Question

Starting salaries of 135 college graduates who have taken a statistics course have a mean of $42,650. Suppose the distribution of this population is approximately normal and has a standard deviation of $8,988. Using a 93% confidence level, find both of the following:

(NOTE: Do not use commas nor dollar signs in your answers.)
(a) The margin of error:
(b) The confidence interval for the mean μ:

Answer 1

Hence, the margin error is
`1415.62189` and the confidence interval for the mean is
`(41234.38,44065.62)`.

Explanation:

Given :

Sample size
`n=135`

Sample mean
`\bar{x}=\$42,650`

Standard deviation
`\sigma =\$8,988`

Confidence level is
`93\%`

(a)

The margin error
`=Z_{(\alpha)/(2)}* (\sigma)/(√(n))`

`\because \alpha =1-`confidence level

`\Rightarrow \alpha=1-93\%=1-0.93`

`\Rightarrow \alpha=0.07\Rightarrow (\alpha)/(2)=0.035`

`Z_{(\alpha)/(2)}=Z_(0.035)=1.83` from the
`z-`table

Now, the margin error
`=1.83* (8988)/(√(135))`

`=1415.62189`

(b)

The confidence interval for the mean
`\mu\\` :

`(\bar{x}-\text{margin error},\bar{x}+\text{margin error})`

`\Rightarrow (\bar{x}-1415.62189,\bar{x}+1415.62189)`

`\Rightarrow (42650-1415.62189,42650+1415.62189)\\\Rightarrow (41234.38,44065.62)`