Question

A fitness company conducted an experiment on multiple volunteers for their 15-day "beach body" weight loss program. Their before and after weights were taken and the results are listed below.Before After247.7 232.9177.8 169.7152.7 151.6115.9 119.1242.5 241121.7 126.2241.3 234.5212.8 217.7182.5 176.7231.3 222.8225.9 217.2155.3 154.4187.9 177.5149.4 139.3187.7 194.1176.4 174.3197.7 191.8Required:a. What is the average before weight of the group?b. What is the average after weight of the group?c. Using an appropriate confidence interval, determine if the program was effective over the 15-days in helping people lose weight.

Answer 1

Answer:

a. Mb = 188.6

b. Ma = 184.8

c. The upper-bound 95% confidence interval for the mean is (-∞, -1.3).

As the upper bound of the confidence interval is negative, we are 95% confident that the true mean difference is negative and the treatment is effective.

Step-by-step explanation:

The matched-pair sample data we have is:

Before After Difference

247.7 232.9 -14.8

177.8 169.7 -8.1

152.7 151.6 -1.1

115.9 119.1          3.2

242.5 241         -1.5

121.7 126.2 4.5

241.3 234.5 -6.8

212.8 217.7 4.9

182.5 176.7 -5.8

231.3 222.8 -8.5

225.9 217.2 -8.7

155.3 154.4 -0.9

187.9 177.5 -10.4

149.4 139.3 -10.1

187.7 194.1 6.4

176.4 174.3 -2.1

197.7 191.8 -5.9

We can calculate the mean and standard deviation of the before weight as:

$M_b=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M_b=\dfrac{1}{17}(247.7+177.8+152.7+. . .+197.7)\\\\\\M_b=\dfrac{3206.5}{17}\\\\\\M_b=188.6\\\\\\s_b=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M_b)^2}\\\\\\s_b=\sqrt{\dfrac{1}{16}((247.7-188.6)^2+(177.8-188.6)^2+(152.7-188.6)^2+. . . +(197.7-188.6)^2)}\\\\\\s_b=\sqrt{\dfrac{27057.6}{16}}\\\\\\s_b=\sqrt{1691.1}=41.1$

We can calculate the mean and standard deviation of the after weight as:

$M_a=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M_a=\dfrac{1}{17}(232.9+169.7+151.6+. . .+191.8)\\\\\\M_a=\dfrac{3140.8}{17}\\\\\\M_a=184.8\\\\\\s_a=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M_a)^2}\\\\\\s_a=\sqrt{\dfrac{1}{16}((232.9-184.8)^2+(169.7-184.8)^2+(151.6-184.8)^2+. . . +(191.8-184.8)^2)}\\\\\\s_a=\sqrt{\dfrac{23958}{16}}\\\\\\s_a=\sqrt{1497.4}=38.7$

As this is a matched-paired data, we can test the effectiveness of the program using the sample data for the difference of each pair.

First, we calculate the mean and standard deviation of the pair differences:

$M_d=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M_d=\dfrac{1}{17}((-14.8)+(-8.1)+(-1.1)+. . .+(-5.9))\\\\\\M_d=\dfrac{-65.7}{17}\\\\\\M_d=-3.9\\\\\\s_d=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M_d)^2}\\\\\\s_d=\sqrt{\dfrac{1}{16}((-14.8-(-3.9))^2+(-8.1-(-3.9))^2+(-1.1-(-3.9))^2+. . . +(-5.9-(-3.9))^2)}\\\\\\s_d=\sqrt{\dfrac{607.7}{16}}\\\\\\s_d=\sqrt{38}=6.2$

We have to calculate a 95% confidence interval for the paired difference.  We are only interested in the upper bound, so the lower bound is not calculated.

If the upper bound is negative, that means that we are 95% confident that the mean difference is negative or significantly smaller than 0, which means that the treatment is effective.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=-3.9.

The sample size is N=17.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{6.2}{\sqrt{17}}=\dfrac{6.2}{4.123}=1.504$

The degrees of freedom for this sample size are:

$df=n-1=17-1=16$

The t-value for an upper-bound 95% confidence interval and 16 degrees of freedom is t=1.746.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=1.746 \cdot 1.504=2.63$

Then, the lower and upper bounds of the confidence interval are:

$UL=M+t \cdot s_M = -3.9+2.63=-1.3$

The upper-bound 95% confidence interval for the mean is (-∞, -1.3).

As the upper bound of the confidence interval is negative, we are 95% confident that the true mean difference is negative and the treatment is effective.