Question

The manager wanted to know the overall quality of a large batch of products. The quality management team selected a sample of 100 items randomly from the whole batch and measured the value of a key index for each item. The mean value of this key index for this sample is 99. The variance of the whole batch is estimated to be 16. (a) (5 points) Construct a 95% confidence level two-tail confidence interval for the mean value of the key index of this batch. (b) (5 points) If we want the sampling error to be no greater than 0.5, what is the minimum sample size to achieve this based on the same confidence level with part (a)

Answer 1

a) The 95% confidence level two-tail confidence interval for the mean value of the key index of this batch is between 98.22 and 99.78

b) The minimum sample size to achieve this is 246.

Explanation:

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

`\alpha = (1-0.95)/(2) = 0.025`

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
`1-0.025 = 0.975`, so
`z = 1.96`

Now, find the margin of error M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population(square root of the variance) and n is the size of the sample. So in this question,
`\sigma = √(16) = 4`

`M = 1.96(4)/(√(100)) = 0.78`

The lower end of the interval is the sample mean subtracted by M. So it is 99 - 0.78 = 98.22

The upper end of the interval is the sample mean added to M. So it is 99 + 0.78 = 99.78.

a) The 95% confidence level two-tail confidence interval for the mean value of the key index of this batch is between 98.22 and 99.78

(b) (5 points) If we want the sampling error to be no greater than 0.5, what is the minimum sample size to achieve this based on the same confidence level with part (a)

We need a sample size of n

n is found when
`M = 0.5`

Then

`M = z*(\sigma)/(√(n))`

`0.5 = 1.96*(4)/(√(n))`

`0.5√(n) = 4*1.96`

`√(n) = (4*1.96)/(0.5)`

`(√(n))^(2) = ((4*1.96)/(0.5))^(2)`

`n = 245.86`

Rounding up

The minimum sample size to achieve this is 246.