3. A sample of 17 BIOS704 students was selected from students at KUMC. The avera

Question

3. A sample of 17 BIOS704 students was selected from students at KUMC. The average height was 169.42 and standard deviation was 9.09. The national average height for adults in the US is normally distributed with a mean of 175.10cm.

A. Conduct a hypothesis test to see if BIOS704 students at KUMC are significantly shorter than the national average. (Use test statistic method) Make sure you clearly state all 5 parts to each hypothesis test.

B. Create a 95% confidence interval for the mean.

C. What is the p-value of your test?

Explanation / Answer

Given that,
population mean(u)=175.1
sample mean, x =169.42
standard deviation, s =9.09
number (n)=17
null, Ho: >175.1
alternate, H1: <175.1
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.746
since our test is left-tailed
reject Ho, if to < -1.746
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =169.42-175.1/(9.09/sqrt(17))
to =-2.576
| to | =2.576
critical value
the value of |t | with n-1 = 16 d.f is 1.746
we got |to| =2.576 & | t | =1.746
make decision
hence value of | to | > | t | and here we reject Ho
p-value :left tail - Ha : ( p < -2.5764 ) = 0.01015
hence value of p0.05 > 0.01015,here we reject Ho
ANSWERS
---------------
null, Ho: =175.1
alternate, significantly shorter than the national average H1: <175.1
test statistic: -2.576
critical value: -1.746
decision: reject Ho
B.
Confidence Interval
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)=169.42
Standard deviation( sd )=9.09
Sample Size(n)=17
Confidence Interval = [ 169.42 ± t a/2 ( 9.09/ Sqrt ( 17) ) ]
= [ 169.42 - 2.12 * (2.205) , 169.42 + 2.12 * (2.205) ]
= [ 164.746,174.094 ]
C.
p-value: 0.01015

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