Question

In: Statistics and Probability

Researchers compared the amount of DDT in the breast milk of a random sample of 12...

Researchers compared the amount of DDT in the breast milk of a random sample of

12 Hispanic women in Yakima Valley in Washington State with the amount of DDT

in breast milk in the general U.S. population. They measured the standard deviation

of the amount of DDT in the general population to be 36.5 parts per billion (ppb).

Assume the population is normally distributed. We are interested in testing whether

the population standard deviation of DDT level in the breast milk of Hispanic women in

Yakima Valley is grater than that of the general population, using level of significance = :01.

(a) Specify appropriate null and alternate hypotheses.

(b) The sample variance is s2 = 119; 025. Calculate X2obs.

(c) Calculate the p-value and compare it with (alpha symbol). State your conclusion.

(d) Find the critical value and compare it with X2obs. State your conclusion.

(e) Construct a 95% confidence interval for the population variance.

Solutions

Expert Solution

We use chi-square test of variance;

(A)

Null and Alternative Hypotheses (stanadrd deviation=36.5,variance=1332.25

The following null and alternative hypotheses need to be tested:

Ho:σ2=1332.25 or there is no difference in the DDT level in the breast milk of Hispanic women in Yakima Valley and that of the general population

Ha:σ2>1332.5 or the DDT level in the breast milk of Hispanic women in Yakima Valley is greater than that of the general population

(b)

The Chi-Squared statistic is computed as follows:

(c)

P-Value is 0.999954>0.01 hence, the null hypothesis is not rejected or there is no difference in the DDT level in the breast milk of Hispanic women in Yakima Valley and that of the general population.

(d)

=24.725

Since it is observed that χ2=0.983≤χU2​=24.725, it is then concluded that the null hypothesis is not rejected.or there is no difference in the DDT level in the breast milk of Hispanic women in Yakima Valley and that of the general population.

(e)

1.lower limit=

=

=

lower limit=6.99516

2.upper limit=

=

=

upper limit=22.4264

6.99516 < < 22.4264

please rate my answer and comment for doubts.

orchestra answered 1 year ago
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