Only about $18 \%$ of all people can wiggle their ears. Is this percent different for millionaires? Of the 384 millionaires surveyed, 88 could wiggle their ears. What can be concluded at the $\alpha=0.10$ level of significance?
a. For this study, we should use $z$-test for a population proportion
b. The null and alternative hypotheses would be:
os
$0^{s}$
$0^{8}$
$\sigma^{s}$
$0^{s}$
$0^{s}$
c. The test statistic $z+\frac{v}{v}=$ (please show your answer to 3 decimal places.)
d. The p-value $=$ (Please show your answer to 3 decimal places.)
Answer
Final Answer: The test statistic \(z\) is approximately \(\boxed{2.508}\) and the p-value is approximately \(\boxed{0.012}\). Since the p-value is less than the level of significance \(\alpha=0.10\), we reject the null hypothesis. Therefore, we can conclude that there is a significant difference between the proportion of millionaires who can wiggle their ears and the general population proportion.
Step 1 ：We are given a population proportion (0.18 or 18%) and a sample proportion (88 out of 384). We are asked to determine if there is a significant difference between these proportions at the 0.10 level of significance. This is a hypothesis testing problem for proportions.
Step 2 ：The null hypothesis (H0) is that the proportion of millionaires who can wiggle their ears is the same as the general population proportion (0.18). The alternative hypothesis (H1) is that the proportion of millionaires who can wiggle their ears is different from the general population proportion.
Step 3 ：We can use a z-test for proportions to test these hypotheses. The test statistic for a z-test for proportions is calculated as: \(z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 * (1 - p_0)}{n}}}\) where \(\hat{p}\) is the sample proportion, \(p_0\) is the population proportion, and n is the sample size.
Step 4 ：The p-value is then calculated using the standard normal distribution. If the p-value is less than the level of significance (0.10), we reject the null hypothesis.
Step 5 ：Given that \(p_0 = 0.18\), n = 384, x = 88, we calculate \(\hat{p} = \frac{88}{384} = 0.22916666666666666\)
Step 6 ：Substituting these values into the z-score formula, we get \(z = \frac{0.22916666666666666 - 0.18}{\sqrt{\frac{0.18 * (1 - 0.18)}{384}}} = 2.5078017380491304\)
Step 7 ：The calculated z-score is approximately 2.508 and the p-value is approximately 0.012. The p-value is less than the level of significance (0.10), so we reject the null hypothesis. This means that there is a significant difference between the proportion of millionaires who can wiggle their ears and the general population proportion.
Step 8 ：Final Answer: The test statistic \(z\) is approximately \(\boxed{2.508}\) and the p-value is approximately \(\boxed{0.012}\). Since the p-value is less than the level of significance \(\alpha=0.10\), we reject the null hypothesis. Therefore, we can conclude that there is a significant difference between the proportion of millionaires who can wiggle their ears and the general population proportion.
