Question

An experimenter was interested in dieting and weight loss among men and women. It was believed that women tend to lose more weight than men in the first 2 weeks of a standard dieting program. A random sample of 15 married coupes were put on the same diet. Their weight loss after 2 weeks is listed below. Conduct a directional t-test of the difference using .05 as your level of significance. Be sure to complete all steps.

Wives (pounds)

2.7,4.4,3.5,3.7,5.6,5.1,3.8,3.5,5.6,4.2,6.3,4.4,3.9,5.1,3.4

Husbands (pounds)

5.0,3.3,4.3,6.1,2.5,1.9,3.2,4.1,4.5,2.7,7.0,1.5,3.7,5.2,1.9

Answer 1

          This is a problem of two sample t test, as husband and wife are not correlated.

          We are given the weight loss of a random sample of 15 married couples who were put on the same diet for 2 weeks. Samples are independent of each other.

          Let us denote the weight loss of wives by A and the weight loss of the husbands by B.

Hypothesis to be tested:

          Here we want to test, H0:μA=μB ag. H1: μA≠μB

Test statistic:

The required test statistic is,

T=(XA-XB)-(μA-μB)SA2n+SB2n follows under H0, tn+n-2,here n=15

Where, XA=1ni=1nXAi, XB=1ni=1nXBi,

SA2=1n-1i=1n(XAi-XA)2, SB2=1n-1i=1n(XBi-XB)2

Rejection criterion:

          We reject the null hypothesis at 5% level of significance iff the |t|>t28;0.05

Calculations:

Now, XA=4.333 and XB=3.587

SA2=1.0323 and SB2=2.574

Now, the test statistic, t=(4.333-3.587)115(1.0323+2.574)=1.521

The tabulated value of t28;0.05=1.701

Conclusion:

          Hence, we reject the null hypothesis H0 at 5% level of significance based on the given sample as |t|>t28;0.05 and conclude that the mean weight loss for the husband and wife is not same.