Question

A random sample of 20 chocolate energy bars of a certain brand has, on average, 210 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. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. Click here to view page 1 of the table of critical values of the t-distribution Click here to view page 2 of the table of critical values of the t-distribution The confidence interval is << (Round to one decimal place as needed.)

Answer 1

Solution :

Given that,

t$\alpha$ /2,df = 2.861

Margin of error = E = t$\alpha$/2,df * (s /$\sqrt$n)

= 2.861 * ( 15 / $\sqrt$ 20)

Margin of error = E = 9.6

The 99% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

210 - 9.6 < $\mu$ < 210 + 9.6

200.4 < $\mu$ < 219.6

The confidence interval is 200.4 < $\mu$ < 219.6