Question 14 of 22 (2.1 points) I Question Attempt: 1 of 3 Moviegoers The average "moviegoer" sees 8.5 movies a year. A moviegoer is defined as a person who sees at least one movie in a theater in a 12 -month period. A random sample of 47 moviegoers from a large university revealed that the average number of movies seen per person was 9.6 . The population standard deviation is 3.2 movies. At the 0.05 level of significance, can it be concluded that this represents a difference from the national average? Part: \( 0 / 5 \) Part 1 of 5 State the hypotheses and identify the claim with the correct hypothesis. \[ \begin{array}{l} H_{0}: \square \text { (Choose one) } \mathbf{V} \\ H_{1}: \square \text { (Choose one) } \mathbf{V} \end{array} \] \[ \square=\square \] \[ \square \neq \square . \] This hypothesis test is a \( \square \) (Choose one) test. \( \mu \) \( \square \) > \( \square \) \( \square \)

The average moviegoer in the nation sees 8.5 movies, so we set our null hypothesis \( H_0: \mu = 8.5 \) to claim there’s no difference. For the alternative hypothesis suggesting a difference in the mean, we express this as \( H_1: \mu \neq 8.5 \). This indicates we are looking for any deviation, either higher or lower. Therefore, this hypothesis test is a two-tailed test, as we’re addressing potential differences in both directions. As for calculating the test statistic, get your pens ready! The formula for the z-test statistic is given by \( z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \). Here, \( \bar{x} \) is your sample mean (9.6), \( \mu \) is the population mean (8.5), \( \sigma \) is the standard deviation (3.2), and \( n \) is your sample size (47). You’ll get to crunch some numbers to see if the difference is significant! Don't forget to compare your computed z-value against critical z-values from the z-table at 0.05 significance level to draw your conclusion!