Question

The television show September Road has been successful for many years. That show recently had a share of 15, meaning that among the TV sets in use, 15% were tuned to September Road. Assume that an advertiser wants to verify that 15% share value by conducting its own survey, and a pilot survey begins with 15 households have TV sets in use at the time of a September Road broadcast. Find the probability that none of the households are tuned to September Road. P(none) = Find the probability that at least one household is tuned to September Road. P(at least one) = Find the probability that at most one household is tuned to September Road. P(at most one) = If at most one household is tuned to September Road, does it appear that the 15% share value is wrong? (Hint: Is the occurrence of at most one household tuned to September Road unusual?)

Answer 1

Let X be the number of households tuned to September Road out of 15 households. Then X ~ Binomial(n = 15, p = 0.15)

P(none) = P(X = 0) =  ¹⁵C₀ * 0.15⁰ * (1 - 0.15)^15-0 = 0.85¹⁵ = 0.08735422

P(at least one) = P(X 1) = 1 - P(X = 0) = 1 - P(none) = 1 - 0.08735422 = 0.9126458

P(at most one) = P(X 1) = P(X = 0) + P(X = 1) = ¹⁵C₀ * 0.15⁰ * (1 - 0.15)^15-0 + ¹⁵C₁ * 0.15¹ * (1 - 0.15)^15-1

= 0.85¹⁵ + 15 * 0.15 * 0.85¹⁴

= 0.318586

Null hypothesis H0: P = 0.15

Alternative hypothesis H0: P > 0.15

Standard error of proportion = = 0.09219544

Test statistic, z = (0.318586 - 0.15) / 0.09219544 = 1.83

P-value = P(z > 1.83) =  0.0336

Since p-value is less than the significance level of 0.05, we reject null hypothesis and it appear that the 15% share value is wrong.