Question

Researchers wondered if there was a difference between males and females in regard to some common annoyances. They asked a random sample of males and females, the following question: “Are you annoyed by people who repeatedly check their mobile phones while having an in-person conversation?” Among the 571 males surveyed, 192 responded “Yes” among the 549 females surveyed, 206 responded “Yes”. Does the evidence suggest a higher proportion of females are annoyed by this behavior? A. Determine the sample proportion for each sample. B. Explain why this study can be analyzed using the methods for conducting a hypothesis test regarding two independent proportions. C. What are the null and alternative hypotheses? Let p1=females who are annoyed by the behavior and p2=males that are annoyed by the behavior in question. D.Describe the sampling distribution of phat-female- phat male. Draw a normal model with the area representing the P-value shaded for this hypothesis test. The sampling distribution is approximately normal with mean_ and standard deviation _ round to four places as needed.

Answer 1

Solution-A:

sample proportion of males=x1/n1=192/571= 0.3362522

sample proportion of females=x2/n2=206/549=0.3752277

Solution-B:

Two independent simple random samples

n1p1^=571*(192/571)=192>=5

n1(1-p1^)=571*(1-192/571)=571-192=379>=5

n2p2^=549*206/549=206>=5

n2(1-p2^)=549*(1-206/549)=549-206=343>=5

For both samples conditions are satsified

Solution-c:

Ho:p1=p2

here was no  difference between males and females in regard to some common annoyances

Ha:p1 not = p2

here was a difference between males and females in regard to some common annoyances

Solution-d:

sample proportion of males=x1/n1=192/571= 0.3362522

sample proportion of females=x2/n2=206/549=0.3752277

difference in proportions=p1^-p2^=0.3362522-0.3752277=-0.0389755

phat=x1+x2/n1+n2

=(192+206)/(571+549)

= 0.3553571

z=p1^-p2^/sqrt(phat*(1-phat)*(1/n1+1/n2))

=-0.0389755/sqrt(0.3553571*(1-0.3553571)*(1/571+1/549))

z= -1.362369

p value in excel for left tail

==NORM.S.DIST(-1.362369,TRUE)

=0.086540731

=2*0.086540731

=0.1730815

p value=0.17308

p>0.05

Fail to reject Ho

Cocnlcuion:

There is no suffcient statistical evidence at 5% level of significance to conclude that there was a difference between males and females in regard to some common annoyances.