Question

In: Statistics and Probability

In a random sample of 62 college students we find that 45 of them eat out...

In a random sample of 62 college students we find that 45 of them eat out at least once a week. Someone tells me that 75% of college students eat out at least once a week. Run a hypothesis test to see if the true proportion of college students that eat out at least once a week is different than 75%

Solutions

Expert Solution

Solution :

Given that,

= 75% = 0.75

1 - = 1 - 0.75 = 0.25

n = 62

x = 45

Level of significance = = 0.05

Point estimate = sample proportion = = x / n = 45 / 62 = 0.73

This is a two tailed test,

Ho: p = 0.75

Ha: p 0.75

Test statistics

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

= ( 0.73 - 0.75) / (0.75*0.25) / 62

= -0.44

p-value = 2*P(Z < -0.44)

= 2*0.3300

= 0.6600

The p-value is p = 0.6600, and since p = 0.6600 ≥ 0.05, it is concluded that the null hypothesis is fails to reject.

Conclusion:

It is concluded that the null hypothesis Ho is fails to reject. Therefore, there is not enough evidence to claim that the

population proportion p is different than at the = 0.05 significance level.


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