Question

General Logs Banana Bombs cereal is sold in 10.40 ounce packages. Because the cereal is sold by weight, the number of pieces of Banana Bombs varies from box to box. A random sample of 19 boxes of Banana Bombs was sampled and the number of pieces in each box was counted. The sample mean for 19 boxes was 696.58. The quality control engineer knows from the experience that population standard deviation is assumed to be 12.33 and that the number of pieces of Banana Bombs arenormally distributed.a.Construct a 99% confidence interval for the true mean number of Banana Bombs in 10.40 ounce packages.

Answer 1

The 99% confidence interval for the true mean number of Banana Bombs in 10.40 ounce packages is between 689.3 and 703.86.

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`.

That 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(12.33)/(√(19)) = 7.28`

The lower end of the interval is the sample mean subtracted by M. So it is 696.58 - 7.28 = 689.3 pieces

The upper end of the interval is the sample mean added to M. So it is 696.58 + 7.28 = 703.86 pieces

The 99% confidence interval for the true mean number of Banana Bombs in 10.40 ounce packages is between 689.3 and 703.86.