Question

The lengths of text messages are normally distributed with a population standard deviation of 6 characters and an unknown population mean. If a random sample of 21 text messages is taken and results in a sample mean of 30 characters, find a 80% confidence interval for the population mean. Round your answers to two decimal places. z0.10 z0.05 z0.04 z0.025 z0.01 z0.005 1.282 1.645 1.751 1.960 2.326 2.576

Answer 1

The 80% confidence interval for the population mean is between 28.32 characters and 31.68 characters.

Explanation:

We have the standard deviation for the population, so we can use the normal distribution. If we had the standard deviation for the sample, we would have to use the t-distribution.

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

`\alpha = (1-0.8)/(2) = 0.1`

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
`1-0.5 = 0.9`, so
`z = 1.282`

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 = 1.282*(6)/(√(21)) = 1.68`

The lower end of the interval is the sample mean subtracted by M. So it is 30 - 1.68 = 28.32 characters.

The upper end of the interval is the sample mean added to M. So it is 30 + 1.68 = 31.68 characters.

The 80% confidence interval for the population mean is between 28.32 characters and 31.68 characters.