Question

43 random movie theaters across the US were surveyed, and each reported its revenue (in $) from the film Gemini Man last weekend. These values were stored in the R vector revenue, and the result of some code is given below. > t.test(revenue, conf.level=.95) $conf.int [1] 5071.03 5748.24 attr(, "conf.level") [1] 0.95

a) Interpret this confidence interval in the context of the story.
b) Calculate the degrees of freedom being used here.
c) Keeping everything else the same, what would happen if the confidence level were increased to 99%?

A. The confidence interval would become wider
B. The confidence interval would become narrower
C. The width of the confidence interval would not change

d) Keeping everything else the same, what would happen if the sample size increased?
A. The confidence interval would become wider
B. The confidence interval would become narrower
C. The width of the confidence interval would not change

e) If we were testing H1: 4 > $5000, what can you say about the p-value?
A. The p-value would be greater than 0.1
B. The p-value would be between 0.1 and 0.05
C. The p-value would be between 0.05 and 0.025
D. The p-value would be below 0.025

Answer 1

Answer:

a) For this case we can conclude with 95% of confidence that the true mean for the revenue (in $) from the film Gemini Man last weekend is between (5071.03;5748.24)

b) $df=n-1=43-1=42$  

c) If we increase the confidence level the value for $t{\alpha/2}$ would increasse and then the margin of error would increase and the interval would be more wider. So then the best answer is:

A. The confidence interval would become wider

d) For this case the margin of error is given by:

$ME= t_{\alpha/2}\frac{s}{\sqrt{n}}$

And if we increase the sample size then we conclude that the interval would be narrower. So then the best answer is

B. The confidence interval would become narrower

e) The mean is given by:

$\bar X= \frac{5071.03+5748.24}{2}= 5409.64$

$ME= \frac{5748.24-5071.03}{2}= 338.61$

And the deviation can be founded :

$\sigma= \frac{ME \sqrt{n}}{t}= \frac{338.61* \sqrt{43}}{2.02} = 1099.198$

And then we can find the statistic:

$t = \frac{5000-5409.64}{\frac{1099.198}{\sqrt{43}}}= -2.44$

And the p value would be:

$p_v =1- P(t_{42}<-2.44) = 0.99$

And the best anwer would be

A. The p-value would be greater than 0.1

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$\bar X$ represent the sample mean  

$\mu$ population mean (variable of interest)  

s represent the sample standard deviation  

n=43 represent the sample size  

Calculate the confidence interval

The confidence interval for the mean is given by the following formula:  

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)  

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:  

$df=n-1=43-1=42$  

Since the confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,42)".And we see that $t_{\alpha/2}=2.02$ and this value is exactly the same for the normal standard distribution and makes sense since the sample size is large enough to approximate the t distribution with the normal standard distribution.  

So on this case the 95% confidence interval is given by (5071.03;5748.24)  

Part a

For this case we can conclude with 95% of confidence that the true mean for the revenue (in $) from the film Gemini Man last weekend is between (5071.03;5748.24)

Part b

$df=n-1=43-1=42$  

Part c

If we increase the confidence level the value for $t{\alpha/2}$ would increasse and then the margin of error would increase and the interval would be more wider. So then the best answer is:

A. The confidence interval would become wider

Part d

For this case the margin of error is given by:

$ME= t_{\alpha/2}\frac{s}{\sqrt{n}}$

And if we increase the sample size then we conclude that the interval would be narrower. So then the best answer is

B. The confidence interval would become narrower

Part e

For this case we are testing $\mu >5000$

The mean is given by:

$\bar X= \frac{5071.03+5748.24}{2}= 5409.64$

$ME= \frac{5748.24-5071.03}{2}= 338.61$

And the deviation can be founded :

$\sigma= \frac{ME \sqrt{n}}{t}= \frac{338.61* \sqrt{43}}{2.02} = 1099.198$

And then we can find the statistic:

$t = \frac{5000-5409.64}{\frac{1099.198}{\sqrt{43}}}= -2.44$

And the p value would be:

$p_v =1- P(t_{42}<-2.44) = 0.99$

And the best anwer would be

A. The p-value would be greater than 0.1