. The shaded area represents the rejection region. Then, set the critical value(s) on the z-axis by moving the slider.
Answer:
For the critical value we know that the significance is 5% and the value for [tex] \alpha/2 = 0.025[/tex] so we need a critical value in the normal standard distribution that accumulates 0.025 of the area on each tail and for this case we got:
[tex] Z_{\alpha/2}= \pm 1.96[/tex]
Since we have a two tailed test, the rejection zone would be: [tex] z<-1.96[/tex] or [tex] z>1.96[/tex]
Step-by-step explanation:
Data given and notation
n represent the random sample taken
[tex]\hat p[/tex] estimated proportion of interest
[tex]p_o=0.15[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the proportion of graduates with degrees in engineering who earn more than $75,000 in their first year of work is not 15%.:
Null hypothesis:[tex]p=0.15[/tex]
Alternative hypothesis:[tex]p \neq 0.15[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
For the critical value we know that the significance is 5% and the value for [tex] \alpha/2 = 0.025[/tex] so we need a critical value in the normal standard distribution that accumulates 0.025 of the area on each tail and for this case we got:
[tex] Z_{\alpha/2}= \pm 1.96[/tex]
Since we have a two tailed test, the rejection zone would be: [tex] Z<-1.96[/tex] or [tex] z>1.96[/tex]
A={a,b,c,1,2,3,octopus,∅,0} B=N C={0} For each of the following statements, select either True or False. a) A∩B={0,1,2,3} Answer 1 b) C−A=∅ Answer 2 c) B∪P(C)=B Answer 3 d) C∈P(C) Answer 4
Answer:a)False b)True c)False d)True
Step-by-step explanation:
Let's consider first that for us the set of the natural numbers is the set of the positive integers and the set of the non-negative numbers is know as the whole number or with notation [tex]N_0[/tex]. Then for
a) the N={1,2,3,...} and therefore the common members with A are only {1,2,3} making the statement false, only if stated to consider the set N as all the non-negative numbers the answer would be true, but otherwise it is standarized to understand N as the positive integers and [tex]N_0[/tex] as the non-negative integers.
b)The difference of sets is taking the elements in the first that do not belong to the second, then it would be to withdraw the only element C has, since 0 belongs to A, and therefore C would turn to be an empty set.
c)The set powers of a given set S, denoted P(S), is a set with sets as elements, every subset of S is an element of P(S). Then P(S) is always non-empty, since at least S belongs to P(S). Here [tex]P(C)=\{C, \emptyset \}[/tex], then [tex]P(C)\cup B=\{C, \emptyset ,1,2,3,\ldots \}\ne B[/tex], therefore the statement is false.
d)As explained in c) [tex]P(C)=\{C, \emptyset \},[/tex] then clearly C is an element of P(C), thus the affirmation is true.
The number of entrees purchased in a single order at a Noodles & Company restaurant has had an historical average of 1.7 entrees per order. On a particular Saturday afternoon, a random sample of 48 Noodles orders had a mean number of entrees equal to 2.1 with a standard deviation equal to 1.01. At the 2 percent level of significance, does this sample show that the average number of entrees per order was greater than expected?
Answer:
[tex]t=\frac{2.1-1.7}{\frac{1.01}{\sqrt{48}}}=2.744[/tex]
[tex]p_v =P(t_{(47)}>2.744)=0.0043[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.02[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 1,7 entrees per order at 2% of signficance.
Step-by-step explanation:
Data given and notation
[tex]\bar X=2.1[/tex] represent the mean
[tex]s=1.01[/tex] represent the sample standard deviation
[tex]n=48[/tex] sample size
[tex]\mu_o =1.7[/tex] represent the value that we want to test
[tex]\alpha=0.02[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is higher than 1.7, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 1.7[/tex]
Alternative hypothesis:[tex]\mu > 1.7[/tex]
If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{2.1-1.7}{\frac{1.01}{\sqrt{48}}}=2.744[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=48-1=47[/tex]
Since is a one side test the p value would be:
[tex]p_v =P(t_{(47)}>2.744)=0.0043[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.02[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 1,7 entrees per order at 2% of signficance.
Answer:
Yes, this sample show that the average number of entrees per order was greater than expected.
