Question

question 5 of 8, step 3 of 3
a hollywood studio believes that a movie that is considered a drama will draw a larger crowd on average than a movie that is considered a comedy. to test this theory, the studio randomly selects several movies that are classified as dramas and several movies that are classified as comedies and determines the box office revenue for each movie. the results of the survey are as follows. do the data substantiate the studios belief that dramas will draw a larger crowd on average than comedies at α = 0.01? let dramas be population 1 and comedies be population 2. assume that the population variances are approximately equal.
box office revenues (millions of dollars)

	n	$\bar{x}$	s
comedy	13	140	20

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step 3 of 3: make the decision and state the conclusion in terms of the original question.
answer
we reject the null hypothesis and conclude that there is sufficient evidence at an α = 0.01 level of significance to say that dramas will draw a larger crowd on average than comedies.
we fail to reject the null hypothesis and conclude that there is sufficient evidence at an α = 0.01 level of significance to say that dramas will draw a larger crowd on average than comedies.
we fail to reject the null hypothesis and conclude that there is insufficient evidence at an α = 0.01 level of significance to say that dramas will draw a larger crowd on average than comedies.

Explanation:

Step1: Identify the null and alternative hypotheses

$H_0:\mu_1\leq\mu_2$, $H_1:\mu_1 > \mu_2$ (where $\mu_1$ is the mean box - office revenue of dramas and $\mu_2$ is the mean box - office revenue of comedies)

Step2: Calculate the pooled variance

$S_p^2=\frac{(n_1 - 1)S_1^2+(n_2 - 1)S_2^2}{n_1 + n_2-2}=\frac{(15 - 1)\times60^2+(13 - 1)\times20^2}{15 + 13-2}=\frac{14\times3600+12\times400}{26}=\frac{50400 + 4800}{26}=\frac{55200}{26}\approx2123.08$

Step3: Calculate the test statistic

$t=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{(180 - 140)-0}{\sqrt{2123.08}\sqrt{\frac{1}{15}+\frac{1}{13}}}=\frac{40}{\sqrt{2123.08}\sqrt{\frac{13 + 15}{15\times13}}}=\frac{40}{\sqrt{2123.08}\sqrt{\frac{28}{195}}}\approx\frac{40}{\sqrt{2123.08\times0.1436}}\approx\frac{40}{\sqrt{304.98}}\approx\frac{40}{17.46}\approx2.29$

Step4: Determine the degrees of freedom and critical value

$df=n_1 + n_2-2=15 + 13-2 = 26$. For $\alpha = 0.01$ and a one - tailed test, the critical value $t_{\alpha,df}=t_{0.01,26}\approx 2.479$

Step5: Make a decision

Since $t = 2.29

Answer:

We fail to reject the null hypothesis and conclude that there is insufficient evidence at an $\alpha = 0.01$ level of significance to say that dramas will draw a larger crowd on average than comedies.