Question

1. for a sample of 25 marathon runners the daily intake of calories was evaluated, yielding the mean as 90 units and the variance as 49 units. find a 95% c. i for the average intake of calories that marathon runners consume. what is the critical value? assume normal distribution for the intake of calories that all marathon runners consume. 2.064, (87.11, 92.89)

Explanation:

Step1: Identify the formula for confidence interval

For a small sample size (\(n < 30\)) and unknown population standard deviation, we use the \(t -\)distribution. The formula for the confidence interval is \(\bar{x}\pm t_{\alpha/2}\frac{s}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(s\) is the sample standard deviation, \(n\) is the sample size, and \(t_{\alpha/2}\) is the critical value.
Given \(n = 25\), \(\bar{x}=90\), \(s=\sqrt{49} = 7\). The confidence level is \(95\%\), so \(\alpha=1 - 0.95=0.05\) and \(\alpha/2=0.025\). The degrees of freedom \(df=n - 1=25 - 1 = 24\).

Step2: Find the critical value \(t_{\alpha/2}\)

Looking up the \(t -\)distribution table with \(df = 24\) and \(\alpha/2=0.025\), we get \(t_{0.025,24}=2.064\).

Step3: Calculate the margin of error \(E\)

\(E=t_{\alpha/2}\frac{s}{\sqrt{n}}=2.064\times\frac{7}{\sqrt{25}}=2.064\times\frac{7}{5}=2.064\times1.4 = 2.8896\)

Step4: Calculate the confidence interval

The lower limit \(LL=\bar{x}-E=90 - 2.8896=87.1104\approx87.11\)
The upper limit \(UL=\bar{x}+E=90 + 2.8896=92.8896\approx92.89\)

Answer:

The critical value is \(2.064\) and the \(95\%\) confidence interval is \((87.11,92.89)\)