Step-by-step explanation:
We are given that the number of entrees purchased in a single order at a Noodles & Company restaurant has had an historical average of 1.7 entrees per order. For this a random sample of 48 Noodles orders had a mean number of entrees equal to 2.1 with a standard deviation equal to 1.01.
We have to test that the average number of entrees per order was greater than expected or not.
Let, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 1.7 {means that the average number of entrees per order was same as expected of 1.7 entrees per order}
Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > 1.7 {means that the average number of entrees per order was greater than expected of 1.7 entrees per order}
The test statistics that will be used here is One sample t-test statistics;
T.S. = [tex]\frac{Xbar-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, Xbar = sample mean number of entrees = 2.1
s = sample standard deviation = 1.01
n = sample of Noodles = 48
So, test statistics = [tex]\frac{2.1-1.7}{\frac{1.01}{\sqrt{48} } }[/tex] ~ [tex]t_4_7[/tex]
= 2.744
Now, at 2% significance level the critical value of t at 47 degree of freedom in t table is given as 2.148. Since our test statistics is more than the critical value of t which means our test statistics will lie in the rejection region. So, we have sufficient evidence to reject our null hypothesis.
Therefore, we conclude that the average number of entrees per order was greater than expected.
A lab network consisting of 20 computers was attacked by a computer virus. This virus enters each computer with probability 0.4, independently of other computers. Find the probability that it entered at least 10 computers
The probability that the virus entered at least 10 computers is 0.7553.
To find the probability that the virus entered at least 10 computers, we can use the complementary probability formula. This formula states that the probability of an event A happening is equal to 1 minus the probability of event A not happening.
In this case, event A is the virus entering at least 10 computers. Event A not happening is the virus entering fewer than 10 computers.
The probability of the virus entering fewer than 10 computers is equal to the sum of the probabilities of the virus entering 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 computers.
We can use the binomial distribution to calculate the probability of the virus entering each of these numbers of computers. The binomial distribution is a probability distribution that describes the probability of getting a certain number of successes in a certain number of trials.
In this case, the trials are the computers, and the success is the virus entering the computer. The probability of success is 0.4, and the probability of failure is 0.6.
To find the probability of the virus entering fewer than 10 computers, we need to add up the probabilities in the table from 0 to 9.
P(virus entering fewer than 10 computers) = 0.36^20 + 20 * 0.36^19 * 0.6 + ... + 167960 * 0.36^11 * 0.6^9
We can use a calculator to evaluate this sum. The result is 0.2447.
Therefore, the probability of the virus entering at least 10 computers is 1 minus the probability of the virus entering fewer than 10 computers.
P(virus entering at least 10 computers) = 1 - 0.2447
P(virus entering at least 10 computers) = 0.7553
Therefore, the probability that the virus entered at least 10 computers is 0.7553.
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Ethan repairs household appliances like dishwashers and refrigerators. For each visit, he charges $25 plus $20 per hour of work. A linear equation that expresses the total amount of money Ethan earns per visit is y = 25 + 20x. What is the independent variable, and what is the dependent variable?
Answer:the independent variable is x, the number of hours of work.
The dependent variable is y, the total charge for x hours of work.
Step-by-step explanation:
A change in the value of the independent variable causes a corresponding change in the value of in dependent variable. Thus, the dependent variable is is output while the independent variable is the input
For each visit, he charges $25 plus $20 per hour of work. The linear expression that represents the total amount of money that Ethan earns per visit is y = 25 + 20x.
Since the total amount charged, y depends on the number of hours of work, x, it means that the dependent variable is y and the independent variable is x
Final answer:
In the equation y = 25 + 20x, x is the independent variable representing hours worked, and y is the dependent variable representing total earnings. The y-intercept is $25, the flat visit charge, and the slope is $20, the hourly charge.
Explanation:
In the equation y = 25 + 20x, which represents the total amount Ethan earns for each visit, the independent variable is the number of hours of work, denoted by x. The dependent variable is the total amount of money Ethan earns, represented by y. This is because Ethan's earnings depend on the amount of time he spends working.
The y-intercept is $25, which is the flat charge for Ethan's visit, regardless of the hours worked. The slope is $20, which represents the amount Ethan charges for each hour of work. Therefore, for each additional hour of work, Ethan will earn an additional $20.
