Question

Ecosystem Response. In the online paper "Changes in Lake Ice: Ecosystem Response to Global Change" (Teaching Issues and Experiments in Enology, time. echoed net, Vol, 3), R. Bohanan et al. questioned whether there is evidence for global warming in long-term data on changes in dates of ice cover in three Wisconsin Lakes. The following table gives data, for a sample of eight years, on the number of days that ice stayed on two lakes in Madison. Wisconsin-Lak Mendota and Lake Monona.

a. Obtain a normal probability plot and boxplot of the paired differences.

b. Based on your results from part (a), is performing a paired $t$-test.

c. At the 10ra significance level, do the data provide sufficient evidence to conclude that a difference exists in the mean length of time that ice stays on these two lakes?

Short answer

a) The graphs can be obtained using software like Minitab. There are no outliers.

b) Yes

c) There is sufficient proof that a difference exists in the mean length of time that ice stays on the two lakes.

Step-by-step solution

Part (a) Step 1: Given Information

$n=$ Sample Size$=8$

$c=$Confidence level $=90\%=0.90$

Part(b) Step 2: Explanation

Here, the difference among each pair of values is needed to calculate as follows,

Normal Probability plot

Box-plot

Part(b) Step 1: Given Information

To calculate on the basis of results from part (a), $t-$procedures are justifiable to use or not.

Part(b) Step 2: Explanation

If the model in the normal probability plot is roughly linear, the population is said to have a normal distribution.

However, because the resulting graph does not show a strong curvature or outliers, it is safe to assume that the data came from a normal population. As a result, the t-procedure can be used on the given data.

Part(c) Step 1: Given Information

To conclude at $10\%$significance level, do the data provide sufficient evidence to conclude that a difference exists in the mean length of time that ice stays on the two lakes.

The mean of males is different from the mean of females.

This claim could be either a null hypothesis or an alternate hypothesis.

$H_0:{\mu }_d=0$

$H_{\alpha }:{\mu }_d\ne 0$

Part(c) Step 2: Explanation

$$\begin{gathered} \overline{\mathrm{d}}=\text{Sample Mean} \\ =\frac{12+7+1+0-11+0-11+0+6-4}{8} \\ \approx 1.375 \end{gathered}$$

$S_d$= Sample standard deviation

$$\begin{gathered} =\sqrt{\frac{(12-1375{)}^2+\dots +(-4-1.375{)}^2}{8-1}} \\ \approx 7.09 \end{gathered}$$

Determine the critical value using$$\begin{gathered} df=n-1 \\ =8-1 \\ =7 \end{gathered}$$

Determine the test statistic as follows:

$$\begin{gathered} t=\frac{\overline{d}-{\mu }_d}{\frac{s_d}{\sqrt{n}}} \\ =\frac{1.375-0}{\frac{7.09}{\sqrt{8}}} \\ \approx 0.549 \end{gathered}$$

The null hypothesis is rejected if the P-value is less than the significance level.

Because the P-value exceeds the significance level, the answer fails to reject the null hypothesis.

There is insufficient evidence to conclude that there is a difference in the average length of time that ice remains on the two lakes.