A film alliance used a random sample of 50 U.S. citizens to estimate that the typical American spent 78 hours watching videos and DVDs last year. The standard deviation of this sample was 9 hours.
(a)
Develop a 95 percent confidence interval for the population mean number of hours spent watching videos and DVDs last year. (Round your answers to 2 decimal places.)
Confidence interval for the population mean is between and .
(b)
How large a sample should be used to be 90 percent confident the sample mean is within 1.0 hour of the population mean?
(a) 95% = mean ± 1.96 SEm
SEm = SD/√n
(b) 90% = mean ± 1.645 SEm
1 = 1.645 SD/√n
(a) To develop a 95 percent confidence interval for the population mean number of hours spent watching videos and DVDs last year, we can use the formula:
Confidence Interval = sample mean ± (critical value * standard deviation / √sample size)
Given:
Sample mean (x̄) = 78 hours
Standard deviation (σ) = 9 hours
Sample size (n) = 50
Confidence level = 95% (which means α = 0.05)
First, we need to find the critical value associated with a 95% confidence level. This critical value can be found using a t-distribution table or a calculator.
For a 95% confidence level and a sample size of 50, the critical value is approximately 2.009 (assuming a two-tailed test).
Now, we can substitute the values into the formula:
Confidence Interval = 78 ± (2.009 * 9 / √50)
Confidence Interval = 78 ± (2.009 * 9 / 7.071) (rounded to 3 decimal places)
Calculating this expression gives us:
Confidence Interval ≈ 78 ± 2.567
Confidence Interval ≈ (75.433, 80.567)
Therefore, the 95 percent confidence interval for the population mean number of hours spent watching videos and DVDs last year is between 75.433 and 80.567 hours.
(b) To determine the sample size required to be 90 percent confident that the sample mean is within 1.0 hour of the population mean, we can use the formula:
Sample size (n) = (z * standard deviation / margin of error)²
Given:
Confidence level = 90% (which means α = 0.10)
Margin of error = 1.0 hour
Standard deviation (σ) = 9 hours
First, we need to find the critical value associated with a 90% confidence level. This critical value can be found using a z-score table or a calculator.
For a 90% confidence level, the critical value (z) is approximately 1.645 (assuming a two-tailed test).
Now, we can substitute the values into the formula:
Sample size (n) = (1.645 * 9 / 1.0)²
Sample size (n) = (14.805 / 1.0)²
Calculating this expression gives us:
Sample size (n) ≈ 217.818
Therefore, the sample size required to be 90% confident that the sample mean is within 1.0 hour of the population mean is approximately 218.