Question

answer neatly and correctly please!

A recent study at a local college claimed that the proportion, p, of students who commute more than fifteen miles to school is no more than 20%. If a random sample of 270 students at this college is selected, and it is found that 58 commute more than fifteen miles to school, can we reject the college's claim at the 0.05 level of significance? Perform a one-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. u op The null hypothesis: XX s The alternative hypothesis: The type of test statistic: (Choose one) O=O oso 020 [+O OO OO x 6 ? The value of the test statistic: (Round to at least three decimal places.) The critical value at the 0.05 level of significance: (Round to at least three decimal places.) Can we reject the claim that the proportion of students who commute more than fifteen miles to school is no more than 20%? Yes No

Answer 1

Solution :

Given that,

$p_{0}$ = 0.20

1 - $p_{0}$ = 0.80

n = 270

x = 58

Level of significance = $\alpha$ = 0.05

Point estimate = sample proportion = $\hat p$ = x / n = 0.215

This a right (One) tailed test.

The null and alternative hypothesis is,

Ho: p = 0.20

Ha: p 0.20

Test statistics

z = ($\hat p$ - $p_{0}$ ) / $\sqrt{}$ $p_{0}$*(1-$p_{0}$) / n

= ( 0.215 - 0.20) / $\sqrt{}$ (0.20*0.80) / 270

= 0.609

Critical value of  the significance level is α = 0.05, and the critical value for a right-tailed test is

$Z_{c}$ = 1.645

Since it is observed that z = 0.609 < $Z_{c}$ = 1.645 , it is then concluded that the null hypothesis is fail to reject.

Conclusion:

No.