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Chapter 9: Problem 592

In the Idaho State Home for Runaway Girls, 25 residents were polled as to what age they ran away from home. The sample mean was 16 years old with a standard deviation of \(1.8\) years. Establish a \(95 \%\) confidence interval for \(\mu\), the mean age at which runaway girls leave home in Idaho.

Short Answer

Expert verified
The 95% confidence interval for the mean age at which runaway girls leave home in Idaho is approximately \( (15.256, 16.744) \) years.

Step by step solution

01

Calculate the degrees of freedom

Since we are using the t-distribution, we need to calculate the degrees of freedom. The degrees of freedom (df) can be calculated with the formula: df = n - 1 In this case, n = 25, so: df = 25 - 1 = 24 The degrees of freedom is 24.
02

Find the t-value

Now we need to find the t-value for a 95% confidence level and 24 degrees of freedom. We can do this using a t-distribution table, or online calculator. We are looking for a t-score that corresponds to an area of 0.975 (since 0.95/2 = 0.475, and we want the one-tailed value). For df = 24, the t-value approximately equals to 2.064.
03

Calculate the margin of error

To calculate the margin of error, we will use the formula: Margin of Error (ME) = t * (s / sqrt(n)) In this case, t = 2.064, s = 1.8, and n = 25, so: ME = 2.064 * (1.8 / sqrt(25)) = 2.064 * (1.8 / 5) 鈮� 0.744 The margin of error is approximately 0.744 years.
04

Establish the 95% confidence interval

Now that we have the margin of error, we can create the 95% confidence interval by adding and subtracting the margin of error from our sample mean: Lower Bound = x虆 - ME = 16 - 0.744 鈮� 15.256 Upper Bound = x虆 + ME = 16 + 0.744 鈮� 16.744 Therefore, the 95% confidence interval for the mean age at which runaway girls leave home in Idaho is approximately (15.256, 16.744) years.

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