According to government data, 74% of employed women have never been married. Rounding to 4 decimal places, if 15 employed women are randomly selected: a. What is the probability that exactly 2 of them have never been married
Answer:
0.000001 = 0.0001% probability that exactly 2 of them have never been married
Step-by-step explanation:
For each employed women, there are only two possible outcomes. Either they have already been married, or they have not. The women are chosen at random, which means that the probability of a woman having been already married is independent from other women. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
74% of employed women have never been married.
This means that [tex]p = 0.74[/tex]
15 employed women are randomly selected
This means that [tex]n = 15[/tex]
a. What is the probability that exactly 2 of them have never been married
This is P(X = 2).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{15,2}.(0.74)^{2}.(0.26)^{13} = 0.000001[/tex]
0.000001 = 0.0001% probability that exactly 2 of them have never been married
Waiting times for an order at Starbucks for all drive-through customers in the US have a uniform distribution from 3 min to 11 min (mean = 7 min, standard deviation = 2.3 min). What distribution would you use to find the probability that a randomly selected Starbucks drive-through customer in the US waits at most 9 minutes to receive their order?
Answer:
[tex] P(X <9) [/tex]
And we can use the cumulative distribution function given by:
[tex] F(X) = \frac{x-a}{b-a}, a\leq X \leq b[/tex]
And using this we got:
[tex] P(X <9) = F(9) = \frac{9-3}{11-3}= 0.75[/tex]
Step-by-step explanation:
For this case we assume that X= represent the waiting times and for this case we have the following distribution:
[tex] X \sim Unif (a= 3, b =11)[/tex]
And the expected value is given by:
[tex]\mu= E(X) = \frac{a+b}{2}= \frac{3+11}{2}=7[/tex]
And the variance is given by:
[tex] Var(X) \sigma^2 = \frac{(b-a)^2}{12} = \frac{(11-3)^2}{12} = 5.333[/tex]
And we can find the deviation like this:
[tex] Sd(X) = \sqrt{5.333}= 2.309[/tex]
And we want to find this probability:
[tex] P(X <9) [/tex]
And we can use the cumulative distribution function given by:
[tex] F(X) = \frac{x-a}{b-a}, a\leq X \leq b[/tex]
And using this we got:
[tex] P(X <9) = F(9) = \frac{9-3}{11-3}= 0.75[/tex]
The probability that a randomly selected Starbucks drive-through customer in the US waits at most 9 minutes to receive their order is 0.75.
What is uniform distribution?Uniform distributions are probability distributions in which all events are equally likely to occur.
Let's assume that X represents the waiting time.
As the distribution is the uniform distribution, we can write a =3 and b =11
Now, the expected value can be written as,
[tex]\mu = E(X) = \dfrac{a+b}{2} = \dfrac{3+11}{2}= 7[/tex]
the variance of the distribution can be written as,
[tex]Var(X) = \sigma^2 = \dfrac{(b-a)^2}{12} = \dfrac{(11-3)^2}{12} =5.34[/tex]
And, the standard deviation can be written as,
[tex]\sigma = \sqrt{5.34}=2.309[/tex]
In order to calculate the probability, we will use the cumulative distribution function. The cumulative function is given by the formula,
[tex]F(X) = \dfrac{x-a}{b-a}, a\leq X\leq b[/tex]
Using the function we can get the probability as,
[tex]F(X < 9) = F(9) = \dfrac{9-3}{11-3} = 0.75[/tex]
Hence, the probability that a randomly selected Starbucks drive-through customer in the US waits at most 9 minutes to receive their order is 0.75.
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A study found that, in 2005, 12.5% of U.S. workers belonged to unions (The Wall Street Journal, January 21, 2006). Suppose a sample of 400 U.S. workers is collected in 2006 to determine whether union efforts to organize have increased union membership.(a) Formulate the hypotheses that can be used to determine whether union membership increased in 2006.(b) If the sample results show that 52 of the workers belonged to unions, what is the p-value for your hypothesis test? Round your answers to four decimal places.
Answer:
There is not enough evidence to support the claim that union membership increased.
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 400
p = 12.5% = 0.125
Alpha, α = 0.05
Number of women belonging to union , x = 52
First, we design the null and the alternate hypothesis
[tex]H_{0}: p = 0.125\\H_A: p > 0.125[/tex]
The null hypothesis sates that 12.5% of U.S. workers belong to union and the alternate hypothesis states that there is a increase in union membership.
