A pizza shop owner wishes to find the 95% confidence interval of the true mean cost of a large cheese pizza. How large should the sample be if she wishes to be accurate to within $0.15 ? A previous study showed that the standard deviation of the price was $0.26. 3. A sample of 8 adult elephants had an average weight of 12,300 pounds. The standard deviation for the sample was 22 pounds. Find the margin of error, E (maximum error of the estimate), for the 95% confidence interval of the population mean for weights of adult elephants.
To find the sample size needed for the 95% confidence interval with an accuracy of $0.15, we will use the formula for sample size calculation:
n = [(z * σ) / E]^2
Where:
n = Sample size
z = Z-score for the desired confidence level (in this case, 95% corresponds to a Z-score of 1.96)
σ = Standard deviation
E = Maximum error or margin of error
Substituting the given values:
n = [(1.96 * 0.26) / 0.15]^2
n ≈ 10.81
Since we cannot have a fractional sample size, the pizza shop owner should round up to the nearest whole number. Therefore, the sample size needed should be at least 11.
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To find the margin of error (E) for the 95% confidence interval of the population mean for weights of adult elephants, we will use the formula:
E = (z * σ) / √n
Where:
E = Margin of error
z = Z-score for the desired confidence level (in this case, 95% corresponds to a Z-score of 1.96)
σ = Standard deviation
n = Sample size
Substituting the given values:
E = (1.96 * 22) / √8
E ≈ 12.71
Therefore, the margin of error for the 95% confidence interval of the population mean for weights of adult elephants is approximately 12.71 pounds.
To find the sample size needed to estimate the true mean cost of a large cheese pizza, we can use the formula:
n = (Z * σ / E)^2
where:
n = sample size
Z = Z-score for the desired confidence level (95% confidence corresponds to Z = 1.96)
σ = standard deviation of the population (previous study showed it to be $0.26)
E = maximum allowable error ($0.15)
Plugging in the values, we get:
n = (1.96 * 0.26 / 0.15)^2
n = 3.41^2
n ≈ 11.59
Therefore, the sample size should be at least 12 (rounded up to the nearest whole number) in order to estimate the true mean cost of a large cheese pizza with a 95% confidence level and an accuracy of within $0.15.
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To find the margin of error (E) for the 95% confidence interval of the population mean for weights of adult elephants, we can use the formula:
E = Z * (σ / sqrt(n))
where:
E = margin of error (maximum error of the estimate)
Z = Z-score for the desired confidence level (95% confidence corresponds to Z = 1.96)
σ = standard deviation of the sample (given as 22 pounds)
n = sample size (given as 8 elephants)
Plugging in the values, we get:
E = 1.96 * (22 / sqrt(8))
E = 1.96 * (22 / 2.83)
E ≈ 15.15
Therefore, the margin of error for the 95% confidence interval of the population mean for weights of adult elephants is approximately 15.15 pounds.