Question

In: Statistics and Probability

While her husband spent 2½ hours picking out new speakers, a statistician decided to determine whether...

While her husband spent 2½ hours picking out new speakers, a statistician decided to determine whether the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment. The population was Saturday afternoon shoppers. Out of 66 men, 25 said they enjoyed the activity. Nine of the 25 women surveyed claimed to enjoy the activity. Interpret the results of the survey. Conduct a hypothesis test at the 5% level. Let the subscript m = men and w = women.

State the distribution to use for the test.

What is the test statistic?

What is the p-value?

Solutions

Expert Solution

The hypotheses are:

H0:pm=pf

H1:pm>pf

Alpha=level of significance=0.05

State the distribution to use for the test.

since samle sizes are are greater then 30.A
According to central limit theorem follows normal distribution.

test statistic:

pm^=x1/n1=25/66=0.378788

pf^=x2/n2=9/25=0.36

Z=0.378788-0.36-0/sqrt(0.378788*(1-0.378788)/66+0.36(1-0.36)/25)

Z=0.166

Test statistc is 0.166

What is the p-value?

p=0.4343

p>0.05

Fail to reject null Hypothesis.

Accept null hypothesis.

There is no sufficient statistical evidence at 5% level of significance to support the claim that thev

he percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment.

orchestra answered 2 years ago
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