Question

The FDA is investigating a claim that drink 1 and drink 2 have the same amount of sugar in them. For drink 1, they sampled 54 drinks, and found the sample mean sugar is 37 grams, while the sample standard deviation is 10 grams. For drink 2 they sampled 46 drinks and find the sample mean sugar is 33 grams, with sample standard deviation of 12 grams. We do not assume two drinks have the same standard deviation.

1.) Calculate the test statistic

2.) Find the critical value at .05 level and .1 level based on the t-table

3.) Make a correct conclusion of the hypothesis test under significance level .05, further will your conclusion change if significance level = .1?

Answer 1

To Test :-

H0 :- µ1 = µ2
H1 :- µ1 ≠ µ2

Test Statistic :-
t = (X̅1 - X̅2) / SP √ ( ( 1 / n1) + (1 / n2))

Sp = 10.964

t = 1.82

Part 2)

For α = 0.05

Reject null hypothesis if | t | > t(α/2, n1 + n2 - 2)
Critical value t(α/2, n1 + n1 - 2) = t(0.05 /2, 54 + 46 - 2) = 1.984
| t | > t(α/2, n1 + n2 - 2) = 1.82 < 1.984
Result :- Fail to Reject Null Hypothesis

For α = 0.10

Test Criteria :-
Reject null hypothesis if | t | > t(α/2, n1 + n2 - 2)
Critical value t(α/2, n1 + n1 - 2) = t(0.1 /2, 54 + 46 - 2) = 1.661
| t | > t(α/2, n1 + n2 - 2) = 1.82 > 1.661
Result :- Reject Null Hypothesis

Part 3)

Yes, decision changes as level of significance level change.