Respuesta :
Answer:
Yes, Hypothesis test represents a statistically significant difference from 50% .
Step-by-step explanation:
We are given that 400 students were randomly sampled from a large university, and 289 said they did not get enough sleep.
Let Null Hypothesis, [tex]H_0[/tex] : p = 0.50 {means from the students sampled from the university, proportion of them who did not get enough sleep is 50%}
Alternate Hypothesis, [tex]H_1[/tex] : p [tex]\neq[/tex] 0.50 {means from the students sampled from the university, proportion of them who did not get enough sleep is different from 50%}
The test statistics we will use here is;
T.S. = [tex]\frac{\hat p - p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion = 289/400 = 0.7225
n = sample of students = 400
So, test statistics = [tex]\frac{0.7225 - 0.50}{\sqrt{\frac{0.7225(1-0.7225)}{400} } }[/tex] = 3.143
Now, at 1% level of significance, z table gives critical value of 2.5758. Since our test statistics is more than the critical value which means test statistics will lie in the rejection region, so we have sufficient evidence to reject null hypothesis.
Therefore, we conclude that from the students sampled from the university, proportion of them who did not get enough sleep is different from 50%.
