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A simple random sample is drawn from a normally distributed population, and when making a statistical inference about the population mean, the margin of error is found to be 5.9 at a [tex]95 \%[/tex] level of confidence. If the mean of the sample is 18.7, what is the [tex]95 \%[/tex] confidence interval for the population mean?

A. [tex]18.7 \pm 5.9[/tex]

B. [tex]18.7 \pm 9.7[/tex]

C. [tex]18.7 \pm 11.6[/tex]

D. [tex]18.7 \pm 15.2[/tex]

Sagot :

GinnyAnsweridn
In order to find the 95% confidence interval for the population mean, we need to use the sample mean and the margin of error. Here's the step-by-step process:

1. Identify the sample mean ( [tex]\(\bar{x}\)[/tex] ) and the margin of error (ME):
- Sample mean [tex]\(\bar{x} = 18.7\)[/tex]
- Margin of error [tex]\(ME = 5.9\)[/tex]

2. Calculate the lower and upper bounds of the confidence interval:
- The lower bound of the confidence interval is given by [tex]\(\bar{x} - ME\)[/tex].
- The upper bound of the confidence interval is given by [tex]\(\bar{x} + ME\)[/tex].

3. Substitute the values into the formulas:
- Lower bound = [tex]\(18.7 - 5.9\)[/tex]
- Upper bound = [tex]\(18.7 + 5.9\)[/tex]

4. Perform the calculations:
- Lower bound = [tex]\(18.7 - 5.9 = 12.8\)[/tex]
- Upper bound = [tex]\(18.7 + 5.9 = 24.6\)[/tex]

Therefore, the 95% confidence interval for the population mean is (12.8, 24.6).

Given the options, the correct representation of this confidence interval is:
[tex]\[ 18.7 \pm 5.9 \][/tex]

So, the correct answer is:
[tex]\[ 18.7 \pm 5.9 \][/tex]
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