Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorable, so a potential customer has decided to go ahead with a purchase arrangement unless it can be conclusively demonstrated that the true average lifetime is smaller than what is advertised. A random sample of 50 bulbs was selected, the lifetime of each bulb determined, and the following information obtained:
Average lifetime = 738.44 hours and a standard deviation of lifetimes equal to 38.2 hours.
Should the customer purchase the light bulbs?
(a) Make the decision on a significance level of .05?
(b) Make the decision on a significance level of .01?

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]

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')

The total cost of the loan, including down payment, PMI, mortgage payments, and yearly taxes and insurance over 30 years, for purchasing a house at $138,000 with specific conditions rounds to approximately $333,400.00.

To calculate the total cost of the loan for purchasing a house at $138,000.00 with a 10% down payment, an interest rate of 4.875%, PMI payments of $25.88 for 77 months, yearly taxes of $2400.00, and insurance of $750.00 per year, the following steps are undertaken:

Rounding to the nearest hundred gives us a total cost of approximately $333,400.00.

Answer:

A B D and E

Step-by-step explanation:

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.

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]

Answer:

D

Step-by-step explanation:

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.

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.

Find out more at: https://brainly.com/question/17280826

Answer:

Calculate the Macaulay duration of Annuity B at the time of purchase is 1.369.

Step-by-step explanation:

First, we use 0.93 to calculate the v which equals 1/(1+i).

[tex]\frac{0+1*v+2v^{2} }{1+v+v^{2} }[/tex] = 0.93

After rearranging the equation, we get 1.07[tex]v^{2}[/tex] + 0.07v - 0.93=0

So, v=0.9

Mac D: [tex]\frac{0+1*v+2*v^{2}+ 3*v^{2} }{1+v+v^{2}+ v^{3} }[/tex]

After substituting the value of v, we get Mac D = 1.369.

The equation of the quadratic function is [tex]y=(x-7)(x+3)[/tex]

Explanation:

The vertex form of the quadratic function is given by

[tex]y=a(x-h)^{2}+k[/tex]

It is given that the quadratic function has a vertex at [tex](2,-25)[/tex]

The vertex is represented by the coordinate [tex](h,k)[/tex]

Hence, substituting [tex](h,k)=(2,-25)[/tex] in the vertex form, we get,

[tex]y=a(x-2)^{2}-25[/tex]

Now, substituting the x - intercept [tex](7,0)[/tex] , we have,

[tex]0=a(7-2)^{2}-25[/tex]

[tex]0=a(5)^{2}-25[/tex]

[tex]25=a(25)[/tex]

 [tex]1=a[/tex]

Thus, the value of a is 1.

Hence, substituting [tex]a=1[/tex], [tex](h,k)=(2,-25)[/tex] in the vertex form [tex]y=a(x-h)^{2}+k[/tex] , we get,

[tex]y=1(x-2)^{2}-25[/tex]

[tex]y=(x-2)^{2}-25[/tex]

[tex]y=x^2-2x+4-25[/tex]

[tex]y=x^2-2x-21[/tex]

[tex]y=(x-7)(x+3)[/tex]

Thus, the equation of the quadratic function is [tex]y=(x-7)(x+3)[/tex]

Answer:

the answer is d

Step-by-step explanation:

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.

Answer:

Step-by-step explanation:

Hello!

X: Cholesterol level of a woman aged 30-39. (mg/dl)

This variable has an approximately normal distribution with mean μ= 190.14 mg/dl

1. You need to find the corresponding Z-value that corresponds to the top 9.3% of the distribution, i.e. is the value of the standard normal distribution that has above it 0.093 of the distribution and below it is 0.907, symbolically:

P(Z≥z₀)= 0.093

-*or*-

P(Z≤z₀)= 0.907

Since the Z-table shows accumulative probabilities P(Z<Z₁₋α) I'll work with the second expression:

P(Z≤z₀)= 0.907

Now all you have to do is look for the given probability in the body of the table and reach the margins to obtain the corresponding Z value. The first column gives you the integer and first decimal value and the first row gives you the second decimal value:

z₀= 1.323

2.

Using the Z value from 1., the mean Cholesterol level (μ= 190.14 mg/dl) and the Medical guideline that indicates that 9.3% of the women have levels above 240 mg/dl you can clear the standard deviation of the distribution from the Z-formula:

Z= (X- μ)/δ ~N(0;1)

Z= (X- μ)/δ

Z*δ= X- μ

δ=(X- μ)/Z

δ=(240-190.14)/1.323

δ= 37.687 ≅ 37.7 mg/dl

I hope it helps!

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.

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  

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]

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:

Using the provided measurements of carbon content, we calculate the uncertainty as follows:

This 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.

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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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.

Answer:

22%

Step-by-step explanation:

In the event of changing a test answer there are three possible outcomes, which should add up to 100%: changing from incorrect to correct (51%),  changing from correct to incorrect (27%) and  changing from incorrect to incorrect (X). Therefore, the percent of changes from incorrect to​ incorrect was:

[tex]100\% = 51\%+27\%+X\\X= 22\%[/tex]

22% of the changes were  from incorrect to​ incorrect.

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.

Answer:

So the probability is P=0.00256.

Step-by-step explanation:

We have 26 letters, so the probability that the first letter in the name is the same is 1/26.

The probability that the second letter in the name is the same is 1/26 and the probability that the third letter in the name is the same is 1/26.

Out of ten people we choose 2.

So the probability is:

[tex]P=C_2^{10}\left(\frac{1}{26}\right)^3\\\\P=\frac{10!}{2!(10-2)!}\cdot \left(\frac{1}{26}\right)^3\\\\P=\frac{45}{17576}\\\\P=0.00256\\[/tex]

So the probability is P=0.00256.

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 .

Answer: a) 2 miles

b) 4 miles

Step-by-step explanation:

There are two right angle triangles formed in the rectangle.

Taking 30 degrees as the reference angle, the length of the side walk, h represents the hypotenuse of the right angle triangle.

The width, w of the park represents the opposite side of the right angle triangle.

The length of the park represents the adjacent side of the right angle triangle.

a) to determine the width of the park w, we would apply

the tangent trigonometric ratio.

Tan θ, = opposite side/adjacent side. Therefore,

Tan 30 = w/2√3

1/√3 = w/2√3

w = 1/√3 × 2√3

w = 2

b) to determine the the length of the side walk h, we would apply

the Cosine trigonometric ratio.

Cos θ, = adjacent side/hypotenuse. Therefore,

Cos 30 = 2√3/h

√3/2 = 2√3/h

h = 2√3 × 2/√3

h = 4

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.

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.

To learn more about equation

https://brainly.com/question/29174899

#SPJ6

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.

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]

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.

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