Question

A recent study found that 51 children who watched a commercial for Walker Crisps featuring a long-standing sports celebrity endorser ate a mean of 36 grams of Walker Crisps as compared toa mean of 25 grams of Walker Crisps for 41 Children who watched a commercial for alternative food snack. Suppose that the sample standard deviation for the children who watched the sports celebrity-endorsed Walker Crisps commercial was 21.4 grams and the sample standard deviation for the children who watched the alternative food snack commercial was 12.8 grams. Assuming the population variances are NOT equal and alpha=.05, is there any evidence that the mean amount of Walker Crisps eaten was significantly higher for the children who watched the sports celebrity endorsed Walker Crisps commercial?

1. What is the claim from the question? What are Null and Alternative Hypothesis for this problem?

2. What kind of test do you want to use? One Sample or Two Sample? Z test or T Test? One-tail or Two-tail test?

3. Calculate Test Statistics

4. Find Critical Value(s) and appropriate degree of freedom if necessary Critical Value(s): Test Statistics:

5. Find P-value

6. What is the conclusion that you could make? Clearly write down the conclusion and business statement and illustrate what type error you could make.

Answer 1

1) Claim : the mean amount of Walker Crisps eaten was higher for the children who watched the sports celebrity endorsed Walker Crisps commercial

H₀:

H₁:

2) We will use independent two-sample t-test. This is a one-tail test.

3) The test statistic t = ()/sqrt(s1^2/n1 + s2^2/n2)

                             = (36 - 25)/sqrt((21.4)^2/51 + (12.8)^2/41)

                             = 3.05

4) DF = (s1^2/n1 + s2^2/n2)^2/((s1^2/n1)^2/(n1 - 1) + (s2^2/n2)^2/(n2 - 1))

          = ((21.4)^2/51 + (12.8)^2/41)^2/(((21.4)^2/51)^2/50 + ((12.8)^2/41)^2/40)

          = 84

At alpha = 0.05, the critical value is t_0.95,84 = 1.663

5) P-value = P(T > 3.05)

                  = 1 - P(T < 3.05)

                  = 1 - 0.9985

                  = 0.0015

6) Since the test statistic value is greater than the critical value (3.05 > 1.663), so we should reject the null hypothesis.

So at 0.05 significance level there is sufficient evidence to support the claim that the mean amount of Walker Crisps eaten was higher for the children who watched the sports celebrity endorsed Walker Crisps commercial .

We could make Type-I error.