(1)(a) We are given a sample of size 5 with a sample mean of `bar(x)=10.0023 ` and a population standard deviation of `sigma=.0002 ` . We want a 98% confidence interval of the population mean.

The interval is given by the sample mean plus or minus the product of the confidence and the standard error.

Since we know the population standard deviation we use a z-interval. With `alpha=.02 ` we find `z_(alpha/2)=2.33 ` . (With the confidence at 98%, 1% of the sample means will lie below and 1% above the interval. Find .9900 in the standard normal table (or the opposite of the z associated with .0100) to get 2.33)

The standard error is the quotient of the standard deviation and the square root of the sample size: `sigma/sqrt(n)=.0002/sqrt(5)~~.00009 `

The interval is given by:

`bar(x)-z_(alpha/2)sigma/sqrt(n)<mu<bar(x)+z_(alpha/2)sigma/sqrt(n) `

`10.0023-(2.33)*.0002/sqrt(5)<mu<10.0023+2.33*.0002/sqrt(5) `

`10.00209<mu<10.00251 `

(b) The error is given by `E=((z_(alpha/2)sigma)/sqrt(n)) ` . Solving for n we get:

`n=((z_(alpha/2)sigma)/E)^2 ` Here `z_(alpha/2)=2.576,sigma=.0002,E=.0001 `

Note that `z_(alpha/2) ` changed from 2.33 to 2.576 as the confidence level changed from 98% 99%.

`n=(2.576*.0002/.0001)^2~~26.543 `

So we need at least 27 measurements.

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