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7. Jose conducts a random sample of 41 people regarding their television viewing habits. He found...
A market researcher for a consumer electronics company wants to study the television viewing habits of...
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 40 respondents is selected, and each respondent is instructed to keep a detailed record of all television viewing in a particular week. The results are as follows: • Viewing time per week: Xbar = 15.3 hours, S = 3.8 hours. • 27 respondents watch the evening news on at least three weeknights. Construct a 95%...
2. In estimating the mean number of television viewing hours per family per week, a random...
2. In estimating the mean number of television viewing hours per family per week, a random sample of 400 families yields a mean of 32.6 hours and a standard deviation of 9.9 hours (a) Find a 95% confidence interval for the average number of viewing hours per family per week in the population. Interpret. (b) Find a 99% confidence interval for the average number of viewing hours per family per week in the population. Interpret. (c) Suppose instead only 25...
1. A market researcher for a consumer electronics company wants to study the TV viewing habits...
1. A market researcher for a consumer electronics company wants to study the TV viewing habits of residents of a particular area. A random sample of 60 respondents is selected & each respondent is instructed to keep a detailed record of all TV viewing in a particular week. The results of the sample indicate the average viewing time is 37 hours & the standard deviation is 4.8 hours. 36 have HD TVs on at least 1 television set. A. Construct...
10. A simple random sample of size n is drawn. The sample mean x is found...
10. A simple random sample of size n is drawn. The sample mean x is found to be 39.1, and the sample standard deviation s is found to be 9.7. a) (2 points) Construct a 90% confidence interval for the population mean u if the sample size n is 41. b) (2 points) Construct a 90% confidence interval for the population mean y if the sample size n is 101. c) (2 points) Construct a 99% confidence interval for the...
10. A simple random sample of size n is drawn. The sample mean x is found...
10. A simple random sample of size n is drawn. The sample mean x is found to be 39.1, and the sample standard deviation s is found to be 9.7. a) (2 points) Construct a 90% confidence interval for the population mean w if the sample size n is 41. b) (2 points) Construct a 90% confidence interval for the population mean 4 if the sample size n is 101. c) (2 points) Construct a 99% confidence interval for the...
In a random sample of 200 people, 154 said that they watched educational television. Find the...
In a random sample of 200 people, 154 said that they watched educational television. Find the 90% confidence interval of the population proportion of people who watched educational television. Which formula will you use for this problem? O3 -1.14( ) <-<a+bala ) On= (*12:0) On=p:4 () Oộ - - - - - - Oz-za-(M) <r<+zar (M)
Suppose a survey research firm selected a simple random sample of 300 people for the purposes...
Suppose a survey research firm selected a simple random sample of 300 people for the purposes of estimating their mean weekly income and found that the sample mean was $1,345.78 with a standard deviation of $257.90 Calculate the 90% confidence interval for the population mean.
A questionnaire about study habits was given to a random sample of high school students. The...
A questionnaire about study habits was given to a random sample of high school students. The sample of 20 students reported that they spent an average of 215 minutes per week studying with a standard deviation of 70 minutes. What is your 95% confidence interval on the population variance?
A researcher surveys middle-school students on their study habits. She finds that in a random sample...
A researcher surveys middle-school students on their study habits. She finds that in a random sample of 28 middle-school students, the mean amount of time that they spend working on the computer each night is 2.4 hours with a standard deviation of 0.92 hours. She uses the sample statistics to compute a 95% confidence interval for the population mean - the the mean amount of time that all middle-school students spend working on the computer each night. What is the...
In a hospital, a random sample of 10 weeks was selected, and it was found that...
In a hospital, a random sample of 10 weeks was selected, and it was found that an average of 495 patients were treated in the emergency room each week. The population standard deviation was 23. Find the 90% confidence interval of the true mean. Assume the variable is normally distributed. (Show Work)
2. In estimating the mean number of television viewing hours per family per week, a random sample of 400 families yields a mean of 32.6 hours and a standard deviation of 9.9 hours (a) Find a 95% confidence interval for the average number of viewing hours per family per week in the population. Interpret. (b) Find a 99% confidence interval for the average number of viewing hours per family per week in the population. Interpret. (c) Suppose instead only 25...
10. A simple random sample of size n is drawn. The sample mean x is found to be 39.1, and the sample standard deviation s is found to be 9.7. a) (2 points) Construct a 90% confidence interval for the population mean u if the sample size n is 41. b) (2 points) Construct a 90% confidence interval for the population mean y if the sample size n is 101. c) (2 points) Construct a 99% confidence interval for the...
10. A simple random sample of size n is drawn. The sample mean x is found to be 39.1, and the sample standard deviation s is found to be 9.7. a) (2 points) Construct a 90% confidence interval for the population mean w if the sample size n is 41. b) (2 points) Construct a 90% confidence interval for the population mean 4 if the sample size n is 101. c) (2 points) Construct a 99% confidence interval for the...
In a random sample of 200 people, 154 said that they watched educational television. Find the 90% confidence interval of the population proportion of people who watched educational television. Which formula will you use for this problem? O3 -1.14( ) <-<a+bala ) On= (*12:0) On=p:4 () Oộ - - - - - - Oz-za-(M) <r<+zar (M)
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