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Unformatted text preview: Solutions for Lecture Problems from Chapter 8 8.6 ALTERNATIVE LANGUAGE TO TEXTBOOK. A major media research organization has studied the weekly television (TV) viewing time during nightly prime time hours (8:00 to 11:00 PM). It is known from previous research that the standard deviation of TV viewing time of all households is 3.5 hours. On the basis of a random sample of 300 households, the mean TV viewing time is 8.5 hours per week. REMARKS: The objective after reading through the parts below is to confidently estimate the true mean TV viewing time per household per week at the 95% confidence level. Note the keywords known and all (referring to the population of households.) As a result, sigma is known: Problem classification: interval estimation is known: use z Given: xbar = 8.5 hours/week = 3.5 hours/week n = 300 a. What is the point estimate for mean weekly TV viewing time of all households? ~ xbar = 8.5 hours/week The sample mean of 8.5 hours/week is the point estimate for . b. What is the margin of error at 95% confidence? At 95% confidence, z = 1.960 Then E = z / n Chapter 8 Lecture Problems 1 2009 Harvey A. Singer = 1.9603.5/300 So at 95% confidence E = 0.396 hours/week c. Calculate a 95 percent confidence interval for the actual mean weekly TV viewing time of all households. xbar E xbar + E with xbar given as 8.5 and E = 0.396 from b, so 8.5 0.396 8.5 + 0.396 so 8.104 8.896 hours/week In a sketch d. What is the chance that the true mean annual household income is contained within the interval calculated in part c? 95% e. What is the chance that the true mean annual household income is not contained within the interval calculated in part c? 5% f. Based on this confidence interval, state a likely (and reasonable) upper bound for the mean annual household income....
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This note was uploaded on 01/26/2011 for the course OM 210 taught by Professor Singer during the Fall '08 term at George Mason.
- Fall '08