Competitive rowers The research in the previous exercise also wondered whether skilled and novice rowers differ in weight or other physical characteristics. Graphs of the data reveal no outliers or strong skewness. Here is the SAS computer output for weight in kilograms:

(a) Calculate the missing t statistic in the computer output. Show your work.

(b) Is there significant evidence of a difference in the mean weights of skilled and novice rowers? Justify your answer. (Note that SAS provides two-sided P-values.)

(c) If the P-value had been computed using the more conservative number of degrees of freedom, would it have been greater than, equal to, or less than $0.6165$Justify your answer without doing any calculations.

The given data is

The given values are

${\overline{x}}_1=70.37$

${\overline{x}}_2=68.45$

$s_1=6.10$

$s_2=9.04$

$n_1=10$

$n_2=8$

Determine the hypothesis

$H_0:{\mu }_1={\mu }_2$

$H_a:{\mu }_1\ne {\mu }_2$

Find the test statistic

localid="1650519634974" $$\begin{gathered} t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ =\frac{70.37-68.45}{\sqrt{\frac{6.{10}^2}{10}+\frac{9.{04}^2}{8}}} \\ \approx 0.5143 \end{gathered}$$.

The given data is

From the given data in part (a)

Find the hypothesis

$H_0:{\mu }_1={\mu }_2$

$H_a:{\mu }_1\ne {\mu }_2$

Find the test statistic

localid="1650519670442" $$\begin{gathered} t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ =\frac{70.37-68.45}{\sqrt{\frac{6.{10}^2}{10}+\frac{9.{04}^2}{8}}} \\ \approx 0.5143 \end{gathered}$$

Find the degree of freedom

localid="1650519689707" $$\begin{gathered} df=\min \left(n_1-1,n_2-1\right) \\ =\min (10-1,8-1) \\ =7 \end{gathered}$$

The P-value is the chance of getting the test statistic's result or a number that is more severe. The P-value is the number (or interval) in Table B's column title that corresponds to the t-value in the row $df=7$

$P>2\times 0.25=0.50$

If the P-value is less than or equal to the significance level, then the null hypothesis is rejected:

$P>0.05\Rightarrow Failtoreject{H}_0$

There is no sufficient evidence to support the claim of a difference.

The given data is

Determine the degrees of freedom using the conservative approach:

$$\begin{gathered} df=\min \left(n_1-1,n_2-1\right) \\ =\min (10-1,8-1) \\ =7 \end{gathered}$$

We note that the degrees of freedom is less than the degrees of freedom in the output of$11.8$

A lower degree of freedom leads to a higher P-value.