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A researcher wanted to determine the mean number of hours per week (Sunday through Saturday) the typical person watches television. Results from the Sullivan Statistics Survey I indicate that \( s = 7.5 \) hours.

(a) How many people are needed to estimate the number of hours people watch television per week within 2 hours with 95% confidence?

(b) How many people are needed to estimate the number of hours people watch television per week within 1 hour with 95% confidence?

(c) What effects does doubling the required accuracy have on the sample size?

(d) How many people are needed to estimate the number of hours people watch television per week within 2 hours with 90% confidence? Compare this result to part (a). How does decreasing the level of confidence in the estimate affect sample size? Why is this reasonable?

Answer :

Answer:

A.) 54 subjects

B.) 216 subjects

C.) doubling the accuracy results in 4 times the sample size

D.) 38 subjects

Decreasing confidence level decreases sample size. For fixed error margin, the lower the confidence level, the lower the sample size

Explanation:

Standard deviation (s) = 7.5 hours

A.)

Error margin 'E' = 2

Confidence level = 0.95

α = 1 - 0.95 = 0.05, α/2 = 0.025

Z - value at α/2 = 0.025 = 1.96

Sample size = [(1.96 × 7.5)/2]^2

Sample size = 7.35^2 = 54.022

54 subjects

B.) E = 1

Sample size = [(1.96 × 7.5)/1]^2

Sample size = 14.7^2 = 216.09

216 subjects

C.) from the above, doubling the accuracy results in 4 times the sample size.

D.) Using a confidence interval of 90%

Error margin 'E' = 2

Confidence level = 0.90

α = 1 - 0.90 = 0.1, α/2 = 0.05

Z - value at α/2 = 0.05 = 1.645

Sample size = [(1.645 × 7.5)/2]^2

Sample size = 6.16875^2 = 38.05

=38 subjects

Decreasing confidence level decreases sample size. For fixed error margin, the lower the confidence level, the lower the sample size

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Rewritten by : Barada

Final answer:

To estimate the mean number of hours people watch television, a formula involving the Z-score, standard deviation, and margin of error is used to calculate the necessary sample size. With a 95% confidence level, 35 people are needed for a 2-hour margin and 137 people for a 1-hour margin. Reducing accuracy or confidence level decreases sample size requirements.

Explanation:

To estimate the mean number of hours a person watches television using confidence intervals, we can use the formula for sample size:
n = (Z·s/E)²

where n is the sample size, Z is the Z-score corresponding to the confidence level, s is the sample standard deviation, and E is the margin of error.

(a) 95% Confidence, 2 hours margin of error

Z = Z-score for 95% confidence approximately equals 1.96. Using the given standard deviation (s = 7.5),

n = (1.96· 7.5/2)² ≈ 34.22

Therefore, approximately 35 people are needed. We round up since the sample size must be a whole number.

(b) 95% Confidence, 1 hour margin of error

Z stays the same, but the margin of error (E) is now 1 hour:

n = (1.96· 7.5/1)² ≈ 136.89

Thus, about 137 people are needed.

(c) Effect of Doubling Accuracy on Sample Size

The required sample size increases by a factor of 4 when the margin of error is halved, demonstrating that precision is costly in terms of sample size.

(d) 90% Confidence, 2 hours margin of error

Z = Z-score for 90% confidence approximately equals 1.645. With the margin of error back at 2 hours:

n = (1.645· 7.5/2)² ≈ 26.49

Approximately 27 people are needed. The sample size decreases with a lower confidence level because there is less certainty that the interval contains the true mean.