Another way to demonstrate the effectiveness of this “Calories Counter” program
ID: 3051531 • Letter: A
Question
Another way to demonstrate the effectiveness of this “Calories Counter” program was to examine the heaviest of its members. Sixteen members were randomly selected and originally, their mean weight was 185 pounds. After two months on the program, their mean weight was 179 pounds with a standard deviation of 10 pounds. With such a large standard deviation and a relatively small aggregate decrease in weight, the volunteers and the public were still skeptical as to whether the diet plan actually worked. So you, as the statistician, step in and analyze the data. So, from these data, test the claim at 99% confidence that the diet actually worked. What are the Null and Alternative Hypotheses? (1) Prepare the PDF and state the Decision/Rejection Rule for this problem (1, 1) Conduct the test (i.e., determine the test value) (2) State the Decision and Interpretation (1, 1) What is the Pvalue? (2) Using this same information, prepare the 99% confidence interval for this diet plan. (2) Does the 99% confidence interval contain the before average weight? Discuss the information / decision made in parts i) and k). (1,1)
Explanation / Answer
Solution:
Part 1
Here, we have to use one sample t test for the population mean. The null and alternative hypothesis for this test is given as below:
Null hypothesis: H0: The mean weight of members is 185 pounds.
Alternative hypothesis: Ha: The mean weight of members is less than 185 pounds.
H0: µ = 185 versus Ha: µ < 185
This is a lower tailed test. (One tailed test)
For this test, we are given
Confidence level = c = 99% = 0.99
Level of significance = = 1 – c = 1 – 0.99 = 0.01
Sample mean = Xbar = 179
Sample standard deviation = S = 10
Sample size = n = 16
Degrees of freedom = n – 1 = 16 – 1 = 15
Lower critical value = -2.6025 (by using t-table)
We reject H0 if absolute value of test statistic is greater than absolute critical value. We reject H0 if p-value is less than given level of significance.
Test statistic formula is given as below:
Test statistic = t = (Xbar - µ) / [S/sqrt(n)]
Test statistic = t = (179 – 185) / [10/sqrt(16)]
Test statistic = t = -6/[10/4] = -6/2.5
Test statistic = t = -2.40
P-value = 0.0149 (by using t-table)
= 0.01
P-value >
So, we do not reject the null hypothesis that the mean weight of members is 185 pounds.
There is insufficient evidence to conclude that the mean weight of members is less than 185 pounds.
There is insufficient evidence to conclude that the diet plan works.
Part 2
Now, we have to find the 99% confidence interval for the population mean.
Confidence interval = Xbar -/+ t*S/sqrt(n)
Critical t value = 2.9467 (by using t-table)
Confidence interval = 179 -/+ 2.9467*10/sqrt(16)
Confidence interval = 179 -/+ 7.3668
Lower limit = 179 - 7.3668 = 171.63
Upper limit = 179 + 7.3668 = 186.37
Given 99% confidence interval contain the before average weight. So, we do not reject the null hypothesis. Both results match with each other.