Question

Begin by re-writing the problem. Minitab18 is required; attach or include your output. Type (use Word) your assignment; you may handwrite equations.

1) In preliminary tests of a vaccine that may help smokers quit by reducing the “rush” from tobacco, 180 subjects who wanted to quit smoking were either given a placebo or the vaccine. Of the 90 in the placebo group, only 10 quit smoking compared with 25 of the vaccine group. (15 points)

a) At the .10 level of significance, determine if the vaccine is effective in reducing smoking. Follow and show the 7 steps for hypothesis testing. (8 points)

b) Is the criterion for normality met? (4 points)

c) Verify with Minitab, by attaching or including the output. (3 points)

Answer 1

Solution:

Part 1

Here, we have to use z test for population proportions.

H₀: p₁ = p₂ versus H_a: p₁ > p₂

This is an upper tailed test.

We are given α = 0.10

Test statistic formula is given as below:

Z = (P1 – P2) / sqrt(P*(1 – P)*((1/N1) + (1/N2)))

Where,

X1 = 25

X2 = 10

N1 = 90

N2 = 90

P = (X1+X2)/(N1+N2) = (25 + 10)/(90 + 90) = 0.1944

P1 = X1/N1 = 25/90 = 0.277777778

P2 = X2/N2 = 10/90 = 0.111111111

Z = (P1 – P2) / sqrt(P*(1 – P)*((1/N1) + (1/N2)))

Z = (0.277777778 – 0.111111111) / sqrt(0.1944*(1 – 0.1944)*((1/90) + (1/90)))

Z = 0.166666667 / sqrt(0.1944*(1 – 0.1944)*((1/90) + (1/90)))

Z = 2.8249

P-value = 0.0024

P-value < α = 0.10

So, we reject the null hypothesis

There is sufficient evidence to conclude that the vaccine is effective in reducing smoking.

Part b

The criterion for normality met because the sample sizes for both samples are adequate and these are greater than 30.

Part c

The Minitab output is given as below:

Test and CI for Two Proportions

Sample      X      N Sample p

1          25     90 0.277778

2          10     90 0.111111

Estimate for p(1) - p(2): 0.166667

95% lower bound for p(1) - p(2): 0.0717987

Test for p(1) - p(2) = 0 (vs > 0): Z = 2.89 P-Value = 0.002