Question

Pharmaceutical companies promote their prescription drugs using television advertising. In a survey of 75 randomly sampled television viewers, 12 indicated that they asked their physician about using a prescription drug they saw advertised on TV.

Develop a 90% confidence interval for the proportion of viewers who discussed a drug seen on TV with their physician. (Round your answers to 3 decimal places.)

Is it reasonable to conclude that 26% of the viewers discuss an advertised drug with their physician?

Answer 1

Given: Sample size = 75

and No. of successes = x = 12

Sample Proportion

100(1-) % CI for proportion is given by the formula,

=

= (0.090,0.230)

90% confidence interval for the proportion of viewers who discussed a drug seen on TV with their physician is (0.090,0.230).

Since, we are confident enough that the true proportion lies within (0.090,0.230), 90% of the time, it may not be reasonable to conclude that 26% of the viewers discuss an advertised drug with their physician.Further to validate,

To test H₀ : p = 0.26 Vs p 0.26

Test Statistic Z =

=

= -1.974

At 10% level of significance since critical Z = 1.645

Here, = 1.974>1.645.

Hence, there is not sufficient evidence to support the null hypothesis, H₀ is rejected.

We may conclude that there is not sufficient evidence to claim that  26% of the viewers discuss an advertised drug with their physician and hence might not be reasonable.