Question

A study was performed to evaluate a potential relationship between heart rate and exercise minutes per week in a random sample of 11 subjects aged 20 to 29 years.

             The results were as follows:

                           Exercise (min/week)           HR (beats/min)

                                        10                                   88

                                        20                                   83

                                        30                                   78

                                        40                                   75

                                        50                                   78

                                        60                                   75

                                        70                                   70

                                        80                                   66

                                        90                                   63

                                        100                                  65

                                        110                                  60

Using the attached SAS file, perform a simple linear regression analysis with the dependent variable, HR, and independent predictor variable, Exmin/week.

Assume normality and equal variance. Use α = 0.05 (two-tailed) and assume 80% power.

1. State the null and alternative hypotheses for testing the slope.

2. List the critical value for testing the slope.

3. List the R².

4. Report your decision based on the critical value (reject Ho and accept Ha OR fail to reject Ho), P-value, and 95% confidence interval.

5. If the decision was to fail to reject Ho, can Ho be accepted?

SAS File Coding:

data;
input hr exmin;
cards;
10 88
20 83
30 78
40 75
50 78
60 73
70 70
80 66
90 63
100 65
110 60
;
proc reg;
model hr=exmin / clb alpha=.05;
run;

Answer 1

From the observed data hypothesis being la) The fested is Ho! B, co H, :B, 70 The critical values are -2.26 and 2.26 R² = 0.qui (c) Id) The p-value is 0.000 the since level (0.05) p-value (0.000) is less than the slgnificance reject the null hypothesis we NO variables coefficients std error + df = q) / p-value Intercept 88.2545 EXERCISE-0-2573 0.0215 -11.993 1774E-07 please is give up a vote it i questions friends. important for to do more