In a simple random sample of size 64 , there were 32 individuals in the category of interest. It is desired to test $H _ { 0 } : p = 0.36$ versus $H _ { 1 } : p

Question

In a simple random sample of size 64 , there were 32 individuals in the category of interest. It is desired to test $H _ { 0 } : p = 0.36$ versus $H _ { 1 } : p < 0.36$. Do you reject $H _ { 0 }$ $0.05$ level?
A) Yes
B) No

Quizplus · Accepted Answer

To determine whether to reject $H_0$, we calculate the test statistic using the formula for a proportion test: $Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$, where $\hat{p}$ is the sample proportion, $p_0$ is the hypothesized population proportion, and $n$ is the sample size. Here, $\hat{p} = \frac{32}{64} = 0.5$, $p_0 = 0.36$, and $n = 64$. Plugging these values into the formula gives $Z = \frac{0.5 - 0.36}{\sqrt{\frac{0.36 \times 0.64}{64}}} = \frac{0.14}{\sqrt{\frac{0.2304}{64}}} = \frac{0.14}{0.06} = 2.33$.Next, we compare the calculated Z-value to the critical Z-value from the Z-table for a one-tailed test at the 0.05 significance level, which is approximately -1.645 (since we are testing for $p < 0.36$, we are interested in the negative critical value). Since 2.33 is not less than -1.645, we do not reject $H_0$ at the 0.05 level. The test statistic does not fall in the critical region that would indicate a significant difference from the hypothesized proportion.

In a Simple Random Sample of Size 64 , There H0:p=0.36H _ { 0 } : p = 0.36H0​:p=0.36

In a Simple Random Sample of Size 64 , There $H _ { 0 } : p = 0.36$