Question

A 7,000-seat theater is interested in determining whether there is a difference in attendance between shows on Tuesday evening and those on Wednesday evening. Two independent samples of 31 weeks are collected for Tuesday and Wednesday. The mean attendance on Tuesday evening is calculated as 5,060, while the mean attendance on Wednesday evening is calculated as 5,390. The known population standard deviation for attendance on Tuesday evening is 540 and the known population standard deviation for attendance on Wednesday evening is 480. Which of the following is the appropriate conclusion at the 5% level of significance?
a. Conclude that the mean attendance differs since the p-value = 0.0067 < 0.05.
b. Conclude that the mean attendance differs since the p-value = 0.0134 < 0.05.
c. Do not conclude that the mean attendance differs since the p-value = 0.0067 < 0.05.
d. Do not conclude that the mean attendance differs since the p-value = 0.0134 < 0.05.

Answer 1

Answer:

b. Conclude that the mean attendance differs since the p-value = 0.0134 < 0.05.

Step-by-step explanation:

Before testing, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Samples of 31 weeks. The mean attendance on Tuesday evening is calculated as 5,060, with a population standard deviation of 540.

This means that:

$\mu_{T} = 5060, s_T = \frac{540}{\sqrt{31}} = 97$

The mean attendance on Wednesday evening is calculated as 5,390, with standard deviation of 480.

This means that:

$\mu_{W} = 5390, s_W = \frac{480}{\sqrt{31}} = 86.21$

A 7,000-seat theater is interested in determining whether there is a difference in attendance between shows on Tuesday evening and those on Wednesday evening.

At the null hypothesis we test if there is no difference, that is:

$H_0: \mu_T - \mu_W = 0$

At the alternate hypothesis, we test if there is difference, that is:

$H_a: \mu_T - \mu_W \neq 0$

The test statistic is:

$z = \frac{X - \mu}{s}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis and s is the standard error.

0 is tested at the null hypothesis:

This means that $\mu = 0$

From the two samples collected:

$X = \mu_T - \mu_W = 5060 - 5390 = -330$

$s = \sqrt{s_T^2+s_W^2} = \sqrt{97^2+86.21^2} = 129.77$

Value of the test statistic:

$z = \frac{X - \mu}{s}$

$z = \frac{-330 - 0}{129.77}$

$z = -2.54$

Pvalue of test and decision:

The pvalue of the test is finding the probability of the sample mean difference differing by 0 by at least 330, that is, P(|z| > 2.54), which is 2 multiplied by the pvalue of z = -2.54.

Looking at the z-table, z = -2.54 has a pvalue of 0.0067.

2*0.0067 = 0.0134

The pvalue of the test is 0.0134 < 0.05, which means that there is evidence that the mean attendance differs. The correct answer is given by option b.