Chapter 9.1, Problem 22E

a.

To determine

State the null and alternate hypotheses.

Answer to Problem 22E

The hypotheses are given below:

Null hypothesis:

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

That is, there is no significant difference between the mean weight loss of two diets (low-carbohydrate diet and low-fat diet).

Alternate hypothesis:

$H_1:{\mu }_1\ne {\mu }_2$

That is, there is a significant difference between the mean weight loss of two diets (low-carbohydrate diet and low-fat diet).

Explanation of Solution

Calculation:

It is given that, the statistics regarding the weight loss of candidates under low-carbohydrate diet is $n_1=77$, ${\overline{x}}_1=4.7$, $s_1=7.16$ and the statistics regarding the weight loss of candidates under low-fat diet is $n_2=79$, ${\overline{x}}_2=2.6$, $s_2=5.9$. Furthermore, it is given that $\alpha =0.01$.

Hypothesis:

Hypothesis is an assumption about the parameter of the population, and the assumption may or may not be true.

Let ${\mu }_1$ be the population mean weight loss subjects under low-carbohydrate diet and ${\mu }_2$ be the population mean weight loss subjects under low-fat diet.

Claim:

Here, the claim is whether the mean weight loss differs between the two diets or not.

The hypotheses are given below:

Null hypothesis:

Null hypothesis is a statement which is tested for statistical significance in the test. The decision criterion indicates whether the null hypothesis will be rejected or not in the favor of alternate hypothesis.

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

That is, there is no significant difference between the mean weight loss of two diets (low-carbohydrate diet and low-fat diet).

Alternate hypothesis:

Alternate hypothesis is contradictory statement of the null hypothesis

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

That is, there is a significant difference between the mean weight loss of two diets (low-carbohydrate diet and low-fat diet).

b.

To determine

Find the value of t-test statistic.

Answer to Problem 22E

The value of t-test statistic is 1.996454.

Explanation of Solution

Calculation:

Requirements for a small sample t-test:

• The samples must be independent and drawn randomly from the populations.
• Either the sample size must be greater than 30 or the population must be approximately normal.

The test statistic for the small sample t is obtained as follows:

$t=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}}}$

Under the null hypothesis, $\left({\mu }_1-{\mu }_2\right)=0$.

Therefore the test statistic is,

$\begin{array}{c}t=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)-0}{\sqrt{\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}}}\\ =\frac{4.7-2.6}{\sqrt{\frac{{7.16}^2}{77}+\frac{{5.9}^2}{79}}}\\ =\frac{2.1}{1.051865}\\ =1.996454\end{array}$

Thus, the test statistic is 1.996454.

c.

To determine

Find the number of degrees of freedom for the test statistic.

Answer to Problem 22E

The number of degrees of freedom for the test statistic is 76.

Explanation of Solution

Calculation:

Degrees of freedom:

The degrees of freedom for t using computer package is,

$\Delta =\frac{{\left[\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)\right]}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s_2^2}{n_2}\right)}^2}{n_2-1}}$

The degrees of freedom, when computing by hand is smaller of $n_1-1$ and $n_2-1$.

The degrees of freedom is,

$\begin{array}{c}d.f=\min \left(n_1-1,n_2-1\right)\\ =\min \left(77-1,79-1\right)\\ =\min \left(76,78\right)\\ =76\end{array}$

Thus, the degree of freedom is 76.

d.

To determine

Check whether the null hypothesis is rejected at $\alpha =0.01$.

Answer to Problem 22E

There is not enough evidence to reject the null hypothesis at $\alpha =0.01$.

There is no significant difference between the mean weight loss of two diets (low-carbohydrate diet and low-fat diet).

Explanation of Solution

P-value:

Step-by-step procedure to obtain the P-value using the MINITAB software:

• Choose Graph > Probability Distribution Plot
• Choose View Probability > OK.
• From Distribution, choose‘t’ distribution.
• In Degrees of freedom, enter 76.
• Click the Shaded Area tab.
• Choose X value and Both Tail for the region of the curve to shade.
• In X-value enter 1.996454.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the P-value is $2\times 0.02473=0.04946$.

Thus, the P-value is 0.04946.

Decision rule based on P-value:

If $P-\text{value}\le \alpha$, then reject the null hypothesis $H_0$.

If $P-\text{value}>\alpha$, then fail to reject the null hypothesis $H_0$.

Here, the level of significance is $\alpha =0.01$.

Conclusion based on P-value approach:

The P-value is 0.04946 and $\alpha$ value is 0.01.

Here, P-value is greater than the $\alpha$ value.

That is, $0.04946\left(=P-\text{value}\right)>0.01\left(=\alpha \right)$.

By the rejection rule, fail to reject the null hypothesis.

Hence, it can be concluded that there is no significant difference between the mean weight loss of two diets (low-carbohydrate diet and low-fat diet).

Thus, it can be concluded that there is sufficient evidence to infer the equality of two population means at $\underset{\_}{\alpha =0.01}$.