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

Answer:

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

Step-by-step 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

$z_{\frac{\alpha}{2}}$ = 2.33

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

                     = $110\pm z_{.01}\frac{7.1}\sqrt{50}$

                      = $110\pm 2.33\frac{7.1}\sqrt{50}$

       First we take  + sign

   $110 + 2.33\frac{7.1}\sqrt{50}$ = 112.34

now  we take  - sign

$110 - 2.33\frac{7.1}\sqrt{50}$ = 107.66

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