Question

Suppose a researcher is interested in understanding the variation in the price of store brand milk. A random sample of 36 grocery stores selected from a population and the mean price of store brand milk is calculated. The sample mean is $3.13 with a standard deviation of $0.23. Construct a 95% confidence interval to estimate the population mean.

Answer 1

Answer: ($3.055, $3.205)

Step-by-step explanation:

Given : Significance level : $\alpha: 1-0.95=0.5$

Critical value : $z_{\alpha/2}=1.96$

Sample size : n= 36

Sample mean : $\overline{x}=\ \ 3.13$

Standard deviation : $\sigma= \ \ 0.23$

The confidence interval for population mean is given by :_

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

$\text{i.e. }\ \ 3.13\pm (1.96)\dfrac{0.23}{\sqrt{36}}\\\\\approx\ \ 3.13\pm0.075\\\\=(\ \ 3.13-0.075,\ \ 3.13+0.075)=(\ \ 3.055,\ \ 3.205)$

Hence, the 95% confidence interval to estimate the population mean = ($3.055, $3.205)