test the hypotheses using the p - value approach with a 5% significance level.\nho: p = 0.72\nha: p≠0.72\nn…

test the hypotheses using the p - value approach with a 5% significance level.\nho: p = 0.72\nha: p≠0.72\nn = 500\nx = 370\nidentify p^, the test statistic, and the p - value. decide if their is convincing evidence to reject the null. (choose 3)\nfail to reject\n0.74\nz = 0.996\nreject\nz = 0.999\nz = 0.991\n0.76\n0.72

test the hypotheses using the p - value approach with a 5% significance level.\nho: p = 0.72\nha: p≠0.72\nn = 500\nx = 370\nidentify p^, the test statistic, and the p - value. decide if their is convincing evidence to reject the null. (choose 3)\nfail to reject\n0.74\nz = 0.996\nreject\nz = 0.999\nz = 0.991\n0.76\n0.72

Answer

Explanation:

Step1: Calculate sample proportion $\hat{p}$

$\hat{p}=\frac{x}{n}=\frac{370}{500} = 0.74$

Step2: Calculate the test - statistic $z$

The formula for the test - statistic in a one - sample proportion test is $z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$, where $p_0 = 0.72$, $\hat{p}=0.74$, and $n = 500$. $z=\frac{0.74 - 0.72}{\sqrt{\frac{0.72\times(1 - 0.72)}{500}}}=\frac{0.02}{\sqrt{\frac{0.72\times0.28}{500}}}=\frac{0.02}{\sqrt{\frac{0.2016}{500}}}=\frac{0.02}{\sqrt{0.0004032}}\approx\frac{0.02}{0.0201}\approx0.995\approx0.996$

Step3: Calculate the p - value

Since $H_a:p\neq0.72$, the p - value is $2\times(1 - P(Z<|z|))$. For $z = 0.996$, $P(Z < 0.996)\approx0.84$, so the p - value is $2\times(1 - 0.84)=2\times0.16 = 0.32$. Since the p - value ($0.32$) is greater than the significance level $\alpha=0.05$, we fail to reject the null hypothesis.

Answer:

Fail to Reject, 0.74, z = 0.996