Question

A study was conducted to evaluate the effectiveness of a weight loss program. Among 36 obese individuals aged 55 to 75 years randomly selected into the study, each individual had his/her BMI computed before and after the program. The decrease in the BMI was recorded into a SAS dataset and proc univariate was used to analyze this dataset. The population m ean of the decrease in the BMI is denoted by μ. Use the SAS output on the following page to answer the questions below. a. What is an estimate of μ? What is its standard error? b. Construct a 95% two-sided confidence interval for A. C. Construct a 99% two-sided confidence interval for μ. d. Consider the hypothesis testing problem H0 : μ 0 against Ha : μ 0 with a significance level of 0.05. What is the decision of whether to reject Ho? Explain. e. Consider the hypothesis testing problem Ho HS1 against Ha > 1 with a significance level of α 0.05. What is the decision of whether to reject Ho? Explain. Moments 24Sum Weights 1.8 Sum Observations Mean Std Deviation Skewness Uncorrected SS Coeff Variation107.28822 Std Error Mean 24 43.2 3.72948696 0.77688058 85.7782 0.39420209 1.93118797 Variance 0.9234592 Kurtosis 163.5382 Corrected SS Basic Statistical Measures Location Variability Mean 1.800000 Std Deviation Median 2.220000 Variance Mode 1.93119 3.72949 7.62000 Interquartile Range 2.52500 Range Basic Confidence Limits Assuming Normality Parameter Mean Estimate 1.80000 Std Deviation 193119 3.72949 95% Confidence Limits 2.61547 1.50095 2.70899 7.33865 0.98453 Variance 2.25284 Tests for Location: Mu0-0 Test Student's t t4.566186Pr>t 0.0001 Sign Signed Rank S Statistic p Value 8PrIMI 0.0015 118 Pr>= ISI 0.0001 Quantiles (Definition 5) Level 100% Max 99% 95% 90% 75% Q3 50% Median 25% Q1 10% 5% 1% 0% Min Quantile 4.690 4.690 4.520 3.750 3.100 2.220 0.575 0.170 2.480 2.930 2.930

Answer 1

a)
sample mean is an estimate of mu = 1.8
standard error = sd/sqrt(n)
= 1.9319/sqrt(24) = 0.394347

b)

95% confidence interval
t = =T.INV.2T(0.05,23) = 2.06865761

(1.8 - 2.06865761 * 0.394347 , 1.8 + 2.06865761 * 0.394347)
= ( 0.98453,2.61547)   {you can see this directly in third table 95% confidence limits of mean

c)

for 99 % CI
t = =T.INV.2T(0.01,23) = 2.807336
(1.8 - 2.807336 * 0.394347 , 1.8 + 2.807336 * 0.394347)
= ( 0.69293,2.9070645)

d)
since 0 is not present in 95 % CI
we reject the null hypothesis

e)
TS = (xbar - mu)/standard error
= (1.8 - 1)/0.394347
= 2.028670

critical value at 0.05 level =
=t.inv(0.95,23)    {right-tailed test}
= 1.713872

TS > critical value

we reject the null hypothesis

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