Question

1. It is known that the average person has 9.5 alcoholic drinks per week. Suppose we take a sample of 25 people and find that the average amount of alcoholic drinks per week they have is 7.8 with a sample standard deviation of 9.6 drinks. We are interested in testing if the average amount of alcoholic drinks per week is actually lower than 9.5 now.
   1. Using the sample of 25 people, what would the 95% confidence interval be for the population mean?
   2. What are the null and alternative hypotheses?
   3. What is the critical value at 95% confidence?
   4. Calculate the test statistic.
   5. Find the p-value
   6. What conclusion would be made here at the 95% confidence level?
   7. Would my conclusion change if I changed alpha to .01? Show reasonin

Answer 1

Here n = 25

Sample mean = = 7.8

Sample standard deviation = s = 9.6

Here Population standard deviation is not known so we will use here t distribution.

a) Confidence level = c = 0.95

95% confidence interval for the population mean is

Where tc is t critical for degrees of freedom = n -1 = 25 -1 = 24 and confidence level =c=0.95

tc = 2.064                                                   (From statistical table of t values)

                               (Round to 4 decimal)

95% confidence interval for the population mean is (3.8371, 11.7629)

b) Null hypothesis

Alternative hypothesis:

Where

c) We have calculated t critical value in part (a).

Critical value = 2.064

d) Test statistic:

t = -0.885                   (Round to 3 decimal)

Test statistic = -0.885

e) P value:

Degrees of freedom = n - 1 = 25 - 1 = 24

P value from excel using command:

=T.DIST(0.885,24,TRUE).

=0.8075                                      (Round to 4 decimal)

P value = 0.8075

f) Confidence level = c = 0.95

alpha = 1 - c = 1 - 0.95 = 0.05

Here P value = 0.8075 > alpha = 0.05

So we fail to reject null hypothesis H0.

Conclusion: There is not sufficient evidence to conclude that the average amount of alcoholic drinks per week is actually lower than 9.5.

g) If alpha = 0.01 then conclusion would not change here since,

P value = 0.8075 > alpha = 0.01

So we fail to reject null hypothesis H0.

Conclusion: There is not sufficient evidence to conclude that the average amount of alcoholic drinks per week is actually lower than 9.5.