Question

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6. Pharmaceutical companies promote their prescription drugs using television advertising. In a survey of 90 randomly sampled television viewers, 18 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. (Hint: Review section Confidence Interval for a Population Proportion)

Answer 1

Solution :

Given that,

n = 90

x = 18

= x / n = 18 / 90 = 0.200

1 - = 1 - 0.200 = 0.800

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 * (((0.200 * 0.800) / 90)

= 0.069

A 90% confidence interval for population proportion p is ,

- E < P < + E

0.200 - 0.069< p < 0.200 + 0.069

0.131 < p < 0.269

(0.131 , 0.269)