Question

It's flu season on campus. A study reported that 10% of students suffered some flu-like symptoms during the first week of finals, versus 7% of faculty & staff suffering flu-like symptoms. Suppose 200 students and 200 faculty & staff responded to the study. Let "students" and "faculty & staff" represent population 1 and population 2, respectively. Use Table 1.

(Note: the automated question following this one will ask you confidence interval questions for this same data, so jot down your work.)

a.	Develop the appropriate null and alternative hypotheses to test whether the proportion of students suffering from flu-like symptoms is greater than the proportion of faculty & staff suffering from flu-like symptoms.
	H0: p1 − p2 = 0; HA: p1 − p2 ≠ 0 H0: p1 − p2 ≤ 0; HA: p1 − p2 > 0 H0: p1 − p2 ≥ 0; HA: p1 − p2 < 0

b.	Calculate the value of the test statistic and the p-value.(Round intermediate calculations to 4 decimal places, "Test statistic" value to 2 decimal places and "p-value" to 4 decimal places.) You do not have to "pool" the proportions.

Test statistic
p-value

c.	At the 5% significance level, what is the conclusion? Do the sample data suggest that students suffer more from flu-like symptoms than faculty & staff?
	Yes, since we reject H0. Yes, since we do not reject H0. No, since we reject H0. No, since we do not reject H0.

Now provide confidence interval information from the previous question. Specifically:

a.    What is the value of the point estimate of the difference between the two population proportions?

b.    What is the margin of error at 90% confidence?

        (± what value; please provide to 4 decimals; e.g. "0.1234")

c.    With that margin of error, what is the low number in the confidence interval?

d.    With that margin of error, what is the high number in the confidence interval?

Answer 1

For Students:

n1 = 200, p̂1 = 0.10

For faculty and staff:

n2 = 200, p̂2 = 0.07

α = 0.05

a). Null and Alternative hypothesis:

H0: p1 − p2 ≤ 0; HA: p1 − p2 > 0

b). Test statistic:

z = (p̂1 - p̂2)/√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ]

= (0.1 - 0.07)/√[(0.1*0.9/200) + (0.07*0.93/200)] = 1.0773 = 1.08

p-value :

p-value = 1- NORM.S.DIST(1.0757, 1) = 0.1410

c) Decision:

p-value > α, Do not reject the null hypothesis

No, Since we do not reject Ho.

---------------------------

a) Point Estimate of difference: p̂1 -p̂2 = 0.10 -0.07 = 0.03

b) At α = 0.10, two tailed critical value, z_c = NORM.S.INV(0.10/2) =1.645

Margin of error , E = z_c*√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ]) = ± 0.0458

c) Low number in confidence interval = p̂1 -p̂2 - E = 0.03 - 0.0458 = -0.0158

d) High number in confidence interval = p̂1 -p̂2 + E = 0.03 + 0.0458 = 0.0758