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:

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Omm2 Omm2 · Answer 1 · 2021-05-15T09:46:09+02:00

Answer:

Hence, the margin error is [tex]1415.62189[/tex] and the confidence interval for the mean is [tex](41234.38,44065.62)[/tex].

Step-by-step explanation:

Given :

Sample size [tex]n=135[/tex]

Sample mean [tex]\bar{x}=\$42,650[/tex]

Standard deviation [tex]\sigma =\$8,988[/tex]

Confidence level is [tex]93\%[/tex]

(a)

The margin error [tex]=Z_{\frac{\alpha}{2}}\times \frac{\sigma}{\sqrt{n}}[/tex]

[tex]\because \alpha =1-[/tex]confidence level

[tex]\Rightarrow \alpha=1-93\%=1-0.93[/tex]

[tex]\Rightarrow \alpha=0.07\Rightarrow \frac{\alpha}{2}=0.035[/tex]

[tex]Z_{\frac{\alpha}{2}}=Z_{0.035}=1.83[/tex] from the [tex]z-[/tex]table

Now, the margin error[tex]=1.83\times \frac{8988}{\sqrt{135}}[/tex]

[tex]=1415.62189[/tex]

(b)

The confidence interval for the mean [tex]\mu\\[/tex] :

[tex](\bar{x}-\text{margin error},\bar{x}+\text{margin error})[/tex]

[tex]\Rightarrow (\bar{x}-1415.62189,\bar{x}+1415.62189)[/tex]

[tex]\Rightarrow (42650-1415.62189,42650+1415.62189)\\\Rightarrow (41234.38,44065.62)[/tex]