= (0.3 + 0.1 + 1.67 + 0.71 + 1.78 + 0.09 + 0.09)/7 = 0.677
sd = sqrt(((0.3 - 0.677)^2 + (0.1 - 0.677)^2 + (1.67 - 0.677)^2 + (0.71 - 0.677)^2 + (1.78 - 0.677)^2 + (0.09 - 0.677)^2 + (0.09 - 0.677)^2)/7) = 0.6936
a) At 95% confidence interval the critical value is t* = 2.447
The 95% confidence interval is
+/- t* * sd/
= 0.677 +/- 2.447 * 0.6936/
= 0.677 +/- 0.6415
= 0.0355, 1.3185
b) The interval doesn't not contain 0. So there is sufficient evidence to conclude that there has been a change in mean triglyceride level.
c) H0: = 0
H1: > 0
The test statistic t = ( - D)/(sd/)
= (0.677 - 0)/(0.6936/)
= 2.582
At 0.05 significance level the critical value is t0.95, 6 = 1.943
Since the test statistic value is greater than the critical value (2.582 > 1.943), so we should reject the null hypothesis.
At 0.05 significance level, there is sufficient evidence to conclude that the mean triglyceride level has been reduced by the exercise program.
d) Yes, the two approaches 95% confidence interval and test at 5% agree in their conclusions.
Question 4 (14 marks) Triglycerides are the main component of body fat in humans and animals....
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