Question

a popular blog reports that 60 % of college students log in to facebook on a daily basis. the dean of students at a certain university thinks that the proportion may be different at her university. she polls a simple random sample of 205 students and 134 of them report that they log in to facebook daily differs from 0.60? use the 0.05 level of significance. find the p value and state a conclusion

a) P-value = 0.0584 b) P-value = 0.9416 c) P-value - 0.0094 d) there is not enough evidence to conclude that the proportion of students who long in to facebook daily differs from 0.60 e) there is enough evidence to conclude that the proportion of students who log in to facebook daily differs from 0.60 f) P-value = 0.1168

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p = 0.6
Alternative Hypothesis, Ha: p ≠ 0.6

Rejection Region
This is two tailed test, for α = 0.05
Critical value of z are -1.96 and 1.96.
Hence reject H0 if z < -1.96 or z > 1.96

Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.6537 - 0.6)/sqrt(0.6*(1-0.6)/205)
z = 1.57

P-value Approach
P-value = 0.1168
As P-value >= 0.05, fail to reject null hypothesis.

there is not enough evidence to conclude that the proportion of students who long in to facebook daily differs from 0.60