Question

ch 21 a dairy scientist compared milk production of cows fed two diets. he randomly divided a set of 10 cows into two groups. one group was fed diet 1 for a month, and the other was fed diet 2 for a month. assume the milk - productions (in lbs/week) are normally distributed. here are the sample results:
a 95% confidence interval for the difference in the mean production of milk is between - 23.395 and 48.395.
true
false

Explanation:

Step1: Calculate the difference in sample means

The difference in sample means \(\bar{x}_1-\bar{x}_2=385.7 - 398.2=- 12.5\)

Step2: Calculate the standard error

The formula for the standard error \(SE=\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}\)
Substitute \(s_1 = 25.7\), \(n_1=10\), \(s_2 = 43.1\), \(n_2 = 10\)
\(SE=\sqrt{\frac{25.7^{2}}{10}+\frac{43.1^{2}}{10}}=\sqrt{\frac{660.49}{10}+\frac{1857.61}{10}}=\sqrt{66.049 + 185.761}=\sqrt{251.81}\approx15.87\)

Step3: Determine the critical value

For a 95% confidence interval and \(n_1=n_2 = 10\), the degrees of freedom \(df=\min(n_1 - 1,n_2 - 1)=9\)
From the t - distribution table, \(t_{0.025,9}=2.262\)

Step4: Calculate the margin of error

The margin of error \(ME=t_{\alpha/2}\times SE\)
\(ME = 2.262\times15.87\approx35.9\)

Step5: Calculate the confidence interval

The confidence interval is \((\bar{x}_1-\bar{x}_2)-ME<\mu_1-\mu_2<(\bar{x}_1-\bar{x}_2)+ME\)
\(-12.5-35.9=-48.4\) and \(-12.5 + 35.9 = 23.4\)

Answer:

False