Q30. Sample mean-11.399 Standard Dev. =13.309 n=991 PE- 11.399 TS-85% CI Alpha= 0.15 Alpha/2=0.075 DF=991-1=990 t-D1.441 SE- 13.309/ sqrt. 991-0.423 ME= 1.441 (0.423) = 0.609543 CI(11.399-1.441 x 13.309/sqrt. 991, 11.399+1.441 x 13.309/sqrt. 991) (10.790, 12.008) 10.790 <µ< 12.008 We are 85% confident that the true mean is contained by the interval.

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Would this be the correct 85% confidence interval?
Q30. Sample mean-11.399 Standard Dev. = 13.309
n-991
PE- 11.399
TS-85% CI
Alpha= 0.15
Alpha/2=0.075
DF= 991-1-990
t-1.441
SE- 13.309/ sqrt. 991-0.423
ME= 1.441 (0.423) = 0.609543
CI(11.399-1.441 x 13.309/sqrt. 991, 11.399+1.441 x 13.309/sqrt. 991)
(10.790, 12.008)
10.790 <µ<12.008
We are 85% confident that the true mean is contained by the interval.
Transcribed Image Text:Q30. Sample mean-11.399 Standard Dev. = 13.309 n-991 PE- 11.399 TS-85% CI Alpha= 0.15 Alpha/2=0.075 DF= 991-1-990 t-1.441 SE- 13.309/ sqrt. 991-0.423 ME= 1.441 (0.423) = 0.609543 CI(11.399-1.441 x 13.309/sqrt. 991, 11.399+1.441 x 13.309/sqrt. 991) (10.790, 12.008) 10.790 <µ<12.008 We are 85% confident that the true mean is contained by the interval.
30.
Construct a 85% confidence interval around the sample mean of the decrease in diastolic
pressure reading of those who took the drug. Show your n, PE, t, SE, ME and interval. Discuss the
interpretation of the Confidence Interval.
Transcribed Image Text:30. Construct a 85% confidence interval around the sample mean of the decrease in diastolic pressure reading of those who took the drug. Show your n, PE, t, SE, ME and interval. Discuss the interpretation of the Confidence Interval.
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