A corrections officer read that 58% of runaways are female. He believes that the percentage is higher than 58%. He selected a random sample of 90 runaways and found that 63 were female. At a=0.05, can we conclude that his belief is correct? **CV Calculation:**CV = invNorm [field for input]Round to Two Decimal PlacesCV = [field for input]---**f) What formula will we use for the test value?**1. \( z = \frac{(\hat{p}_1 - \hat{p}_2) - (p_1 - p_2)}{\sqrt{\hat{p} \cdot \hat{q} \cdot \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \)2. \( z = \frac{\hat{p} - p}{\sqrt{\frac{p \cdot q}{n}}} \) [Selected option]3. \( t = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \)4. \( t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \)5. \( z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \)[The selected option has a check mark next to it.]**Explanation of the Formulas:**- **Z-Test for Two Proportions:** The first option is for comparing two population proportions, using their sample estimates and size.- **Z-Test for Single Proportion:** The selected option is for checking a sample proportion against a known population proportion.- **T-Test for Two Means:** The third option compares means from two independent samples, accounting for variance.- **T-Test for Single Mean (with sample standard deviation):** The fourth option tests a sample mean against a known population mean using the sample's standard deviation.- **Z-Test for Single Mean (with population standard deviation):** The final option utilizes the population standard deviation to test the sample mean against the population mean.

Name: A corrections officer read that 58% of runaways are female. He believe
Uploaded: 2024-03-20T07:06:41.000Z
Channel: Bartleby
Description: A corrections officer read that 58% of runaways are female. He believes that the percentage is higher than 58%. He selected a random sample of 90 runaways and found that 63 were female. At a=0.05, can we conclude that his belief is correct? **CV Calc

The null hypothesis for testing significance of difference between sample proportion (p) and population proportion (P) is H0: p=P Against the corresponding alternative, H1:p≠P The null and the alternative hypothesis may be right or left-tailed as according to question The Test statistic for large sample test for single proportion is Z=p-PPQn, where Q=1-P, n=sample sizeEvidently, Z~N(0,1). So, the significant values of Z at 5% and 1% level of significance are respectively 1.96 and 1.58.Hence, we can conclude at a 95% level of significance that his belief is correct.