If a 0.01 significance level is used does the claim of an advertiser that among the households with tv sets in use, less than 20% were tuned into 60 minutes.

A recently televised broadcast of 60 minutes had a 15 share, meaning that among 5000 monitored households with tv sets in use, 15% of them were tuned into 60 minutes.

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Of the 5000 households with a TV set that were monitored 15% had tuned in to watch the show. The percentage not tuned in was 85%. The confidence interval with a 0.01 significance level is given by `2.575*sqrt((0.15*0.85)/5000) = 0.013` . With a 0.01 significance level the upper level of the percentage of the population watching the show is 0.15 + 0.013 = 16.3%

**This shows that the advertiser's claim that less than 20% of the TV viewing population was tuned in holds.**

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