Question

Problem 6 (10 marks) Based on the BBM TV ratings, the Sudbury local television claims its 11:00 PM newscast reaches at least 41% of the viewing audience in the area. In a survey of 100 viewers, 36 indicated that they watch the late evening news on this Sudbury local television. At a significance level a = 0.05, can you infer that the Sudbury local television claim is false?

Answer 1

Given : n=100 , X=36

The estimate of the sample proportion is , $\hat{p}=\frac{X}{n}=\frac{36}{100}=0.36$

The null and alternative hypothesis is ,

$H_0:p\geq 0.41 ; H_1:p<0.41$

The test is left-tailed test.

The critical value is ,

$Z_{\alpha}=Z_{0.05}=-1.64$ ; From Z-table

The decision rule is ,

Reject Ho , if Z-stat<-1.64 , otherwise fail to reject Ho.

The test statistic is ,

$Z_{stat}=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} =\frac{0.36-0.41}{\sqrt{\frac{0.41(1-0.41)}{100}}}=-1.02$

Decision : Here , Z-stat=-1.02>-1.64

Therefore , fail to reject Ho.

Conclusion : No. You can not infer that the Sudbury local television claim is false.