Question

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

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

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.

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.

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) ≈ 0.744 The margin of error is approximately 0.744 years.

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 = x̅ - ME = 16 - 0.744 ≈ 15.256 Upper Bound = x̅ + ME = 16 + 0.744 ≈ 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.