Question

Does this diet help? A group of 78 people enrolled in a weight loss program that involved adhering to a special diet and to a daily exercise program. After six months, their mean weight loss was 27 pounds, with a sample standard deviation of 10 pounds. A second group of 43 people went on the diet but didn't exercise. After six months, their mean weight loss was 15 pounds, with a sample standard deviation of 8 pounds. Construct a 99.9% confidence interval for the mean difference in weight losses. Let denote the mean weight loss of the group with daily exercise. Use tables to find the critical value and round the answers to the nearest Integer A 99.9% confidence interval for the difference in the mean difference in weight loss, in pounds, i <H- H

Answer 1

Here we have given that

Xi: Weight loss in the first group in pounds

Yi: Weight loss in the second group in pounds

n1=number of people in the first group = 78

$\bar x$ =sample mean of weight loss of first group = 27 pounds

S1=sample standard deviation of first group = 10 pounds

n2=number of people of the second group = 43

$\bar y$ =sample mean of weight loss of second group = 15 pounds

S2=sample standard deviation of second group = 8 pounds

Now we want to find the 95% confidence interval for the difference in the population mean the difference in weight loss$\mu 1- \mu 2$).

For finding this CI 1st we want to check the two population variance equal or not.

i.e. if the population variances are equal then we use the pooled sample standard deviation approach to find Confidence intervals else we use the unpooled standard deviation approach.

Claim: To check whether the two population variance equal or not.

The hypothesis is

$Ho: \sigma 1^2 = \sigma 2^2$

Versus

$H1: \sigma 1^2 \neq \sigma 2^2$

Test statistics is

F-statistitics= $\frac{S1^2}{S2^2}$ =$\frac{10^2}{8^2}=1.56$

Now, we can find the p-value

Degrees of freedom 1=n1-1=78-1=77

Degrees of freedom 2= n2-1= 43-1=42

P-value=0.0588 Using Excel software =FDIST(F-statistics= 1.56, D.F1=77, D.F2=42)

Decision:

C=confidnece level =0.999

$\alpha$=level of significance=1-c=1-0.999=0.001

Here, P-value (0.0588) greater than (>) 0.001 $\alpha$

Conclusion:

Then we fail to reject Null hypothesis

We conclude that the population variances are equal.

Now For equal population variances, we find the confidence interval

Now we can find the 95% confidence interval for the difference in the population mean the difference in weight loss$\mu 1- \mu 2$.

The formula is as follows,

$(\bar x1 - \bar x2 ) - t_{critical}*Sp \sqrt{ \frac{1}{n1}+\frac{1}{n2}} < \mu 1 -\mu 2 < (\bar x1 - \bar x2 ) + t_{critical} *Sp \sqrt{\frac{1}{n1}+\frac{1}{n2}}$

Where,

Sp=Pooled Sample standard deviation = $\sqrt \frac{(n1-1)S1^2+(n2-1)S2^2}{n1+n2-2}$

Now we find the S-pooled

Sp =$\sqrt \frac{(n1-1)S1^2+(n2-1)S2^2}{n1+n2-2}$

     = $\sqrt \frac{(78-1) 10^2+(43-1)8^2}{78+43-2}$

     = $\sqrt {87.2941}$

= 9.3431

now, we can find the critical value,

degrees of freedom= n1+n2-78+43-2= 119

C=confidnece level =0.999

$\alpha$=level of significance=1-c=1-0.999=0.001

This is two tailed confidence interval

$\large t_{critical}$=3.374 Using EXCEL software =TINV(probability = 0.001, D.F=119)

We get the 95% confidence interval is

$(\bar x1 - \bar x2 ) - t_{critical}*Sp \sqrt{ \frac{1}{n1}+\frac{1}{n2}} < \mu 1 -\mu 2 < (\bar x1 - \bar x2 ) + t_{critical} *Sp \sqrt{\frac{1}{n1}+\frac{1}{n2}}$

$\small (27 - 15 ) - 3.374* 9.3431 \sqrt{ \frac{1}{78}+\frac{1}{43}} < \mu 1 -\mu 2 < (27 - 15 ) + 3.374* 9.3431 \sqrt{ \frac{1}{78}+\frac{1}{43}}$

$\small (27 - 15 ) - 5.9875 < \mu 1 -\mu 2 <(27 - 15 ) + 5.9875$

$\small 6.012 < \mu 1 -\mu 2 < 17.988$

A 99.9% confidence interval for the difference in the mean difference in weight loss in pounds is $\small 6.012 < \mu 1 -\mu 2 < 17.988$

Interpretation:

This confidence interval shows the we are 99.9% confidence that the difference in the population mean the difference in weight loss$\mu 1- \mu 2$. will falls within that interval.