Find the Critical Chi-Square Values: a. df = 28: SL = 99% Answer: ___ b. df = 14

Question

Find the Critical Chi-Square Values: a. df = 28: SL = 99% Answer: ___ b. df = 14: SL = 99.5% Answer: ___ c. df = 70: Sl = 10% Answer: ___ d. df = 40: SL = 90% Answer____ e. df = 8: SL = 10% Answers: ___ The City of Anchor banks, Alaska has three major television stations, each with its own evening news program fro 6 pm to 7pm. The NOONI GROUP is hired to determine to determine if there is a significance difference among the viewing audience for any station. A random sample of 150 viewers revealed that 50 watched the evening news on KENI-TV, 30 viewers watched their news on KTUU-TV and 70 viewers watched their news on KTVA- TV. At a 2.5 significance level is there sufficient evidence to show that the three TV stations have equal shares of the evening news audience.

Explanation / Answer

Ans:1)Critical chi square values

Using Excel function CHIINV:

a. CHIINV(0.99,28)=13.565

b.CHIINV(0.995,14)=4.075

c.CHIINV(0.1,70)=85.527

d.CHIINV(0.9,40)=29.051

e.CHIINV(0.1,8)=13.362

2)

H0:Three TV stations have equal shares of the eveing new audience.

H1:Three TV stations do not have equal shares of the evening new audience.

Expected count=150/3=50

(as three channels are equally likely to be watched)

Test statistic:

Calculated chi square score=16

Critical chi square score(for 0.025 significance level and df=3-1=2)=7.38

As,calculated chi square score >critical chi square score,we reject the null hypothesis,H0.

p-value approach:

p-value=CHIDIST(16,2)=0.0003

As,p-value<0.025,we reject the null hypothesis,H0.

We have sufficient evidence to conclude that three TV stations do not have equal shares of the evening new audience.

KENI-TV KTUU-TV KTVA-TV Total Observed (O) 50 30 70 150 Expected(E) 50 50 50 150 (O-E)^2/E 0 8 8 16

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