Homework SourseA market researcher for a consumer electronics company wants
| 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: | |||||||||||
Solution
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
