Question

you are told that in a sample of size 225 the mean is 48.5 and the standard deviation is 1.8. the study is reported with 90% confidence level. explain how to determine if 48.8 is within the confidence interval.

Explanation:

Calculate the standard error

Using the Standard Error of the Mean knowledge point

$$ SE = \frac{s}{\sqrt{n}} = \frac{1.8}{\sqrt{225}} = \frac{1.8}{15} = 0.12 $$

Calculate the margin of error

Using the Margin of Error Calculation knowledge point

$$ ME = z^* \times SE = 1.645 \times 0.12 = 0.1974 $$

Determine the confidence interval

Using the Confidence Interval Interpretation knowledge point

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Evaluate the value 48.8

Using the Confidence Interval Interpretation knowledge point

$$ 48.8 > 48.6974 $$

Answer:

To determine if \(48.8\) is within the confidence interval, follow these steps:

1. Calculate the Standard Error (\(SE\)):

$$ SE = \frac{s}{\sqrt{n}} = \frac{1.8}{\sqrt{225}} = 0.12 $$

1. Calculate the Margin of Error (\(ME\)) for a \(90\%\) confidence level (\(z^* = 1.645\)):

$$ ME = z^* \times SE = 1.645 \times 0.12 = 0.1974 $$

1. Find the Confidence Interval Limits:

• Lower limit: \(\bar{x} - ME = 48.5 - 0.1974 = 48.3026\)
• Upper limit: \(\bar{x} + ME = 48.5 + 0.1974 = 48.6974\)
• The interval is \([48.3026, 48.6974]\).

1. Compare \(48.8\) to the Interval:

Since \(48.8\) is greater than the upper limit of \(48.6974\), it is not within the confidence interval.