Question

A student records the repair cost for 17 randomly selected stereos. A sample mean of $52.33 and standard deviation of $21.44 are subsequently computed. Determine the 90% confidence interval for the mean repair cost for the stereos. Assume the population is approximately normal. Step 2 of 2 : Construct the 90% confidence interval. Round your answer to two decimal places.

Answer 1

Answer:

90% confidence interval is (43.25, 61.41)

Step-by-step explanation:

Given that,

n = 17

Mean, $\bar{x}$ = 52.33

Standard deviation, $\sigma$ = 21.44

∝ = 0.10

Now,

Confidence interval = $\bar{x}$  ± $t_{\frac{\alpha }{2}, n-1} [\frac{\sigma}{\sqrt{n} } ]$

                                 = 52.33 ± 1.7646 [ 21.44 / √17 ]

                                 = 52.33 ± 9.0791

So,

52.33 + 9.0791  = 61.4091

52.33 - 9.0791  = 43.2509

So,

90% confidence interval is (43.25, 61.41)

Also,

Lower end-point = 43.25

Upper end-point = 61.41