Question

According to a report, the proportion of Lancaster University students who reported insufficient rest or sleep during each of the preceding 30 days is

8.0%, while this proportion is 8.8% for University of Cumbria students. These data are based on simple random samples of 11,545 Lancaster and 4,691 Cumbria students.

(a) Calculate a 95% confidence interval for the difference between the proportions of Lancaster and Cumbria students who are sleep deprived and interpret it in the context of the data.

(b) Conduct a hypothesis test to determine if these data provide strong evidence the rate of sleep deprivation is different for the two Universities. (Reminder: Check conditions.)

(c) It is possible the conclusion of the test in part (b) is incorrect. If this is the case, what type of error was made?

Answer 1

Given : p1 = 8% = 0.08 , p2 = 8.8% = 0.088 , n1=11545, n2 =4691

In Lancaster Univerity 8% of them reportes insufficient sleep which is =11545*8/100 = 923.6

In Cumbria University 8.8% of them reportes insufficient sleep which is =4691*8.8/100 = 412.81

a) 95% Confidence interval for the difference between the proportions ={ (p1 - p2) - Z_{p1 - p2},(p1 - p2) + Z_{p1 - p2}}

where _p1-p2 = {p1(1-p1)/n1 + p2(1-p2)/n2} = 0.0048

CI_95% = {(0.08-0.088)-1.96*0.0048,(0.08-0.088) + 1.96*0.0048} = {-0.0174 , 0.0014}

b) Null hypothesis,H₀ : p1=p2 against Alternative Hypothesis,H₁ : p1 p2

Pooled sample proportion, p = (p1 * n1 + p2 * n2) / (n1 + n2) = 0.0823

SE = sqrt{ p*(1-p)*[(1/n1) + (1/n2)]} = 0.0048

Test statistic, z = (p1 - p2) / SE = (0.08 - 0.088) / 0.0048 = -1.67

Calculated z-statistic is less than the z table value at 5% level of significance

Hence we accept H₀ and conclude that the rate of sleep deprivation is same for both the universities.

c) It is possible that the test in part (b) is incorrect. The Type II error was made.