In: Statistics and Probability
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%
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.