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A simple random sample of 90 is drawn from a normally distributed population, and the mean is found to be 138 with a standard deviation of 34. What is the [tex]$90 \%$[/tex] confidence interval for the population mean? Use the table below to help you answer the question.

\begin{tabular}{|c|c|c|c|}
\hline Confidence Level & [tex]$90 \%$[/tex] & [tex]$95 \%$[/tex] & [tex]$99 \%$[/tex] \\
\hline [tex]$z^\ \textless \ em\ \textgreater \ $[/tex]-score & 1.645 & 1.96 & 2.58 \\
\hline
\end{tabular}

Remember, the margin of error, [tex]$ME$[/tex], can be determined using the formula [tex]$ME = \frac{z^\ \textless \ /em\ \textgreater \ \cdot s}{\sqrt{n}}$[/tex].

A. 128.75 to 147.25
B. 130.98 to 145.02
C. 132.10 to 143.90
D. 137.38 to 138.62



Answer :

GinnyAnswer
To determine the 90% confidence interval for the population mean, we will follow these steps:

1. Identify the given values:
- Sample size ([tex]\( n \)[/tex]) = 90
- Sample mean ([tex]\( \bar{x} \)[/tex]) = 138
- Sample standard deviation ([tex]\( s \)[/tex]) = 34
- Z-score for 90% confidence level ([tex]\( z^* \)[/tex]) = 1.645 (from the given table)

2. Calculate the standard error of the mean (SE):
[tex]\[ SE = \frac{s}{\sqrt{n}} = \frac{34}{\sqrt{90}} \][/tex]
Plugging in the values:
[tex]\[ SE \approx 3.5839 \][/tex]

3. Calculate the margin of error (ME):
[tex]\[ ME = z^* \times SE = 1.645 \times 3.5839 \][/tex]
Plugging in the values:
[tex]\[ ME \approx 5.8955 \][/tex]

4. Calculate the confidence interval:
- Lower bound = Sample mean - Margin of error
[tex]\[ Lower\ Bound = 138 - 5.8955 \approx 132.1045 \][/tex]
- Upper bound = Sample mean + Margin of error
[tex]\[ Upper\ Bound = 138 + 5.8955 \approx 143.8955 \][/tex]

Therefore, the 90% confidence interval for the population mean is approximately [tex]\( (132.10, 143.90) \)[/tex].

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