fter a losing season, there is a great uproar to fire the head football coach. I

Question

fter a losing season, there is a great uproar to fire the head football coach. In a random sample of 300 college alumni, 138 favor keeping the coach. Test at the .05 level of significance whether the proportion of alumni who support the coach is less than 50 percent.

State the null hypothesis and the alternate hypothesis. (Round your answers to 2 decimal places.)

State the decision rule for .05 significance level. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)

Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)

Test at the .05 level of significance whether the proportion of alumni who support the coach is less than 50 percent.

fter a losing season, there is a great uproar to fire the head football coach. In a random sample of 300 college alumni, 138 favor keeping the coach. Test at the .05 level of significance whether the proportion of alumni who support the coach is less than 50 percent.

Explanation / Answer

Answer in full with calculation and details :

p = x/n = 138/300 = .46 ( sample prop.)
p0 = .50 ( pop. prop)

a.

The claim always goes to the alternative hypothesis.

H0: p >= .50
Ha: p<.50

b.

The decision rule is that :

Reject Ho: if Z is < -1.645
For this we have gone to the Z tables and taken out Z equivalent of alpha = .05 ( the critical value in the question)

c.

Test statistic , Z = (p-p0)/sqrt(p0*p0'/n) = (.46-.50)/sqrt(.5*.5/300) = -1.3856

d.

Since Z statistic is more than critical value of -1.645 we conclude that we can't reject null hypothesis. Hence, p>=.50

So, the claim that  the proportion of alumni who support the coach is less than 50 percent is FALSE

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.