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# Estimating the mean difference in weight gain

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Based on a survey of 1,000 adults by Greenfield Online and reported in a May 2009 USA Today Snapshot, adults 24 years of age and under spend a weekly average of \$35 on fast food. If 200 of the adults surveyed were in the age category of 24 and under and they provided a standard deviation of \$14.50, construct a 95% confidence interval for the weekly average expenditure on fast food for adults 24 years of age and under. Assume fast food weekly expenditures are normally distributed.

Question 2
An experiment was designed to estimate the mean difference in weight gain for pigs fed ration A as compared with those fed ration B. Eight pairs of pigs were used. The pigs within each pair were litter mates. The rations were assigned at random to the two animals within each pair. The gains (in pounds) after 45 days are shown below:
RationA RationB
65 58
37 39
40 31
47 45
49 47
65 55
53 59
59 51
Assuming weight gain is normal, find the 95% confidence interval estimate for the mean of the differences μd where d= ration A - ration B.

https://brainmass.com/statistics/descriptive-statistics/estimating-the-mean-difference-in-weight-gain-558623

#### Solution Preview

1.
mean, X = \$35
sample size, n =200
Standard deviation, s = \$14.50
confidence interval = 95%

degree of freedom, df = n -1 = 199
t(199, 0.05) = 1.972

mu = X +/- s/sqrt(n) = 35 +/- 1.972*14.5/sqrt(200) = 32.978, 37.022
Hence, weekly average expenditure = (\$32.978, \$37.022)

2.
Mean:
X = sum(x)/n
Here, x: data value, X = mean of sample
X1 = ...

#### Solution Summary

Two problems: 1st is estimating 95% confidence interval of expenditure on fast food and second is difference in weight of pigs fed different rations.

\$2.19