This is a one-tailed(right) test.
Formula:
[tex]\hat{p} = \dfrac{x}{n} = \dfrac{52}{400} = 0.13[/tex]
[tex]z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
Putting the values, we get,
[tex]z = \displaystyle\frac{0.13-0.125}{\sqrt{\frac{0.125(1-0.125)}{400}}} = 0.3023[/tex]
Now, we calculate the p-value from the table.
P-value = 0.3812
Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept the null hypothesis.
Conclusion:
Thus, there is not enough evidence to support the claim that union membership increased.
Final answer:
To determine whether union membership increased in 2006, the null hypothesis states the proportion is 12.5% or less, and the alternative hypothesis states it is greater than 12.5%. Based on the sample data (where 52 out of 400 workers are union members), the p-value is calculated to be approximately 0.22. Since the p-value is greater than 0.05, we fail to reject the null hypothesis, indicating insufficient evidence of an increase in union membership.
Explanation:
To determine whether union membership increased in 2006, we start by formulating the hypotheses for our hypothesis test:
Hypotheses
Null hypothesis (H0): The proportion of U.S. workers belonging to unions in 2006 is equal to or less than the 2005 level of 12.5% (p ≤ 0.125).
Alternative hypothesis (H1): The proportion of U.S. workers belonging to unions in 2006 is greater than the 2005 level of 12.5% (p > 0.125).
Next, we calculate the test statistic and the p-value based on the sample results:
The sample proportion is the number of workers belonging to unions divided by the total number of workers in the sample. Therefore:
Sample proportion = 52/400 = 0.13 or 13%
To find the p-value, we assume the null hypothesis is true. The test statistic for a one-sample Z-test for proportions is calculated as:
Z = (Sample proportion - Hypothesized proportion) / Standard error of the sample proportion
Standard error = sqrt((Hypothesized proportion * (1 - Hypothesized proportion)) / sample size)
Z = (0.13 - 0.125) / sqrt((0.125 * (1 - 0.125)) / 400) = 0.7727
Since this is a one-tailed test, the p-value is the probability that the standard normal variable is greater than the calculated Z value. We find the p-value using standard normal distribution tables or software. Assuming the Z value is 0.7727, the corresponding p-value would be approximately 0.22.
As the p-value is greater than typical significance levels like 0.05, we fail to reject the null hypothesis. This means there is not enough evidence at the 0.05 significance level to conclude that union membership has increased in 2006.
In 2001, one county reported that, among 3132 white women who had babies, 94 were multiple births. There were also 20 multiple births to 606 black women. Does this indicate any racial difference in the likelihood of multiple births? Test an appropriate hypothesis and state your conclusion in context.
Hypothesis:
The ratio of ladies giving multiple birth to total number of women for any race will be the same.
Test:
Ratio of white women giving multiple births = 94 / 3132 = 0.0300
Ratio of black women giving multiple births = 20 / 606 = 0.0330
Conclusion:
There is no racial difference in the likelihood of multiple births. Although we do see a difference in the ratios calculated above, the difference is small enough to be due to sample size difference of white and black women. The smaller number of total black women makes the ratio calculated from this sample have a higher probability to deviate from what is expected. This deviation will account for the difference in probability between both races.
We can see the effects of this small sample size by increasing or decreases the numerator by 1 for black women:
21 / 606 = 0.0347
19 / 606 = 0.0313
This change in the data of one woman produces a very large percentage change in our ratio for black women (5%). Thus despite inaccuracy due to small sample size, our hypothesis is correct.
Final answer:
A hypothesis test for the difference in proportions can be used to assess if there is a racial difference in the likelihood of multiple births between white and black women, based on the given data. The null hypothesis is no difference, and if the test statistic is significant, it may indicate a racial difference.
Explanation:
To evaluate any racial differences in the likelihood of multiple births between white women and black women based on the given data, we can perform a hypothesis test. Specifically, this would be a test for the difference between two proportions.
For white women:
Multiples: 94
Total births: 3132
For black women:
Multiples: 20
Total births: 606
Using a chi-squared distribution table or calculator, at α=0.05 and 1 degree of freedom, the critical value is approximately 3.841.
Since our calculated chi-squared value (0.2094) is less than the critical value (3.841), we fail to reject the null hypothesis.
Therefore, there is no significant evidence to conclude that there is a racial difference in the likelihood of multiple births among white and black women in this county.
