Question

For each exercise, answer the following along with any additional questions. Assume group variances are equal (unless the problem is ran via statistical software). |  Provide the null and alternative hypotheses in formal and plain language for appropriate two-tailed test (viz., dependent or independent) at the 0.05 significance level  Do the math and reject/accept null at a=.05. State your critical t value.  Explain the results in plain language.  Calculate the 95% confidence interval for the difference in means and state both formally and in plain language if appropriate.|

1. The State of Florida severely cut funding for the TRUTH campaign (advertising which is aimed at teenagers to reduce smoking). Advocates claim that TRUTH reduces teen smoking. To demonstrate this, they provide two separate samples of the state’s high school students reporting their number of cigarettes smoked per day two years before (nine students) and two years after the start of TRUTH (eight students). (C11PROB1.SAV) Before TRUTH: 5, 5, 8, 0, 0, 10, 0, 4, 10 After TRUTH: 6, 0, 6, 7, 0, 0, 2, 5

Answer 1

The analysis is done in Minitab.

The output is:

Two-sample T for Before vs After

N Mean StDev SE Mean
Before 9 4.67 4.09 1.4
After 8 3.25 3.06 1.1

Difference = μ (Before) - μ (After)
Estimate for difference: 1.42
95% lower bound for difference: -1.69
T-Test of difference = 0 (vs >): T-Value = 0.80 P-Value = 0.218 DF = 15
Both use Pooled StDev = 3.6469

Hypothesis:

Let μ (Before) and μ (After) denote the mean no. of cigarettes smoked before and after the truth.

The test statistic = 0.80 (from the output)

Critical value =

Since test statistic < critical value, we fail to reject H₀.

Conclusions: The data does not provide enough evidence to say that TRUTH reduces teen smoking.

95% C.I. for difference of means is:

s = pooled standard deviation = 3.6469

=(-2.38, 5.18)

Hence it is 95% sure that the true mean difference of the cigarettes smoked will lie within -2.38 and 5.18