Question

An article in Obesity Research “Impaired pressure natriuresis in obese youths,” (2003, Vol. 11, pp. 745–751) described a study in which all meals were provided for 14 lean boys for three days followed by one stress test (with a video-game task). The average systolic blood pressure (SBP) during the test was 118.3 mm HG with a standard deviation of 9.9 mm HG. Construct a 97.5% one-sided upper confidence interval for mean SBP. Assume population is approximately normally distributed. Round your answer to 3 decimal places.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 118.3

sample standard deviation = s = 9.9

sample size = n = 14

Degrees of freedom = df = n - 1 = 14 - 1 = 13

At 97.5% confidence level

= 1 - 975%

=1 - 0.975

= 0.025

t,df = t_0.025,13 = 2.160

Margin of error = E = t_/2,df * (s /n)

= 2.160 * ( 9.9 / 14)

Margin of error = E = 5.715

The 97.5% one-sided upper confidence interval estimate of the population mean is,

< + E

   < 118.3 + 5.715

   < 124.015

upper confidence interval = 124.015