Answer to Question #227843 in Statistics and Probability for Carol

Question

A random sample of 100 bags of potatoes found that the sample mean X for the weight of the bags

was 10.2 kilograms. Assume that the population variance of the weights is known to be 2.5 kg.

(a) Compute an unbiased point estimate for the population mean : (3)

(b) Find the standard error of the mean. (3)

(c) Find the 90% and 99% confidence intervals for the population mean, : (10)

(d) Explain what the 99% confidence interval tells us about the population mean, : (4)

Answer 1

$n=100 \\ Mean = 10.2 \;kg \\ SD \; \sigma = 2.5 \;kg$

(a) Population mean = Sample mean = 10.2 kg

(b) Standard error

$SE= \frac{\sigma}{\sqrt{n}} \\ = \frac{2.5}{\sqrt{100}} = 0.25$

(c) For 90 % confidence interval Z = 1.645

$CI = Mean ±Z \times SE \\ CI_{90 \;\%} = 10.2 ± 1.645 \times 0.25 \\ = (9.8, 10.6)$

For 99 % confidence interval Z = 2.578

$CI_{99 \;\%} = 10.2 ± 2.578 \times 0.25 \\ = (9.6, 10.8)$

(d) Confidence interval tells that about 99 % of values fall between 9.6 and 10.8 only.