Match the scenario with its distribution type. Group of answer choices Sammy takes a sample of 200 students at her college in an attempt to estimate the average time students spend studying at her university. She plots her data using a histogram, the histogram shows what type of distribution? Phil wants to estimate the average driving time to work for individuals in Corvallis, Oregon. He takes a simple random sample of 100 individuals and records their drive time. Phil knows that a different sample will yield a different sample average. Suppose Phil was able to repeatedly sample a different group of Corvallis residents until he obtained every combination of 100 residents. If he plots the sample means from each sample this distribution will represent which type of distribution? An instructor wants see the scores of her entire class after the midterm. The distribution of scores represents what type of distribution?
Answer:
Scenario 1: Sample distributionScenario 2: Sampling distribtuionScenario 3: Population distributionStep-by-step explanation:
Scenarios are: Sample Distribution, sampling distribution, Population distribution
sample distribution: A sample distribution contains the data of individuals of sample collected from a population.
Sampling distribution: distribution of multiple sample statistics
Population distribution: Statistics of entire population without drawing sny sample
In scenario 1, only one saple is taken from a population and that sample is plotted
In Scenario 2, multiple samples are taken from a population and means of each sample is plotted
In Scenario 3, scores of entire population is considered.
Among the equation students taking a graduate statistics class, equation are master students and the other equation are doctorial students. A random sample of equation students is going to be selected to work on a class project. Use equation to denote the number of master students in the sample. Keep at least 4 decimal digits if the result has more decimal digits.
Answer:
a) P=0.2861
b) P=0.0954
c) P=0.3815
d) P=0.6185
Step-by-step explanation:
The question is incomplete:
Among the N=16 students taking a graduate statistics class, A=10 are master students and the other N-A=6 are doctorial students. A random sample of n=5 students is going to be selected to work on a class project. Use X to denote the number of master students in the sample. Keep at least 4 decimal digits if the result has more decimal digits.
a) The probability that exactly 4 master students are in the sample is closest to?
b) The probability that all 5 students in the sample are master students is closest to?
c) The probability that at least 4 students in the sample are master students is closest to?
d) The probability that at most 3 students in the sample are master students is closest to?
We use a binomial distribution with n=5, with p=10/16=0.625 (proportion of master students).
a)
[tex]P(k=4)=\binom{5}{4}p^4q^1=5*0.625^4*0.375=5*0.1526*0.3750\\\\P(k=4)=0.2861[/tex]
b)
[tex]P(k=5)=\binom{5}{5}p^5q^0=1*0.625^5*1=\\\\P(k=4)=0.0954[/tex]
c)
[tex]P(k\geq4)=P(k=4)+P(k=5)=0.2861+0.0954=0.3815[/tex]
d)
[tex]P(k\leq3)=1-P(x\geq4)=1-0.3815=0.6185[/tex]
Consider the given function and the given interval. f(x) = 6 sin(x) − 3 sin(2x), [0, π]
(a) Find the average value fave of f on the given interval.
(b) Find c such that fave = f(c). (Round your answers to three decimal places.)
Answer:
(a) The average value of the given function is 12/π
(b) c = 1.238 or 2.808
Step-by-step explanation:
The average value of a function on a given interval [a, b] is given as
f(c) = (1/(b - a))∫f(x)dx;
from x = b to a
Now, given the function
f(x) = 6sin(x) - 3sin(2x), on [0, π]
The average value of the function is
1/(π-0) ∫(6sinx - 3sin2x)dx
from x = 0 to π
= (1/π) [-6cosx + (3/2)cos2x]
from 0 to π
= (1/π) [-6cosπ + (3/2)cos 2π - (-6cos0 + (3/2)cos0)]
= (1/π)(6 + (3/2) - (-6 + 3/2) )
= (1/π)(12) = 12/π
f(c) = 12/π
b) if f_(ave) = f(c), then
6sinx - 3sin2x = 12/π
2sinx - sin2x = 4/π
But sin2x = 2sinxcosx, so
2sinx - 2sinxcosx = 4/π
sinx - sinxcosx = 2/π
sinx(1 - cosx) = 2/π
This equation can only be estimated to be x = 1.238 or 2.808
Solving sin(x)(1 - cos(x)) = 2/π on [0, π] yields x = π/2 as the only solution. This corresponds to approximately 1.5708, satisfying the given equation.
