1 The systolic blood pressures of college students follow a

1. The systolic blood pressures of college students follow a normal distribution. 20 college students are selected randomly and their systolic blood pressures are measured during their final exam week. Their readings are as follows: 139 127 141 124 129 134 144 127 133 79 110 124 126 140 136 141 133 129 123 98

A) Find the point estimate for µ, the mean systolic blood pressure of college students? ____________________________________

B) Construct a 95% confidence interval for the population mean systolic blood pressure. Show your work or calculator commands. ____________________________________

C) Interpret this interval in the context of the problem. Use complete sentences. _________________________________________________________________________________________ _________________________________________________________________________________________ _________________________________________________________________________________________

D) What is the margin of error? ____________________________________ E) A college professor claims that the mean systolic blood pressure of college students can be as high as 133.5 during the final exam week. Does this contradict your findings above? Justify your answer. __________________________________________________________________________________________ __________________________________________________________________________________________

Solution

a)
Point of estimate of mean (x) =126.85
b)
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=126.85
Standard deviation( sd )=15.682
Sample Size(n)=20
Confidence Interval = [ 126.85 ± t a/2 ( 15.682/ Sqrt ( 20) ) ]
= [ 126.85 - 2.093 * (3.507) , 126.85 + 2.093 * (3.507) ]
= [ 119.511,134.189 ]

c)
Interpretations:
1) We are 95% sure that the interval [ 119.511,134.189 ] contains the
true population mean(u)
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contain true
population mean(u)

d)
Margin of Error = t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
Mean(x)=126.85
Standard deviation( sd )=15.682
Sample Size(n)=20
Margin of Error = t a/2 * 15.682/ Sqrt ( 20)
= 2.093 * (3.507)
= 7.339