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The average price of a gallon of unleaded regular gasoline was reported to be $2.34 in northern Kentucky (The Cincinnati Enquirer, January 21, 2006). Use this price as the population mean, and assume the population standard deviation is $.20.
1) Calculate the sample size necessary to guarantee at least .95 probability that the sample mean is within $.03 of the population mean
Formula:
n = [(z-value * sd)/E]^2
z-value = 1.96
sd = .20
E = .03
Plug the values into the formula and finish the calculation. Round your answer to the next highest whole number.
170
To calculate the sample size necessary to guarantee at least a 0.95 probability that the sample mean is within $0.03 of the population mean, we can use the formula for sample size estimation for means in a normal distribution:
n = (Z * σ / E)^2
Where:
n = sample size
Z = Z-score corresponding to the desired level of confidence (in this case, a 95% confidence level, which corresponds to a Z-score of 1.96)
σ = population standard deviation
E = margin of error
In this case, the population standard deviation (σ) is given as $0.20 and the margin of error (E) is $0.03.
Substituting these values into the formula:
n = (1.96 * 0.20 / 0.03)^2
Calculating the expression within the parentheses:
n = (3.92 / 0.03)^2
n = 1304.44
Rounding up to the nearest whole number since you can't have a fraction of a sample:
n = 1305
Therefore, a sample size of at least 1305 would guarantee at least a 0.95 probability that the sample mean is within $0.03 of the population mean.