Answer to Question #234502 in Statistics and Probability for napi

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

(b) Find the standard error of the mean.

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

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

Answer 1

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

(a) An unbiased point estimate for the population mean

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 $z_c = 1.6449$

$$CI = (\bar{X}-z_c\times\frac{\sigma}{\sqrt{n}} , \bar{X}+z_c\times\frac{\sigma}{\sqrt{n}})$$ $$=(10.2-1.6449\times0.25, 10.2+1.6449\times0.25)$$

$$=(9.788775, 10.611225)$$

For 99 % confidence interval $z_c = 2.5758$

$$CI = (\bar{X}-z_c\times\frac{\sigma}{\sqrt{n}} , \bar{X}+z_c\times\frac{\sigma}{\sqrt{n}})$$ $$=(10.2-2.5758\times0.25, 10.2+2.5758\times0.25)$$

$$=(9.55605, 10.84395)$$

(d) Confidence interval tells that we are 99% confident that the true population mean $\mu$  is contained by the interval $(9.55605, 10.84395).$