Questions: Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. Complete parts (a) through (c) TInterval below. Click the icon to view a t distribution table. (13.046,22 15) x=17.598 S x=16.01712719 n=50 a. What is the number of degrees of freedom that should be used for finding the critical value talpha / 2 ? df= (Type a whole number.)

Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. Complete parts (a) through (c) TInterval below. Click the icon to view a t distribution table. (13.046,22 15) x=17.598 S x=16.01712719 n=50 a. What is the number of degrees of freedom that should be used for finding the critical value talpha / 2 ? df= (Type a whole number.)
Transcript text: Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. Complete parts (a) through (c) TInterval below. Click the icon to view a t distribution table. (13.046,22 15) \[ \begin{array}{l} x=17.598 \\ S x=16.01712719 \\ n=50 \end{array} \] a. What is the number of degrees of freedom that should be used for finding the critical value $\mathrm{t}_{\alpha / 2}$ ? \[ \mathrm{df}= \] (Type a whole number.)
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Solution

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Solution Steps

Step 1: Calculate Degrees of Freedom

To find the degrees of freedom (\(df\)) for the t-distribution, we use the formula: \[ df = n - 1 \] where \(n\) is the sample size. Given \(n = 50\): \[ df = 50 - 1 = 49 \]

Step 2: Calculate the Confidence Interval

We calculate the confidence interval for the mean of a single population with unknown variance and a large sample size at a 95% confidence level using the formula: \[ \bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} \] where:

  • \(\bar{x} = 17.598\) (sample mean)
  • \(s = 16.01712719\) (sample standard deviation)
  • \(n = 50\) (sample size)
  • \(t_{\alpha/2} \approx 1.96\) for a 95% confidence level (from the t-distribution table for \(df = 49\))

Substituting the values into the formula: \[ 17.598 \pm 1.96 \cdot \frac{16.01712719}{\sqrt{50}} \]

Calculating the margin of error: \[ \text{Margin of Error} = 1.96 \cdot \frac{16.01712719}{\sqrt{50}} \approx 1.96 \cdot 2.264 = 4.44 \]

Thus, the confidence interval is: \[ (17.598 - 4.44, 17.598 + 4.44) = (13.158, 22.038) \]

Final Answer

The degrees of freedom is \( \boxed{49} \) and the confidence interval is \( \boxed{(13.16, 22.04)} \).

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