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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.

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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
See Also This Related BrainMass Solution

Cumulative weight loss: propranolol treatment

The following table presents the mean cumulative weight loss (in grams) for 30 patients receiving propranolol (treatment) and for 60 control patients following sweating during insulin-induced hypoglycemia.

Mean loss weight compared between treatment and control for 90 patients during insulin-induced hypoglycemia
Intervention N Mean Weight Loss
(in grams) Standard Deviation
Propranolol 30 173 27.0
Control 60 153 26.0

(Note: you can assume equal variance for the two groups.)

a) Provide a 95% confidence interval for the mean weight loss for the propranolol group.
b) Provide a 95% confidence interval for the mean weight loss for the control group.
c) Provide the interpretation of the 95% confidence interval for the mean weight loss for the propranolol group.
d) Using a two-sided test, conduct a test of hypothesis and provide the details of this test. Do we have sufficient evidence to conclude that the mean cumulative weight loss is different for the two groups? (Use a type I error level of = 0.05)
e) Write your conclusion from (d) in words, as you would present them in a report of the findings to a group of biomedical researchers (who might or might not be as well-versed in biostatistics as you are :-).
f) Compute the mean weight loss difference between the propranolol and control group.
g) Provide a 95% confidence interval for the difference in mean weight loss between the propranolol and control group.
h) Provide the interpretation of the 95% confidence interval for the difference in mean weight loss between the propranolol and control group.
Perform the same hypothesis test as in (d) and (e), but change the type I error level to 0.01. How does your main result / conclusion change? (Do not forget your one-two sentence interpretation.)

a) What is the possible alternative?

b) Why do you think we suggested using a two-sided test instead of that alternative?

c) Conduct the test with the possible alternative.

d) Please interpret your findings.

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