The standard deviation of test scores on a certain achievement test is 11.3. A random sample of 50 scores on this test had a mean of 75.9. Based on this sample, find a 95% confidence interval for the true mean of all scores. Then give its lower limit and upper limit.Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.)Lower limit: Upper limit:
Question
The standard deviation of test scores on a certain achievement test is 11.3. A random sample of 50 scores on this test had a mean of 75.9. Based on this sample, find a 95% confidence interval for the true mean of all scores. Then give its lower limit and upper limit.Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.)Lower limit: Upper limit:
Solution
To calculate the 95% confidence interval for the true mean of all scores, we will use the formula for the confidence interval:
CI = x̄ ± Z * (σ/√n)
where: x̄ = sample mean = 75.9 σ = standard deviation = 11.3 n = sample size = 50 Z = Z-score (for a 95% confidence interval, Z = 1.96)
First, calculate the standard error (SE):
SE = σ/√n = 11.3/√50 = 1.597
Then, calculate the margin of error (ME):
ME = Z * SE = 1.96 * 1.597 = 3.130
Finally, calculate the confidence interval:
Lower limit = x̄ - ME = 75.9 - 3.130 = 72.770 Upper limit = x̄ + ME = 75.9 + 3.130 = 79.030
So, the 95% confidence interval for the true mean of all scores is (72.770, 79.030).
Rounding to one decimal place, the lower limit is 72.8 and the upper limit is 79.0.
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