Question

The average weight of 25 chocolates bars selected from a normally distributed population is 200 g with a standard deviation of 10 g. Fight the point and the interval estimates using 95% confidence level

Answer 1

A sample mean is a point estimate of a population mean: $\bar{x}=200\ g.$

A confidence interval is the most common type of interval estimate.

a. The critical value for $\alpha = 0.05$ and $df = n-1 = 24$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.063899.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(200-2.063899\times\dfrac{10}{\sqrt{25}},$$

$$200+2.063899\times\dfrac{10}{\sqrt{25}})$$

$$=(195.8722, 204.1278)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $195.8722<\mu<204.1278,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(195.8722, 204.1278).$

b. The test statistic is t-test to calculate the interval.

c. The level of confidence is 0.95.