A sports writer for local paper asserts that the proportion

A sports writer for local paper asserts that the proportion of games won by the bears over
the past 40 years is 72%. You want to test if the proportion of won games may be different
than the value the writer has claimed. You gather the results for a random sample of 80
games played by the Bears within the past 40 years.
(a) In your sample of 80 games, you found that the Bears have won 55 games. You decide
to use a significance level of 0.04. What is the conclusion from this hypothesis testing?
Can you conclude that the sports writer was incorrect?
(b) For what values of the sample proportion would the null hypothesis be rejected?
(c) Calculate the probability of type II error if the true proportion is 78%.

Solution

(a) In your sample of 80 games, you found that the Bears have won 55 games. You decide
to use a significance level of 0.04. What is the conclusion from this hypothesis testing?
Can you conclude that the sports writer was incorrect?

The test hypothesis:

Ho: p=0.72 (i.e. null hypothesis)

Ha: p not equal to 0.72 (i.e. alternative hypothesis)

The test statistic is

Z=(phat-p)/sqrt(p*(1-p)/n)

=(55/80-0.72)/sqrt(0.72*(1-0.72)/80)

=-0.65

It is a two-tailed test.

Given a=0.04, the critical values are Z(0.02) =-2.05 or 2.05 (from standard normal table)

The rejection regions are if Z<-2.05 or Z>2.05, we reject the null hypothesis.

Since Z=-0.65 is between -2.05 and 2.05, we do not reject the null hypothesis.

So we can not conclude that the sports writer was incorrect

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(b) For what values of the sample proportion would the null hypothesis be rejected?

(phat-0.72)/sqrt(0.72*(1-0.72)/80) =-2.05

--> phat= 0.72-2.05*sqrt(0.72*(1-0.72)/80) =0.617

(phat-0.72)/sqrt(0.72*(1-0.72)/80) =2.05

--> phat= 0.72+2.05*sqrt(0.72*(1-0.72)/80) =0.823

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(c) Calculate the probability of type II error if the true proportion is 78%.

Type II error is

P(0.617<phat<0.823)

=P((0.617-0.78)/sqrt(0.78*(1-0.78)/80) <Z<(0.823-0.78)/sqrt(0.78*(1-0.78)/80))

=P(-3.52<Z<0.93)

=0.8236 (from standard normal table)