In a random sample of males it was found that 25 write with

In a random sample of males, it was found that 25 write with their left hands and 210 do not.

In a random sample of females, it was found that 64 write with their left hands and 468 do not.

Use a 0.05 significance level to test the claim that the rate of left-handedness among males is less than that among females.

Complete parts (a) through (c) below.

a.?Test the claim using a hypothesis test.

Consider the first sample to be the sample of males and the second sample to be the sample of females. What are the null and alternative hypotheses for the hypothesis test?

A. H0: p1=p2 H1: p1>p2

B. H0: p1?p2 H1: p1?p2

C. H0: p1=p2 H1: p1?p2

D. H0: p1=p2 H1: p1<p2

E. H0: p1?p2   H1: p1=p2

F. H0: p1?p2   H1:p1?p2

Identify the test statistic.

Identify the test statistic.

z=_____

(Round to two decimal places as needed.)

Identify the P-value.

P-value=______

(Round to three decimal places as needed.)

What is the conclusion based on the hypothesis test?

The P-value greater than////or ////not in the significance level of ?=0.05,?

so?reject//// or ////fail to reject the null hypothesis. There is sufficient/// or// is not sufficient//// evidence to support the claim that the rate of left-handedness among males is less than that among females.

b.?Test the claim by constructing an appropriate confidence interval.

The 90% confidence interval is?

_____<(p1?p2)< _______

(Round to three decimal places as needed.)?

Solution

D. H0: p1=p2 H1: p1<p2

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p1=25/(25+210) = 0.106383

p2=64/(64+468) = 0.1203008

Z=(p1-p2)/sqrt(p1*(1-p1)/n1+p2*(1-p2)/n2)

=(0.106383-0.1203008)/sqrt(0.106383*(1-0.106383)/235+0.1203008*(1-0.1203008)/532)

=-0.57

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It is a left-tailed test.

So the p-value = P(Z<-0.57) = 0.284 (from standard normal table)

The P-value greater than in the significance level of ?=0.05

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so fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the rate of left-handedness among males is less than that among females.

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Given a=0.1, Z(0.05) = 1.645 (from standard normal table)

So the lower bound is

(p1-p2)-Z*sqrt(p1*(1-p1)/n1+p2*(1-p2)/n2)

=(0.106383-0.1203008)-1.645*sqrt(0.106383*(1-0.106383)/235+0.1203008*(1-0.1203008)/532)

=-0.054

So the upper bound is

(p1-p2)+Z*sqrt(p1*(1-p1)/n1+p2*(1-p2)/n2)

=(0.106383-0.1203008)+1.645*sqrt(0.106383*(1-0.106383)/235+0.1203008*(1-0.1203008)/532)

=0.026