Answer to Question #197267 in Statistics and Probability for jack

Question

I was found that 85.6% of students at IUL worldwide are enrolling to undergraduate program. A random sample of 50 students from IUL Morocco revealed that 42 of them were enrolled in undergraduate program. Is there evidence to state that the proportion of IUL Morocco differs from the IUL Morocco proportion? Use α = 0.05

Answer 1

Given that,

The following information is provided: The sample size is n = 50

sample proportion is $\hat{P} = 0.84$ and the significance level is α=0.05

NULL AND ALTERNATIVE HYPOTHESES:-

The following null and alternative hypotheses need to be tested:

$H_0 : p = 0.856 \\ H_1 : p ≠ 0.856$

This corresponds to a two-tailed test, for which a z-test for one population proportion needs to be used.

REJECTION REGION :-

Based on the information provided, the significance level is α=0.05, and the critical value for a two-tailed test is

$Z_{critical}=Z_(\frac{α}{2})= Z_(\frac{0.05}{2})= 1.96$

The rejection region for this right-tailed test is ,

R={Z : | Z| > 1.96}

TEST STATISTICS:-

The z-statistic is computed as follows:

$Z = \dfrac{(\hat{P}-P_0)}{\sqrt{(\frac{(P_0) (1-P_0))}{N})}}$

$Z = \dfrac{(0.84- 0.856)}{\sqrt{(\frac{(0.856)(1-0.856))}{50})}}=-1.748$

Test-Statistics

Z_STAT= 1.748

DECISION ABOUT THE NULL HYPOTHESIS :-

1)Since it is observed that$Z_{stat}=|0.322| < Z_{critical}=1.96,$

it is then concluded that the null hypothesis is not rejected.

2)Using the P-value approach:

$P(-1.748<z<1.748)=0.91863$

The p-value is p=0.91863,

since p=0.91863 > 0.05, it is concluded that the null hypothesis is not rejected.

CONCLUSION :-

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population proportion p is different than to $p_0$ , at the α=0.05 significance level.