Answer to Question #333676 in Statistics and Probability for Catherine

Question

A random sample of 10 chocolate energy bars of a certain brand has, on average, 230 calories per bar, with a standard deviation of 15 calories. Construct a 99% confidence interval for the true mean calorie content of this brand of energy bar. Assume that the distribution of the calorie content is approximately normal.

Answer 1

The critical value for $\alpha = 0.01$ and $df = n-1 = 9$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 3.249823.$

The corresponding confidence interval is computed as shown below:

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

$$=(230-3.249823\times\dfrac{15}{\sqrt{10}},$$

$$230+3.249823\times\dfrac{15}{\sqrt{10}})$$

$$=(214.5847,245.4153)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $214.5847<\mu<245.4153,$ which indicates that we are 99% confident that the true population mean $μ$  is contained by the interval $(214.5847,245.4153).$