Final Answer:
a. The 95% confidence interval for the population mean μ based on a sample of size n = 11 is (105.64, 120.36).
b. the 95% confidence interval for the population mean μ based on a sample of size n = 20 is (106.91, 119.09).
c. The 99% confidence interval for the population mean μ based on a sample of size n = 11 is (103.76, 122.24).
d. No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed.
Step-by-step explanation:
(a) To construct a 95% confidence interval for the population mean μ based on a sample of size n = 11, we can use the formula:
± z*α/2 * se
where:
- is the sample mean, which is 113
- zα/2 is the two-tailed z-score corresponding to the desired confidence level, which is 1.96 for a 95% confidence level
- se is the standard error of the mean, which is calculated as s/√n, which is 10/√11 ≈ 3.09
Plugging in these values, we get:
113 ± 1.96 * 3.09 = (105.64, 120.36)
Therefore, the 95% confidence interval for the population mean μ based on a sample of size n = 11 is (105.64, 120.36).
(b) To construct a 95% confidence interval for the population mean μ based on a sample of size n = 20, we can use the same formula as in part (a):
± z*α/2 * se
where:
- is the sample mean, which is 113
- z α/2 is the two-tailed z-score corresponding to the desired confidence level, which is 1.96 for a 95% confidence level
- se is the standard error of the mean, which is calculated as s/√n, which is 10/√20 ≈ 2.24
Plugging in these values, we get:
113 ± 1.96 * 2.24 = (106.91, 119.09)
Therefore, the 95% confidence interval for the population mean μ based on a sample of size n = 20 is (106.91, 119.09).
(c) To construct a 99% confidence interval for the population mean μ based on a sample of size n = 11, we can use the same formula as in part (a):
± z*α/2 * se
where:
- is the sample mean, which is 113
- zα/2 is the two-tailed z-score corresponding to the desired confidence level, which is 2.58 for a 99% confidence level
- se is the standard error of the mean, which is calculated as s/√n, which is 10/√11 ≈ 3.09
Plugging in these values, we get:
113 ± 2.58 * 3.09 = (103.76, 122.24)
Therefore, the 99% confidence interval for the population mean μ based on a sample of size n = 11 is (103.76, 122.24).
d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed. The formula for the standard error of the mean, se = s/√n, assumes that the population is normally distributed. If the population is not normally distributed, then the standard error of the mean is not a good estimate of the true standard error, and the confidence intervals will not be accurate.