A market researcher for a consumer electronics company wants to study the televi

Question

A market researcher for a consumer electronics company wants to study the television viewing habits of residents of a particular area. A random sample of 65 respondents is selected and each respondent is instructed to keep a detailed record of all television viewing in a particular week. The results are below: http://www.tutor-homework.com/statistics_tables/statistics_tables.html The average viewing time per week was 23.4 hours. The standard deviation was 3.5 hours. 18 respondents watch the evening news on at least 5 weeknights. 1 construct a 99% confidence interval for the mean amount of TV watched per week in this area and explain what it means Upper _________ Lower _________ Conclusion: 2 construct a 95% confidence interval for the population proportion who watch the evening news on at least five weeknights and explain what it means Upper _________ Lower _________ Conclusion: A market researcher for a consumer electronics company wants to study the television viewing habits of residents of a particular area. A random sample of 65 respondents is selected and each respondent is instructed to keep a detailed record of all television viewing in a particular week. The results are below: http://www.tutor-homework.com/statistics_tables/statistics_tables.html The average viewing time per week was 23.4 hours. The standard deviation was 3.5 hours. 18 respondents watch the evening news on at least 5 weeknights. 1 construct a 99% confidence interval for the mean amount of TV watched per week in this area and explain what it means Upper _________ Lower _________ Conclusion: 2 construct a 95% confidence interval for the population proportion who watch the evening news on at least five weeknights and explain what it means Upper _________ Lower _________ Conclusion:

Explanation / Answer

1)

standard error

= 3.5/sqrt(65)

confidence interval = 23.4 +/- 2.575 * 3.5 / sqrt(65)

= 23.4 +/- 1.118

upperlimit = 24.518 hrs

lower limit = 22.282

hence 99% percent of people watch between 22.282 and 24.518 hours

2)

standard error = sqrt( 5/18 * 13/18 * 1/18) = 0.1056

confidence interval = 5/18 +/- 2.2575 * 0.1056

upper limit = 0.5162

lower limit = 0.0394

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