Question

An episode of 60 Minutes had a 15% share of 5000 sample households (750 out of 5000). Use an α = 0.01 significance level to test this proportion claim that the total number of households nationwide tuned in was different than (≠) 20%.

Select the correct P-value and appropriate conclusion answer from the list below:

TIP: Conduct the appropriate Hypothesis Test on your TI 83/84 calculator and consider the P-value it provides

A. My P-value greater than α Alpha, so I Reject Null Hypothesis

B. My P-value greater than α Alpha, so I Accept Null Hypothesis

C. My P-value less than α Alpha, so I Accept Null Hypothesis

D. My P-value less than α Alpha, so I Reject Null Hypothesis

Answer 1

null Hypothesis:              Ho:   p=		0.200
alternate Hypothesis:    Ha: p ≠		0.200

sample success x   =		750
sample size          n    =		5000
std error σp =√(p*(1-p)/n) =		0.0057
sample prop p̂ = x/n=750/5000=		0.1500
z =(p̂-p)/σp=(0.15-0.2)/0.006=		-8.839
p value                          =		0.0000	(from excel:2*normsdist(-8.839)

option D is correct

D. My P-value less than α Alpha, so I Reject Null Hypothesis