Ecesystem Response. Refer to Problem 10, and find a 90\% confidence interval for the difference in the mean lengths of time that ice stays on the two lakes. Interpret your result.

We need to find the $90\%$confidence interval for the difference in the mean lengths of tme that ice stays on the two lakes.

The following data shows the data for a sample of 8 years, on the number of days that ice stayed on two lakes in MENDOTA and MONONA.

Since the sample size is small and the two samples are related and can be paired according to the years, we have to consider the paired $t$-test.

The significance level is $10\%$

Here we need to compare two pairs of small samples for the difference of the means. So we have to consider the paired $t$-test. Let ${\mu }_1$and ${\mu }_2$denote, respectively the lengths of times that ice stays on the two lakes. To perform the hypothesis test the null and alternative hypothesis are as follows.

$H_0:{\mu }_1={\mu }_2$(Mean lengths of times that ice stays on the two lakes are same)

$H_a:{\mu }_1\ne {\mu }_2$(Mean lengths of times that ice stays on the two lakes are not same)

Use MINITAB software to perform the Paired $t$-test as follows.

1) Choose Stat >basic Statistics>Paired t ...

2) Select samples in columns.

3) Click in first text box and click specifyMENDOTA.

4) Click in second text box and click specifyMONONA.

5) Click the Options .... Button

6) Click in the Confidence level text box and type $90$

7) Click in thetest difference text box and type $0$

8) Click arrowbutton at the right of the Alternative drop-down list box and select not equal

9) Click Ok twice

The following is the Minitab out put for the hypothesis test and the confidence interval.

From the above out put, we have the confidence interval for difference of mean lengths of times that ice stays on the two lakes is $(-3.37412,6.12412)$

Therefore

We can say with $90\%$confidence that mean lengths of times that ice stays on the two lakes will be in between $-3.4$and $6.1$days.