A golf club manufacturer measures the impact elasticity of a new material. This is done by randomly selecting a driver head made from the new material, firing a golf ball at it from an air cannon (under carefully controlled conditions), and making certain measurements of the ball's motion. These measurements are used to calculate a number called the coefficient of restitution of the driver head. Engineers use this procedure 30 times. The sample mean coefficient of restitution is 0.835. Calculate a 99% confidence interval for the true mean coefficient of restitution, assuming the population standard deviation of such coefficients is 0.019. Round your answers to the fourth decimal digit.

Respuesta :

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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