A recent broadcast of a television show had a 10 share meaning that among 5000 m

Question

A recent broadcast of a television show had a 10 share meaning that among 5000 monitored households with TV sets in use. 10 percentage of them were tuned to this program Use a 0 01 significance level to test the claim of an advertiser that among the households with TV sets in use less than 15% were tuned into the program Identify the null hypothesis. alternative hypothesis test statistic P-value. conclusion about the null hypothesis, and final conclusion that addresses the original claim Use the P-value method Use the normal distribution as an approximation of the binomial attribution Identify the null and alternative hypotheses Choose the correct answer below.

Explanation / Answer

Solution

Let X = number of TV sets in use which were tuned to the programme. Then, X ~ B(n, p), where n = sample size (5000 in the given question) and p = Probability that a TV set in use is tuned to the programme = proportion of TV sets in use which were tuned to the programme.

Claim:

Less than 15% (i.e., 0.15) of TV sets in use were tuned to the programme

Hypotheses:

Null: H0: p = p0 = 0.15   Vs Alternative: HA: p < 0.15. Option B ANSWER

Test Statistic:

Z = {(x/n) – p0}/[{ p0 (1 - p0)/n}] = {0.1 – 0.15}/[{(0.15 x 0.85)/5000}]

= - 0.05/0.0000255 = - 0.05/0.00504975 = - 9.90 ANSWER

p-value:

Under H0, test statistic is approximately distributed as N(0, 1).

So, p-value = P(Z < - 9.90 ) = 0 (approximated to 4 places of decimal)

‘<’ is used above since the alternative is left-one-sided.

Conclusion:

Since 0 (p-value) < 0.01 (given level of significance), H0 is rejected which implies that there is sufficient evidence to support the claim that less than 15% (i.e., 0.15) of TV sets in use were tuned to the programme

DONE

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