Question

1. If we increase our food intake, we generally gain weight. 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, so they consumed a total of 56,000 calories. According to theory, 3500 extra calories will translate into a weight gain of 1 pound. Therefore, we expect each of these subjects to gain 56,000/3500 = 16 pounds (lb), or equivalently, 7.26 kilograms. Use the weight.sas7bdat data (saved as kilograms) to complete the following:

a. Download the data weight.sas7bdat from Canvas-Module 8 to your computer.

Here is the file

The SAS System
id	wtb	wta	wtdiff
1	55.7	61.7	6
2	54.9	58.8	3.9
3	59.6	66	6.4
4	62.3	66.2	3.9
5	74.2	79	4.8
6	75.6	82.3	6.7
7	70.7	74.3	3.6
8	53.3	59.3	6
9	73.3	79.1	5.8
10	63.4	66	2.6
11	68.1	73.4	5.3
12	73.7	76.9	3.2
13	91.7	93.1	1.4
14	55.9	63	7.1
15	61.7	68.2	6.5
16	57.8	60.3	2.5

b. Read this downloaded permanent SAS dataset (weight.sas7bdat) into SAS using data step (you can give your own dataset name). In this data step, I want you to create a new variable “wtdiff” based on the variable “wtb” (weight before) and “wta” (weight after), the expression is given below:

wtdiff = wta - wtb

c. Using proc means, generate summary statistics for weight change variable: wtdiff. Summary statistics includes: mean, median, min, max, standard deviation, variance, 95% confidence limits of means.

d. Report the mean and standard deviation for the weight change “wtdiff”. <ANSWER>

e. Report the 95% confidence interval for weight change in a sentence that explains the meaning of the 95%. <ANSWER>

f. Using PROC TTEST, test the null hypothesis that the mean weight gain is 7.26 kilograms. Hint: this is a one-sample t-test. Be sure to specify the null and alternative hypotheses, the test statistic with degrees of freedom, and the p-value. What do you conclude? <ANSWER>

Answer 1

ANSWER

C.

Mean:	4.73125
Median:	5.05
Range:	5.7
Mode:	6, 3.9, each appeared 2 times
Largest:	7.1
Smallest:	1.4

Variance (s²): 3.047625

standard deviation s = 1.75

E.

The 95% CI is (3.8010, 5.6615)

F.

H0: µ = 7.26 against Ha: µ ≠ 7.26

OR

Ha: µ < 7.26 or Ha: µ > 7.26

Since we are testing a two-sided hypothesis, the significance level shown above is appropriate for our decision-making. Based on this outcome, we reject H0in favor of Ha, (p < .001).We can see that the mean of this sample has a weight change 4.73 kg. Here, thet-statistics is -5.794 with df=15 and 95% confidence interval is -3.4590 to -1.5985

ANSWERED

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