Questions: The management of the local zoo wants to know if all of their animal exhibits are equally popular. If there is significant evidence that some of the exhibits are not being visited frequently enough, then changes may need to take place within the zoo. A tally of visitors is taken for each of the following animals throughout the course of a week, and the results are contained in the following table. At α=0.025, determine whether there is sufficient evidence to conclude that some exhibits are less popular than others. Animal Exhibits at the Zoo Elephants Lions/Tigers Giraffes Zebras Monkeys Birds Reptiles Number of visitors 179 149 155 126 145 128 178 Step 1 of 4: State the null and alternative hypotheses in terms of the expected proportion for each animal exhibit. Enter your answer as a fraction or a decimal rounded to six decimal places, if necessary. H0: pi= Ha: Some exhibits are less popular than others.

Solution Steps

The null hypothesis \( H_0 \) states that all exhibits are equally popular, meaning the expected proportion of visitors for each exhibit is the same. The alternative hypothesis \( H_a \) states that some exhibits are less popular than others.

\[ H_0: p_i = \frac{1}{7} \quad \text{for each exhibit} \] \[ H_a: \text{Some exhibits are less popular than others.} \]

First, compute the total number of visitors across all exhibits: \[ \text{Total visitors} = 179 + 149 + 155 + 126 + 145 + 128 + 178 = 1060 \]

Under the null hypothesis, the expected number of visitors for each exhibit is: \[ \text{Expected visitors per exhibit} = \frac{\text{Total visitors}}{\text{Number of exhibits}} = \frac{1060}{7} \approx 151.428571 \]

The chi-square test statistic is calculated using the formula: \[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \] where \( O_i \) is the observed number of visitors and \( E_i \) is the expected number of visitors for each exhibit.

Calculate for each exhibit: \[ \begin{align_} \text{Elephants} & : \frac{(179 - 151.428571)^2}{151.428571} \approx 4.98 \\ \text{Lions/Tigers} & : \frac{(149 - 151.428571)^2}{151.428571} \approx 0.04 \\ \text{Giraffes} & : \frac{(155 - 151.428571)^2}{151.428571} \approx 0.08 \\ \text{Zebras} & : \frac{(126 - 151.428571)^2}{151.428571} \approx 4.24 \\ \text{Monkeys} & : \frac{(145 - 151.428571)^2}{151.428571} \approx 0.28 \\ \text{Birds} & : \frac{(128 - 151.428571)^2}{151.428571} \approx 3.63 \\ \text{Reptiles} & : \frac{(178 - 151.428571)^2}{151.428571} \approx 4.66 \\ \end{align_} \]

Sum these values to get the chi-square test statistic: \[ \chi^2 \approx 4.98 + 0.04 + 0.08 + 4.24 + 0.28 + 3.63 + 4.66 = 17.91 \]

The critical value for a chi-square test with \( \alpha = 0.025 \) and \( df = 6 \) (since there are 7 exhibits) is approximately 14.449. Compare the test statistic to the critical value: \[ \chi^2 = 17.91 > 14.449 \]

Since the test statistic exceeds the critical value, reject the null hypothesis. There is sufficient evidence to conclude that some exhibits are less popular than others.

Final Answer