Question

A research center estimates that no more than 30% of college students eat breakfast every day. In a random sample of 250 college students say, half of the sample say they eat breakfast every day. At α=0.05, is there enough evidence to reject the researcher's claim?

Answer 1

Solution :

Given that,

= 030

1 - = 0.70

n = 250

x = 125

Level of significance = = 0.05

Point estimate = sample proportion = = x / n = 0.5

This a right (One) tailed test.

The null and alternative hypothesis is,

Ho: p 0.30

Ha: p 0.30

Test statistics

z = ( - ) / *(1-) / n

= ( 0.5 - 0.30 ) / ( 0.30 * 0.70 ) / 250

= 6.901

P-value = P(Z>z)

= 1 - P(Z <z )

= 1- P(Z < 6.901)

= 1 - 1

= 0.000

The p-value is p = 0.0, and since p = 0. < 0.05, it is concluded that the null hypothesis is rejected.

Conclusion:

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the researcher's claim. at the α = 0.05 significance level.