Question

In the 1920s, about 97% of U.S. colleges and universities required a physical education course for graduation. Today, about 40% require such a course. A recent study of physical education requirements included 352 institutions: 125 private and 227 public. Among the private institutions, 101 required a physical education course, while among the public institutions, 60 required a course

(a) What are the statistics ( ±0.0001)? X1 = , n1 = , 1p^1 = X2 = , n2 = , 2p^2 =

(b) Use a 95% confidence interval to compare the private and the public institutions with regard to the physical education requirement ( ±±0.0001) The 95% confidence interval if from to Conclusion

We have no evidence to conclude that private institutions are more likely to require physical education

We have evidence to conclude that private institutions are more likely to require physical education

(c) Use a significance test to compare the private and the public institutions with regard to the physical education requirement ( ±0.0001) p^ = z = P-value = Conclusion

We have evidence to conclude that public institutions are more likely to require physical education

We have no evidence to conclude that public institutions are more likely to require physical education Check Syntax

Answer 1

Part a

Required Statistics are given as below:

X1 = 101

n1 = 125

P1_hat = X1/n1 = 101/125 = 0.808

x2 = 60

n2 = 227

P2_hat= X2/n2 = 60/227 = 0.264317181

Part b

Confidence interval = (P1_hat – P2_hat) ± Z*sqrt[(P1_hat*(1 – P1_hat)/N1) + (P2_hat*(1 – P2_hat)/N2)]

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Confidence interval = (0.808 – 0.264317181) ± 1.96*sqrt[(0.808*(1 – 0.808)/N1) + (0.264317181*(1 – 0.264317181)/227)]

Confidence interval = 0.543682819 ± 1.96*0.0458

Confidence interval = 0.543682819 ± 0.0898

Lower limit = 0.543682819 - 0.0898 = 0.4539

Upper limit = 0.543682819 + 0.0898 = 0.6335

Confidence interval = (0.4539, 0.6335)

We have evidence to conclude that private institutions are more likely to require physical education, because confidence interval do not contain zero and both limits are positive.

Part c

Here, we have to use two sample z test for difference in population proportions.

H₀: p₁ = p₂ versus H_a: p₁ > p₂

α = 0.05

Z = (p̂1 - p̂2) / sqrt(p̄*(1 - p̄)*((1/n1)+(1/n2)))

p̄ = (X1+X2)/(N1+N2) = (101 + 60) / (125 + 227) = 0.4574

Z = (0.808 - 0.264317181) / sqrt(0.4574*(1 - 0.4574)*((1/125)+(1/227)))

Z = 0.543682819 / sqrt(0.4574*(1 - 0.4574)*((1/125)+(1/227)))

Z = 9.7984

P-value = 0.0000

(by using z-table)

Critical Z value = 1.6449

(by using z-table)

P-value < α = 0.05

So, we reject the null hypothesis

We have evidence to conclude that public institutions are more likely to require physical education.