To solve the equation sin(x)(1 - cos(x)) = 2/π, we can use the double-angle identity for sine, which states that sin(2x) = 2sin(x)cos(x). Rewrite the equation in terms of sin(2x):
sin(x)(1 - cos(x)) = 2/π
sin(x) - sin(x)cos(x) = 2/π
Now, substitute sin(x) = 2/π into the equation:
(2/π) - (2/π)cos(x) = 2/π
Multiply both sides by π to simplify:
2 - 2cos(x) = 2
Subtract 2 from both sides:
-2cos(x) = 0
Divide by -2:
cos(x) = 0
Now, find the values of x where cos(x) = 0, which occurs at x = π/2 and 3π/2. Since we are looking for solutions in the interval [0, π], x = π/2 is the only valid solution.
So, the solution to the equation sin(x)(1 - cos(x)) = 2/π on the interval [0, π] is x = π/2, which is approximately 1.5708.
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Choose all of the equivalent expressions.
A. 300e^−0.0577t
B. 300(1/2)^t/12
C. 252.290(0.9439)^t
D. 300(0.9439)^t
E. 252.290(0.9439)^t−3
Answer:
A B D and E
Step-by-step explanation:
Find the surface area of the triangular prism
The surface area of the triangular prism is 1664 square inches.
Explanation:
Given that the triangular prism has a length of 20 inches and has a triangular face with a base of 24 inches and a height of 16 inches.
The other two sides of the triangle are 20 inches each.
We need to determine the surface area of the triangular prism.
The surface area of the triangular prism can be determined using the formula,
[tex]SA= bh+pl[/tex]
where b is the base, h is the height, p is the perimeter and l is the length
From the given the measurements of b, h, p and l are given by
[tex]b=24[/tex] , [tex]h= 16[/tex] , [tex]l=20[/tex] and
[tex]p=20+20+24=64[/tex]
Hence, substituting these values in the above formula, we get,
[tex]SA= (24\times16)+(64\times20)[/tex]
Simplifying the terms, we get,
[tex]SA=384+1280[/tex]
Adding the terms, we have,
[tex]SA=1664 \ square \ inches[/tex]
Thus, the surface area of the triangular prism is 1664 square inches.
Suppose that the mean weight for men 18 to 24 years old is 170 pounds, and the standard deviation is 20 pounds. In each part, find the value of the standardized score (z-score) for the given weight.a. 200 pounds.b. 140 pounds.c. 170 pounds.d. 230 pounds.
Answer:
a) [tex]Z = 1.5[/tex]
b) [tex]Z = -1.5[/tex]
c) [tex]Z = 0[/tex]
d) [tex]Z = 3[/tex]
Step-by-step explanation:
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
In this problem, we have that:
[tex]\mu = 170, \sigma = 20[/tex]
a. 200 pounds.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{200 - 170}{20}[/tex]
[tex]Z = 1.5[/tex]
b. 140 pounds.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{140 - 170}{20}[/tex]
[tex]Z = -1.5[/tex]
c. 170 pounds.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{170 - 170}{20}[/tex]
[tex]Z = 0[/tex]
d. 230 pounds.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{230 - 170}{20}[/tex]
[tex]Z = 3[/tex]
The article "Calibration of an FTIR Spectrometer" (P. Pankratz, Statistical Case Studies for Industrial and Process Improvement, SIAM-ASA, 1997: 19–38) describes the use of a spectrometer to make five measurements of the carbon content (in ppm of a certain silicon wafer whose true carbon content was known to be 1.1447 ppm. The measurements were 1.0730, 1.0825, 1.0711.1.0870, and 1.0979.
a. Is it possible to estimate the uncertainty .in these measurements? If so, estimate it. If not, explain why not.
b. Is it possible to estimate the bias in these measurements? If so. estimate it. If not. explain why not.
Answer:
a) 0.011
b) -0.0624
Step-by-step explanation:
See attached pictures.
Final answer:
Uncertainty in the FTIR spectrometer measurements can be estimated as the standard deviation of the measurements, yielding 0.0109 ppm. Bias is estimated as the difference between the mean of the measurements (1.0823 ppm) and the true value (1.1447 ppm), resulting in a bias of -0.0624 ppm.
