Question

In a test of the quality of two television commercials, each commercial was shown in a separate test area six times over a one-week period. The following week a telephone survey was conducted to identify individuals who had seen the commercials. Those individuals were asked to state the primary message in the commercials. The following results were recorded.

Commercial A Commercial B

Number who saw the commercial: 155 Number who saw the commercial: 204

Number who recalled the message: 64 Number who recalled the message: 63

Use a=.05 and test the hypothesis that there is no difference in the recall proportions for the two commercials.

Formulate the null and the alternative hypotheses.

What is the value of the test statistic?

What is the p-value( round to 4 decimals)

Compute a 95% confidence interval for the difference between the recall proportions for the two populations (to 4 decimals).

( , )

Answer 1

For commercial A:

n1 = 155, x1 = 64

p̂1 = x1/n1 = 0.4129

For commercial B:

n2 = 204, x2 = 63

p̂2 = x2/n2 = 0.3088

α = 0.05

Null and Alternative hypothesis:

Ho : p1 = p2

H1 : p1 ≠ p2

Pooled proportion:

p̄ = (x1+x2)/(n1+n2) = (64+63)/(155+204) = 0.3538

Test statistic:

z = (p̂1 - p̂2)/√ [p̄*(1-p̄)*(1/n1+1/n2)] = (0.4129 - 0.3088)/√[0.3538*0.6462*(1/155+1/204)] = 2.0429

p-value = 2*(1-NORM.S.DIST(ABS(2.0429), 1)) = 0.0411

Decision:

p-value < α, Reject the null hypothesis

95% Confidence interval for the difference between the recall proportions for the two population:

At α = 0.05, two tailed critical value, z_c = NORM.S.INV(0.05/2) = 1.960

Lower Bound = (p̂1 - p̂2) - z_c*√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ]

= (0.4129 - 0.3088) - 1.96*√[(0.4129*0.5871/155) + (0.3088*0.6912/204)] = 0.0039

Upper Bound = (p̂1 - p̂2) + z_c*√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ]

= (0.4129 - 0.3088) + 1.96*√[(0.4129*0.5871/155) + (0.3088*0.6912/204)] = 0.2042

0.0039 < p1 -p2 < 0.2042