91Ó°ÊÓ

$n=$ Sample Size$=8$

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

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

Normal Probability plot

Box-plot

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

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.

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.

$H_0:{\mathrm{μ}}_d=0$

$H_{\mathrm{Î\pm }}:{\mathrm{μ}}_d≠0$

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

$S_d$= Sample standard deviation

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

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

Determine the test statistic as follows:

$$\begin{gathered} t=\frac{\stackrel{¯}{d}-{\mathrm{μ}}_d}{\frac{s_d}{\sqrt{n}}} \\ =\frac{1.375-0}{\frac{7.09}{\sqrt{8}}} \\ ≈0.549 \end{gathered}$$

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.