Question

Data taken from a random sample of 60 students chosen from the student population of a large urban high school indicated that 36 of them planned to pursue post-secondary education. An independent random sample of 50 students taken at a neighboring large suburban high school resulted in data that indicated that 31 of those students planned to pursue post-secondary education. Do these data provide sufficient evidence at the 5% level to reject the hypothesis that these population proportions are equal

Answer 1

No, these data do not provide sufficient evidence at the 5% level to reject the hypothesis that these population proportions are equal.

Explanation:

We are given that data taken from a random sample of 60 students chosen from the student population of a large urban high school indicated that 36 of them planned to pursue post-secondary education.

An independent random sample of 50 students taken at a neighboring large suburban high school resulted in data that indicated that 31 of those students planned to pursue post-secondary education.

Let
`p_1` = population proportion of students of a large urban high school who pursue post-secondary education.

`p_2` = population proportion of students of a large suburban high school who pursue post-secondary education.

So, Null Hypothesis,
`H_0` :
`p_1-p_2` = 0 {means that these population proportions are equal}

Alternate Hypothesis,
`H_A` :
`p_1-p_2\\eq` 0 {means that these population proportions are not equal}

The test statistics that would be used here Two-sample z proportion statistics;

T.S. =
`\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } }` ~ N(0,1)

where,
`\hat p_1` = sample proportion of students of a large urban high school who pursue post-secondary education =
`(36)/(60)` = 0.60

`\hat p_2` = sample proportion of students of a large urban high school who pursue post-secondary education =
`(31)/(50)` = 0.62

`n_1` = sample of students of a large urban high school = 60

`n_2` = sample of students of a large suburban high school = 50

So, the test statistics =
`\frac{(0.60-0.62)-(0)}{\sqrt{(0.60(1-0.60))/(60)+(0.62(1-0.62))/(50) } }`

= -0.214

The value of z test statistics is -0.214.

Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that these population proportions are equal.