Noise levels at 4volcanoes were measured in decibels yielding the following data:152,157,157,158Construct the 90%confidence interval for the mean noise level at such locations. Assume the population is approximately normal. Step 3 of 4:Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

t critical = 2.353

CI = (152.82 , 159.18)

Step-by-step explanation:

We are required to find the t critical value and also confidence interval

The sample mean is = 152+157+157+158/4

= 156

The standard deviation is 2.7

At a confidence level of 90%, we have α = 1 - 0.90 = 0.1

α/2 = 0.1/2 = 0.05.

Degree of freedom = 4-1 = 3

When we look this up in the t critical value table,

t critical = 2.353

Next we have to find the confidence interval

CI = (156 - 2.3530 x 2.7/√4 , 156 + 2.353 x 2.7/√4)

= 156-3.17655, 156+3.17655

CI = (152.82 , 159.18)

Upper bound = 159.18

Lower bound = 152.82

126 maybe. sorry