Explanation:
To address the student's question regarding the calibration of an FTIR spectrometer and the estimation of uncertainty and bias in measurements, we shall consider the given data.
Uncertainty in measurements can be estimated using the standard deviation of the measurements, which provides an indication of the spread of the data around the mean. To calculate the uncertainty:
Find the mean (μ) of the measurements.Subtract the mean from each measurement to find the deviation of each measurement.Square each deviation.Sum all the squared deviations.Divide by the number of measurements minus one to find the variance.Take the square root of the variance to find the standard deviation (SD), which represents the uncertainty.Using the provided measurements of carbon content, we calculate the uncertainty as follows:
μ = (1.0730 + 1.0825 + 1.0711 + 1.0870 + 1.0979) / 5 = 1.0823 ppmDeviations: [-0.0093, 0.0002, -0.0112, 0.0047, 0.0156]Squared deviations: [8.649E-05, 4.00E-08, 1.254E-04, 2.209E-05, 2.436E-04]Sum of squared deviations = 4.758E-04Variance = 4.758E-04 / (5-1) ≈ 1.190E-04 ppm²SD = √(1.190E-04) ≈ 0.0109 ppmThis standard deviation represents the uncertainty in the measurements.
Estimation of Bias
Bias in the measurements can be estimated as the difference between the mean of the measurements and the true value. Thus, the bias is calculated by subtracting the true carbon content from the mean measurement:
Bias = Mean - True value = 1.0823 ppm - 1.1447 ppm = -0.0624 ppm
The negative sign indicates that the measurements are, on average, lower than the true value.
Medical tests were conducted to learn about drug-resistant tuberculosis. Of cases tested in New Jersey, were found to be drug-resistant. Of cases tested in Texas, were found to be drug-resistant. Do these data suggest a statistically significant difference between the proportions of drug-resistant cases in the two states?
Answer:
Yes at the level of 0.02 significance
Step-by-step explanation:
we want to compare if P₁ = P₂
P1 = 9/142= 0.0634
P2 = 5/268 = 0.0187
P = 14/410 = 0.03414
significance level, α = 0.02
Test statistic, z = [tex]\frac{p1 - p2}{\sqrt{(p * (1 - p} )) * (\frac{1}{142} * \frac{1}{268})}}[/tex]
Test Statistic, z = [tex]\frac{0.0634 - 0.0187}{\sqrt{({0.0341 * ({1 - 0.0341}})) * ({\frac{1}{142} + \frac{1}{268} ) }} } = 2.373[/tex]
Test statistic, z = 2.373
p-value = 2*p(z<|z₀|) = 2*p(z<2.37) = 0 .0176
Answer: Since p-value (0.0176) is less than the significance level, α (0.02), the null hypothesis can not hold. we can therefore say that at 0.02 level of significance, there is sufficient evidence, statistically, that p₁ is different from p₂
For f(x) = 9x and g(x) = x + 3, find the following functions.
a. (f o g)(x);
b. (g o f )(x);
c. (f o g )(2);
d. (g o f )(2)
Answer:
a) 9*x + 27
b) 9*x+3
c) 45
d) 21
Step-by-step explanation:
since (f o g)(x) = f (g(x)) , then
a) (f o g)(x) = f (x + 3) = 9*(x+3) = 9*x + 27
similarly
b) (g o f)(x) = g (f(x)) = g ( 9x) = (9*x)+3 = 9*x+3
c) for x=2
(f o g)(2) = 9*2 + 27 = 45
d) for x=2
(g o f )(2) = 9*2 +3 = 21
thus we can see that the composition of functions is not necessarily commutative
From a plot of ln(Kw) versus 1/T (using the Kelvin scale), estimate Kw at 37°C, which is the normal physiological temperature. Kw = What is the pH of a neutral solution at 37°C?
