The confidence interval for the true mean coefficient of restitution is (29.9911,30.0089).
Given sample mean coefficient of restitution 0.835,population standard deviation of 0.019 and the confidence level of 0.835.
We have to find the confidence interval for the true mean coefficient of restitution.
We can easily find the confidence interval through the formula of margin of error.
Margin of error is the difference between the calculated values and the real values.
Margin of error=z*σ/[tex]\sqrt{n}[/tex]
where z is the z value of p value,
σ is population mean or sample mean,
n is the sample size.
In our problem the sample size is 30.
z value for the p value of 0.99 is 2.576.
Margin of error=2.576*0.019/[tex]\sqrt{30}[/tex]
=0.048944/5.4772
=0.0089
Upper level=mean +margin of error
=30+0.0089
=30.0089
Lower level=Mean-margin of error
=30-0.0089
=29.9911.
Hence the confidence interval for the true mean coefficient of restitution is (29.9911,30.0089).
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