A survey was done at U/Mass. Boston to see if women students at the University w

Question

A survey was done at U/Mass. Boston to see if women students at the University were more likely to have a cat than were men students. The researcher selected 300 men at random and independently selected 500 women at random. Of the men selected 93 had a cat, while of the women selected 180 had a cat. Find the percentage of cat owners in each sample. Is the difference in the percentages of the men and the women real or can it be explained as a chance variation? Formulate the null and alternative hypotheses in terms of a box model before answering the question. Then show the statistical test chosen, the statistic used, its standard error, the resulting standard units, the P-value generated from the data, and the decision indicated by that P-value.

Explanation / Answer

Sol:

find sample proportion of men and women

sample proporton=x/n=successes/total

p1=%catowners in women=180/500*100=36%

p2=% cat owners in men=93/300*100=31%

hypothesis test for difference in proportions

Null hypothesis:

H0:p1=p2

Alternative Hypothesis

H1:p1>p2

level of significance=0.05

z=0.36-0.31/sqrt[0.36(1-0.36)/500+0.31(1-0.31)/300]

z=1.46

The P-Value is 0.072145.

The result is not significant at p < 0.05.

Fail to reject Null hypothesis

Decision:

Accept Null hypothesis:

Conclusion:

there is no sufficient statistical evidence at 5% level of significance to support the claim that' women students are more like to have cat than men.

ANSWER A:

CONCLUSION HOLDS ABOUT POPULATION

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