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According to the census data, the average number of languages spoken by American

ID: 3222324 • Letter: A

Question

According to the census data, the average number of languages spoken by American citizens is 1.25. A poll of 30 randomly selected students at a state university shows that the average number of languages the student in this sample can speak is 1.5, with the sample standard deviation of 0.8. The responses have a right-skewed distribution. Does this sample provide significant evidence that students at this university can speak more languages, on average, than Americans in general? (a) State the null and alternative hypotheses. (b) Find the value of the test statistic. (c) Find the P-value of the statistic. (d) Explain in words what this P-value means, in terms of probability. (e) Do we have strong evidence in favor of the alternative hypothesis? Can we reject. the null hypothesis at the 1% significance level? At the 0.1% significance level?

Explanation / Answer

Answer to part a)

The null hypothesis: Population Mean = 1.25

Alternate hypohtesis: Population Mean > 1.25 ..........[right tailed test]

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Answer to part b)

Formula of tests statistics is :

Z = (Xbar - Mean) / (SD/n)

Z = (1.5 -1.25 ) / (0.8/30)

Z = 1.712

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Answer to part c)

P value for Z = 1.712 can be obtained from the Z table

P value = 0.0430

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Answer to part d)

P value means the probability or the chance of the null hypothesis to be true. The higher the P value , this implies there are more chances for the null hypothesis to be true. On the other hand a smaller P value implies , that there are chances for the null to be "not True"

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Answer to part e)

At significance level 5% , we got P value 0.0430 < 0.050 , thus we reject the null hypothesis , and consider that there is strong evidence in favor of the alternate hypothesis

At significance level 1% , we got Pvalue 0.0430 > 0.010 , in this case we "fail to reject" the null hypothesis , and does do not consider that there is strong evidence in favor of the alternate hypothesis

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