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.

1. A point estimate is a single value estimate of a parameter. For instance, a sample mean is a point estimate of a population mean.

$\bar{x}=200\ g$

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

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 $(195.8722, 204.1278).$