Final answer:
a) The confidence interval is (108.889, 121.111).
b) The confidence interval is (110.379, 119.621).
c) For a 99% confidence level and a sample size of 13 is (108.214, 121.786).
d) The population had not been normally distributed.
Step-by-step explanation:
To construct a confidence interval for a population mean with a known standard deviation, we use the formula:
(x - EBM, x + EBM), where EBM is the estimated error bound for a population mean and x is the sample mean. EBM is calculated as follows:
EBM = (z * s) / sqrt(n), where z is the z-score corresponding to the desired confidence level, s is the sample standard deviation, and n is the sample size.
(a) For a 98% confidence level and a sample size of 13:
The z-score for a 98% confidence level is approximately 2.326.
Plugging the values into the formula:
EBM = (2.326 * 10) / sqrt(13) = 6.111
The confidence interval is therefore (115 - 6.111, 115 + 6.111)
= (108.889, 121.111)
(b) For a 98% confidence level and a sample size of 26:
The z-score remains the same, but the sample size doubles:
EBM = (2.326 * 10) / sqrt(26) = 4.621
The confidence interval is (115 - 4.621, 115 + 4.621)
= (110.379, 119.621)
(c) For a 99% confidence level and a sample size of 13:
The z-score for a 99% confidence level is approximately 2.576:
EBM = (2.576 * 10) / sqrt(13) = 6.786
The confidence interval is (115 - 6.786, 115 + 6.786)
= (108.214, 121.786).
(d) If the population had not been normally distributed, we could not have computed the confidence intervals in parts (a)-(c) as the formula assumes a normal distribution.
Since the formula assumes a normal distribution, we would not have been able to compute the confidence intervals in sections (a)–(c) if the population had not been normally distributed.