Question

A recent study stated that if a person chewed gum, the average number of sticks of gum he or she chewed
daily was 8. To test the claim, a researcher selected a random sample of 36 gum chewers and found the
mean number of sticks of gum chewed per day was 9. The standard deviation of the sample was 1. Use the level of significance of 0.05
, test the claim that the number of sticks of gum a person chews per day is actually greater than 8? Use any method, however, follow the PHANTOMS acronym to answer the question.

P - Parameter Statement

H - Hypotheses

A - Assumptions & Conditions

N - Name the Test and state the curve you're using

T - Test Statistic - Round your value to TWO decimals and state the command you used to find this value

O - Obtain the P-Value or Critical Value . State the command you are using to find these values

M - Make a Decision about the Null Hypothesis and explain why

S - State Your Conclusion About the Claim

Answer 1

Parameter : The mean no of gums chewed by gum chewers

Hypothesis :

Ho : = 8

Ha:   > 8

Null hypothesis states that the average no of gums chewed by a gum chewer is 8

Alternative hypothesis states that the average no of gums chewed by a gum chewer is greater than 8

Assumptions & test

Assuming that the data is normally distributed. Also as the population standard deviation is not given, we will calculate t stat. Hence we assume that the data has bell shaped curve.

Test statistics

sample mean = 9

sample sd = 1

n = 36

t = 6.00

P value & Critical value

p value = TDIST(6.00, 35, 1) = 0.00

t critical (right tailed test) = 1.68957245

Decision

As the t stat falls in the rejection area, we reject the Null hypothesis.

Also as the p value is less than level of significance (0.05), we reject the Null hypothesis.

Conclusion

Hence we have sufficient evidence to believe that the average no of gums chewed by a gum chewer is greater than 8