Question

A nutritionist has developed a diet that she claims will help people lose weight. Twelve people were randomly selected to try the diet. Their weights were recorded prior to beginning the diet and again after 6 months. Here are the original weights, in pounds, with the weight after 6 months in parentheses. Before 192 212 171 215 180 207 165 168 190 184 200 196 After 183 196 174 211 160 191 162 175 190 179 189 195 Test the claim that the diet is effective at the 0.05 level of significance.

Answer 1

The diet which is developed by the nutritionist is effective at the 0.05 level of significance. Claim is agreed.

When do we use two-sample t-test?

The two-sample t-test is used to determine if two population means are equal.

A nutritionist has developed a diet that she claims will help people lose weight. In this,

• Twelve people were randomly selected to try the diet.
• Their weights were recorded prior to beginning the diet and again after 6 months.

Here are the original weights, in pounds, with the weight after 6 months in parentheses.

• Before 192 212 171 215 180 207 165 168 190 184 200 196
• After    183 196  174 211 160 191   162 175 190  179  189 195

The mean of the weights before 6 moths is,

$\overline X_1=\dfrac{192+ 212 +171 +215 +180 +207 +165 +168 +190 +184 +200 +196 }{12}\\\overline X_1=190$

The mean of the weights after 6 months is,

$\overline X_2=\dfrac{ 183 +196 +174 +211 +160 +191 +162 +175 +190 +179 +189 +195 }{12}\\\overline X_1=183.75$

Standard deviation of both the data is 16.9 and 14.7.

1. Null and Alternative Hypotheses.

The following null and alternative hypotheses need to be tested:

$\begin{array}{ccl} H_0: \mu_1 & = & \mu_2 \\\\ \\\\ H_a: \mu_1 & > & \mu_2 \end{array}$

This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

• (2) Rejection Region

Based on the information provided, the significance level is α=0.05 and the degrees of freedom are df = 22. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

$df_{Total} = df_1 + df_2 = 11 + 11 = 22$

Hence, it is found that the critical value for this right-tailed test is

$t_c=1.717$, for α=0.05 and df=22

The rejection region for this right-tailed test is,

$R = \{t: t > 1.717\}$

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$t = \displaystyle \frac{\bar X_1 - \bar X_2}{\sqrt{ \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}(\frac{1}{n_1}+\frac{1}{n_2}) } }$

$\displaystyle \frac{ 190 - 183.75}{\sqrt{ \frac{(12-1)16.9^2 + (12-1)14.7^2}{ 12+12-2}(\frac{1}{ 12}+\frac{1}{ 12}) } } = 0.967$

• (4) Decision about the null hypothesis

Since it is observed that t=0.967≤tc=1.717 it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.1721 and since p=0.1721≥0.05p = 0.1721  it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis H₀ is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is greater than μ2​, at the α=0.05 significance level.

Confidence Interval

The 95% confidence interval is −7.16<μ<19.66

Thus, the diet which is developed by the nutritionist is effective at the 0.05 level of significance. Claim is agreed.

Learn more about the two-sample t-test here;

brainly.com/question/27198724

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