Question

4. a fitness magazine advertises that the mean monthly cost of joining a health club is less than $50. you work for a consumer advocacy group and find that a random sample of 30 clubs has a mean monthly cost of 48.25$ and a standard deviation of $5.23

1. construct a 99% confidence interval for the mean monthly cost of joining a health club

2. at a 0.1 level fo significance, do you have enough evidence to reject the advertisement's claim? your answer should include both hypotheses, the rejection region, the test statistic and a conclusive statement in non-technical terms.

3. based on your conclusion, what type of error are you possibly making (type 1 or 2)? can the probability of this error be easily measured?

4. if the number of fitness clubs you surveyed had been 10 instead of 30. with the same sample mean and standard deviation, would the conclusion of your test in (2) have been different? if so, is it surprising? explain. (assume the distribution of the population to be close to normal

Answer 1

1.

Degree of freedom = n-1 = 30-1 = 29

Critical value of t at df = 29 and 99% confidence interval is 2.756

Standard error of mean = 5.23 / = 0.954863

99% confidence interval  for the mean monthly cost of joining a health club is,

(48.25 - 2.756 * 0.954863 ,  48.25 + 2.756 * 0.954863)

(45.62 ,  50.88)

2.

H0: = $50

Ha: < $50

For 0.01 significance level and left tail test, Rejection region is t < -2.756

Test statistic, t = (48.25 - 50) / 0.954863 = -1.833

Since the observed test statistic (-1.833) does not lie in the rejection region , we fail to reject H0 and conclude that there is a no significant evidence that the mean monthly cost of joining a health club is less than $50.

3.

As, we fail to reject the null hypothesis, we may commit Type 2 error in case the null hypothesis is false. The probability of this error cannot be easily measured as it needs the value of true mean monthly cost .

4.

H0: = $50

Ha: < $50

Standard error of mean = 5.23 / = 1.653871

Degree of freedom = n-1 = 10-1 = 9

For 0.01 significance level , df = 9 and left tail test, Rejection region is t < -3.25

Test statistic, t = (48.25 - 50) / 1.653871 = -1.06

Since the observed test statistic (-1.06) does not lie in the rejection region , we fail to reject H0 and conclude that there is a no significant evidence that the mean monthly cost of joining a health club is less than $50.Thus, the conclusion of test in (2) have not changed.