Question

A pet food producer fills 25-pound bags of dog food on two different production lines located in separate cities. In an effort to determine whether differences exist between the average fill rates for the two lines, a random sample of 18 bags from line 1 and a random sample of 23 bags from line 2 were recently selected. Each bag's weight was measured and the accompanying table reports the summary measures from the samples. Assume the two lines are normally distributed with equal variances. Complete parts a through c below. Click the icon to view the table of summary statistics from the samples. a. Calculate the point estimate for the difference between the population means of the two lines. Point estimate=0

B. Develop a 90%,95%,98%,99​% confidence interval estimate of the true mean difference between the two lines.

Answer 1

solution:-

a. point estimate = (x1-x2) = (25.08-25.26) = -0.18

B. degrees of freedom df = (n1+n2)-2 = (18+23)-2 = 39

90% confidence

the value of 90% confidence with df from t table is t = 1.685

confidence interval formula

=> point estimate +/- t * sqrt(s1^2/n1 + s2^2/n2)

=> -0.18 +/- 1.685 * sqrt((0.05^2/18) + (0.08^2/23))

=> (-0.2144 , -0.1456)

95% confidence

the value of 95% confidence with df from t table is t = 2.022

confidence interval formula

=> point estimate +/- t * sqrt(s1^2/n1 + s2^2/n2)

=> -0.18 +/- 2.022 * sqrt((0.05^2/18) + (0.08^2/23))

=>(-0.2213 , -0.1387)

98% confidence

the value of 98% confidence with df from t table is t = 2.426

confidence interval formula

=> point estimate +/- t * sqrt(s1^2/n1 + s2^2/n2)

=> -0.18 +/- 2.426 * sqrt((0.05^2/18) + (0.08^2/23))

=> (-0.2295 , -0.1305)

99% confidence

the value of 99% confidence with df from t table is t = 2.708

confidence interval formula

=> point estimate +/- t * sqrt(s1^2/n1 + s2^2/n2)

=> -0.18 +/- 2.708 * sqrt((0.05^2/18) + (0.08^2/23))

=> (-0.2353, -0.1247) (rounded to four decimal places)