Question

Well it’s time to put up…. You promised as a New Year’s resolution that you would lose some weight and as a result, you joined a group known as the “Calorie Counter”. The reason for joining is that they boasted of significant weight loss in two months for all their members. Well it’s almost the end of February and you have been diligent and have worked hard. Now is the time to put up….

For eight randomly chosen members of your Calorie Counter group, their results are as follows:

	Weight (LBs)
Person ID	Before	After
1	125	120
2	98	96
3	123	127
4	112	115
5	187	161
6	154	168
7	127	116
8	245	220
Sums	1,171	1,123

It seems that almost 50 pounds were lost in aggregate by these eight people and that the program actually does something. But you, as the statistician, are not convinced. So you use these data to test at 95% confidence and the dependent sample approach, that there is indeed a significant weight loss.

a)    Determine d (the mean of the differences) (5)

b)    Determine Sd (the standard deviation of the differences) (5)

c)    State the Null and Alternative hypothesis for this problem. (1)

d)    Prepare the PDF and state the Rejection/Decision rule. (2)

e)    Perform the test. (5)

f)     What are the decision and the interpretation? (1, 1)

Answer 1

For part a) and b) look the following excel output

The formulae used on the above excel-sheet are as follows:

Here we want to test "whether the weight loss after the treatment or not "

Let's write null hypothesis (H₀) and alternative hypothesis(H₁)

Where is the population mean of the difference Before - After.

d)    Prepare the PDF and state the Rejection/Decision rule.

Let's assume that the weights follows normal distribution.

Here we use sample standard deviation so we need to use t distribution to test the above hypothesis.

Let's find critical value

degrees of freedom = n - 1 = 7

level of significance = 1 - 0.95 = 0.05

critical t = "=TINV(2*0.05,7)" = 1.895

Here we reject null hypothesis if the test statistic value > 1.895

e)    Perform the test.

Let's find t test statistic value.

f)     What are the decision and the interpretation?

Here at 5% level of significance, we fail to reject the null hypothesis because t = 1.207 < 1.895

Interpretation: From these data , with 95% confidence , we do'nt have sufficient evidence to say that the weight loss is significant.