Answer:
A) Kw (37°C) = 2.12x10⁻¹⁴
B) pH (37°C) = 6.84
Step-by-step explanation:
The following table shows the different values of Kw in the function of temperature:
T(°C) Kw
0 0.114 x 10⁻¹⁴
10 0.293 x 10⁻¹⁴
20 0.681 x 10⁻¹⁴
25 1.008 x 10⁻¹⁴
30 1.471 x 10⁻¹⁴
40 2.916 x 10⁻¹⁴
50 5.476 x 10⁻¹⁴
100 51.3 x 10⁻¹⁴
A) The plot of the values above gives a straight line with the following equation:
y = -6218.6x - 11.426 (1)
where y = ln(Kw) and x = 1/T
Hence, from equation (1) we can find Kw at 37°C:
[tex] ln(K_{w}) = -6218.6 \cdot (1/(37 + 273)) - 11.426 = -31.49 [/tex]
[tex] K_{w} = e^{-31.49} = 2.12 \cdot 10^{-14} [/tex]
Therefore, Kw at 37°C is 2.12x10⁻¹⁴
B) The pH of a neutral solution is:
[tex] pH = -log([H^{+}]) [/tex] (2)
The hydrogen ion concentration can be calculated using the following equation:
[tex] K_{w} = [H^{+}][OH^{-}] [/tex] (3)
Since in pure water, the hydrogen ion concentration must be equal to the hydroxide ion concentration, we can replace [OH⁻] by [H⁺] in equation (3):
[tex] K_{w} = ([H^{+}])^{2} [/tex]
which gives:
[tex] [H^{+}] = \sqrt {K_{w}} [/tex]
Having that Kw = 2.12x10⁻¹⁴ at 37 °C (310 K), the pH of a neutral solution at this temperature is:
[tex] pH = -log ([H^{+}]) = -log(\sqrt {K_{w}}) = -log(\sqrt {2.12 \cdot 10^{-14}}) = 6.84 [/tex]
Therefore, the pH of a neutral solution at 37°C is 6.84.
I hope it helps you!
a. Endothermic (Kw increases with temperature).
b. pH = 7 for pure water at 50.8°C.
c. Plot ln(Kw) vs. 1/T to estimate Kw at 37.8°C.
d. pH = 7 for neutral solution at 37.8°C.
Here's the step-by-step solution:
a. Since Kw increases with temperature, indicating more ionization, the process is endothermic.
b. At 50.8°C, Kw ≈ 5.47 × 10^(-14). Since pure water is neutral, pH = 7.
c. To estimate Kw at 37.8°C:
- Plot ln(Kw) against 1/T (in Kelvin).
- Find the y-intercept of the plot. This intercept corresponds to ln(Kw) at 1/T = 0, which is at the temperature where T = 1/(37.8 + 273.15).
- Take the exponential of this value to find Kw.
d. At 37.8°C, Kw ≈ 1.3 × 10^(-14). Since pure water is neutral, pH = 7.
The correct question is:
Values of Kw as a function of temperature are as follows: Temp (8C) Kw 0 1.14 3 10215 25 1.00 3 10214 35 2.09 3 10214 40. 2.92 3 10214 50. 5.47 3 10214
a. Is the autoionization of water exothermic or endothermic?
b. What is the pH of pure water at 50.8C?
c. From a plot of ln(Kw) versus 1/T (using the Kelvin scale), estimate Kw at 378C, normal physiological temperature. d. What is the pH of a neutral solution at 378C?
A publisher reports that 49I% of their readers own a personal computer. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 200200 found that 42B% of the readers owned a personal computer. Determine the P-value of the test statistic. Round your answer to four decimal places.
Answer:
P-value of test statistics = 0.9773
Step-by-step explanation:
We are given that a publisher reports that 49% of their readers own a personal computer. A random sample of 200 found that 42% of the readers owned a personal computer.
And, a marketing executive wants to test the claim that the percentage is actually different from the reported percentage, i.e;
Null Hypothesis, [tex]H_0[/tex] : p = 0.49 {means that the percentage of readers who own a personal computer is same as reported 63%}
Alternate Hypothesis, [tex]H_1[/tex] : p [tex]\neq[/tex] 0.49 {means that the percentage of readers who own a personal computer is different from the reported 63%}
The test statistics we will use here is;
T.S. = [tex]\frac{\hat p -p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }[/tex] ~ N(0,1)
where, p = actual % of readers who own a personal computer = 0.49
[tex]\hat p[/tex] = percentage of readers who own a personal computer in a
sample of 200 = 0.42
n = sample size = 200
So, Test statistics = [tex]\frac{0.42 -0.49}{\sqrt{\frac{0.42(1- 0.42)}{200} } }[/tex]
= -2.00
Now, P-value of test statistics is given by = P(Z > -2.00) = P(Z < 2.00)
= 0.9773 .