Question

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. assume that the population variances are approximately equal. box office revenues (millions of dollars) n x s drama 19 160 50 comedy 14 130 20 copy data calculate a 99% confidence interval for the difference in mean revenue at the box office for drama and comedy movies. let dramas be population 1 and comedies be population 2. write your answer using interval notation and round the interval endpoints to two decimal places. answer

Explanation:

Step1: Calculate degrees of freedom

The degrees of freedom formula for two - sample t - test with equal variances is $df=n_1 + n_2-2$. Here, $n_1 = 19$ (drama), $n_2=14$ (comedy), so $df=19 + 14-2=31$.

Step2: Find the t - critical value

For a 99% confidence interval and $df = 31$, looking up in the t - distribution table, the two - tailed t - critical value $t_{\alpha/2}= 2.744$.

Step3: Calculate the pooled variance

The pooled variance $s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}$. Substitute $n_1 = 19$, $s_1 = 50$, $n_2=14$, $s_2 = 20$:
\[

$$\begin{align*} s_p^2&=\frac{(19 - 1)\times50^2+(14 - 1)\times20^2}{19+14 - 2}\\ &=\frac{18\times2500+13\times400}{31}\\ &=\frac{45000+5200}{31}\\ &=\frac{50200}{31}\approx1619.35 \end{align*}$$

\]

Step4: Calculate the standard error

The standard error $SE=s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$. Since $s_p=\sqrt{s_p^2}\approx\sqrt{1619.35}\approx40.24$, then
\[

$$\begin{align*} SE&=40.24\sqrt{\frac{1}{19}+\frac{1}{14}}\\ &=40.24\sqrt{\frac{14 + 19}{19\times14}}\\ &=40.24\sqrt{\frac{33}{266}}\\ &=40.24\times\sqrt{0.124}\\ &\approx40.24\times0.352\approx14.16 \end{align*}$$

\]

Step5: Calculate the difference in sample means

$\bar{x}_1-\bar{x}_2=160 - 130=30$.

Step6: Calculate the confidence interval

The confidence interval for $\mu_1-\mu_2$ is $(\bar{x}_1-\bar{x}_2)\pm t_{\alpha/2}\times SE$.
Lower limit: $30-2.744\times14.16=30 - 38.85=- 8.85$.
Upper limit: $30 + 2.744\times14.16=30+38.85 = 68.85$.

Answer:

$(-8.85,68.85)$