Question

The USGA specifications for golf balls is that a ball must weigh no more than 1.62 oz. The manufacturing process for one company’s golf balls produces a normally distributed distribution of ball weights. To determine if a lot of golf balls produces acceptable balls, a random sample of 25 balls is selected from each lot and tested. One particular lot has a mean weight of 1.625 oz with a standard deviation of .011 oz. Create a 99% confidence interval for the actual weight of a golf ball. Using this interval, is the company making acceptable balls? Explain your answer.

Answer 1

The 99% confidence interval for the actual weight of a golf ball is between 1.619 oz and 1.631 oz.

The almost entirity of values in this interval are values higher than the specification of 1.62 oz, which means that the company is not making acceptable balls.

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.99)/(2) = 0.005`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.005 = 0.995`, so
`z = 2.575`

Now, find the margin of error M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 2.575*(0.011)/(√(25)) = 0.006`

The lower end of the interval is the sample mean subtracted by M. So it is 1.625 - 0.006 = 1.619 oz

The upper end of the interval is the sample mean added to M. So it is 1.625 + 0.006 = 1.631 oz

The 99% confidence interval for the actual weight of a golf ball is between 1.619 oz and 1.631 oz.

The almost entirity of values in this interval are values higher than the specification of 1.62 oz, which means that the company is not making acceptable balls.