The confidence interval or the mean noise level is (108.17,150.23) and sample standard deviation is 12.52.
Given noise levels:108,133,140,129,136. and confidence level 98%.
We have to find the confidence level and the sample standard deviation of the given data.
Confidence level=98%.
n=5
Sample standard deviation=[tex]\sqrt{(x-x bar)^{2} /n-1}[/tex]
=[tex]\sqrt{626.8/4}[/tex]
=[tex]\sqrt{156.7}[/tex]
=12.52
critical t value at 98% confidence level with degree of freedom(5-1) is 3.747
Standard error= critical t*s/[tex]\sqrt{n}[/tex]
Standard error is the difference between the real values and calculated values.
=3.747*12.52/[tex]\sqrt{5}[/tex]
=46.91/2.23
=21.03
Upper confidence interval=Mean +standard error
=129.2+21.03
=150.23
Lower confidence interval=Mean - standard error
=129.2-21.03
=108.17
Hence the confidence interval is (108.17,150.23) ,sample standard deviation is 12.52 and critical t value is 3.747.
Learn more about confidence interval at https://brainly.com/question/15712887
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