Question

The caffeine content (in mg) was examined for a random sample of 50 cups of black coffee dispensed by a new machine. The mean and the standard deviation were 110 mg and 7.1 mg respectively. Use the data to construct a 98% confidence interval for the mean caffeine content for cups dispensed by the machine. Interpret the interval!

Answer 1

We are 98% confident interval for the mean caffeine content for cups dispensed by the machine between 107.66 and 112.34 mg .

Explanation:

Given -

The sample size is large then we can use central limit theorem

n = 50 ,

Standard deviation
`(\sigma)` = 7.1

Mean
`\overline{(y)}` = 110

`\alpha =` 1 - confidence interval = 1 - .98 = .02

`z_{(\alpha)/(2)}` = 2.33

98% confidence interval for the mean caffeine content for cups dispensed by the machine =
`\overline{(y)}\pm z_{(\alpha)/(2)}\frac{\sigma}√(n)`

=
`110\pm z_(.01)\frac{7.1}√(50)`

=
`110\pm 2.33\frac{7.1}√(50)`

First we take + sign

`110 + 2.33\frac{7.1}√(50)` = 112.34

now we take - sign

`110 - 2.33\frac{7.1}√(50)` = 107.66

We are 98% confident interval for the mean caffeine content for cups dispensed by the machine between 107.66 and 112.34 .