Question

known to be normally distributed with a standard deviation of $2.50. The last seven times John has taken a taxi from Logan to downtown Boston, the fares have been $22.10, $23.25, $21.35, $24.50, $21.90, $20.75, and $22.65. Construct a 95% confidence interval for the population mean.

Answer 1

The 95% confidence interval for the population mean is between $20.51 and $24.21.

Explanation:

The first step to find this question is find the sample mean

`\mu_(x) = (22.10 + 23.25 + 21.35 + 24.50 + 21.90 + 20.75 + 22.65)/(7) = 22.36`

Confidence interval

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 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 = 1.96*(2.5)/(√(7)) = 1.85`

The lower end of the interval is the sample mean subtracted by M. So it is 22.36 - 1.85 = $20.51

The upper end of the interval is the sample mean added to M. So it is 22.36 + 1.85 = $24.21

The 95% confidence interval for the population mean is between $20.51 and $24.21.