Question

A university financial aid office polled an SRS of undergraduate students to study their summer employment. Not all students were employed in the previous summer. Among 797 students, 518 of them were employed.

a. Is there evidence to support that the proportion of students employed during the summer is greater than 60%? Using a significance level 0.05.

b. Construct a 95% confidence interval for the proportion of students who were employed during the summer.

Answer 1

(a ) Solution :

Givan that

This is the right tailed test .

H0 : p = 0.60

Ha : p > 0.60

= x / n = 797 / 518 = 0.6499

P0 = 0.60

Test statistic = z

= - P0 / [P0 * (1 - P0 ) / n]

= 0.6499 - 0.60 / [0.60 ( 1-0.60 ) / 797 ]

= 2.878

The test statistic = 2.878

P-value = 0.002

= 0.05    

0.002 < 0.05

P-value <

Reject the null hypothesis .

There is sufficient evidence to the claim

( b )

Given that,

n = 797

x = 518

Point estimate = sample proportion = = x / n = 797 / 518 = 650

1 - = 0.350

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025 = 1.960

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.96 ( ( 0.650 * 0.350 ) / 797)

= 0.033

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.650 - 0.033 < p < 0.650 - 0.033

0.617 < p < 0.683

The 95% confidence interval for the population proportion p is : ( 0.617 , 0.683 )