Question

4. Nutrition scientists can calculate the amount of weight gain that would be associated with a given increase in calories. In one study, 16 non-obese adults, aged 25 to 36 years, were fed 1000 calories per day in excess of the calories needed to maintain a stable body weight. The subjects maintained this diet for 8 weeks, thus consuming a total of 56,000 calories extra. According to theory, 3500 extra calories will translate to weight gain of 1 pound. Therefore, we expect each of these subjects to gain 56,000/3500 = 16 pounds (lbs). The weights before and after the 8 week period, listed in the file STAT240 Hwk14_P4_Weight Gain.xls, are expressed in kilograms (kg). Note that there are 2.2 lbs per kg. (a) At the 5% significance level, test the null hypothesis that the mean weight gain is 16 lbs versus the alternative hypothesis that it is less than 16 lbs. To do this, you will need to first subtract the weight before from the weight after, and then convert the differences from kilograms to pounds. (This data processing should be done in Excel and does not need to be included with your homework.) Be sure to include all steps of the hypothesis test: i. State the hypotheses in proper statistical notation, defining the population parameter in the context of the problem. ii. Compute the test statistic with degrees of freedom. iii. Report the exact p-value from Excel, as well as an interval for the p-value using the t critical value table. iv. State the conclusion in the context of the problem. (b) If instead of converting the data from kilograms to pounds, we converted po = 16 from pounds to kilograms, would the p-value you computed in part (a) increase, decrease, or stay the same. Explain your answer.

Answer 1

(a) Ho : The mean weight gain is 16 lbs i.e. $\hat{\mu }$ = 16

Ha : The mean weight gain is less than 16 lbs i.e. $\hat{\mu }$ < 16

Level of significance (l.o.s.) : $\alpha$ = 0.05

Test statistic : Paired two sample t-test ( since population variance is not known, we make use of t-test)

and the degrees of freedom = n - 1 = 15

Decision Criteria : Reject Ho at 5% l.o.s. if t cal > t tab , where t tab = t$\alpha$, n-1 = t0.05 , 15 = 1.753

Calculations : Using the t-test: Paired Two Sample for Means function from the data analysis package in Excel, following results are obtained.

t-Test: Paired Two Sample for Means
	weight_before	weight_after
Mean	144.63625	155.045
Variance	515.511425	459.5902667
Observations	16	16
Pearson Correlation	0.986496442
Hypothesized Mean Difference	16
df	15
t Stat	-27.50454039
P(T<=t) one-tail	1.49643E-14
t Critical one-tail	1.753050356
P(T<=t) two-tail	2.99286E-14
t Critical two-tail	2.131449546

t cal = -27.5045 and t tab = 1.753 or p-value = 1.49643E-14

Conclusion : Since t cal > t tab or p-value < 0.05 , we reject Ho at 5% l.o.s. and thus conclude that the mean weight gain is less than 16 lbs i.e. $\hat{\mu }$ < 16

(b) Neither the p-value and nor the t cal would change when we compute the same while considering the kilograms as compared to lbs. This is because the t cal is a fractional equation and thus the conversion rate gets cancel out from both the numerator and the denominator.

This can be proved as the same analysis as shown above is done on the table with kg values and thus, same results are obtained and they are follows :

t-Test: Paired Two Sample for Means
	weight_before	weight_after
Mean	65.74375	70.475
Variance	106.510625	94.95666667
Observations	16	16
Pearson Correlation	0.986496442
Hypothesized Mean Difference	7.27273
df	15
t Stat	-27.50454664
P(T<=t) one-tail	1.49642E-14
t Critical one-tail	1.753050356
P(T<=t) two-tail	2.99285E-14
t Critical two-tail	2.131449546

Hope this answers your query!