1. A sample of 49 adults was selected from students at KUMC. The average height
ID: 3174267 • Letter: 1
Question
1. A sample of 49 adults was selected from students at KUMC. The average height was 173.45cm and standard deviation was 7.47. 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 students at KUMC are significantly different 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
a.
Given that,
population mean(u)=175.1
sample mean, x =173.45
standard deviation, s =7.47
number (n)=49
null, Ho: =175.1
alternate, H1: !=175.1
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.011
since our test is two-tailed
reject Ho, if to < -2.011 OR if to > 2.011
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =173.45-175.1/(7.47/sqrt(49))
to =-1.546
| to | =1.546
critical value
the value of |t | with n-1 = 48 d.f is 2.011
we got |to| =1.546 & | t | =2.011
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -1.5462 ) = 0.1286
hence value of p0.05 < 0.1286,here we do not reject Ho
ANSWERS
---------------
null, Ho: =175.1
alternate, H1: !=175.1
test statistic: -1.546
critical value: -2.011 , 2.011
decision: do not 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)=173.45
Standard deviation( sd )=7.47
Sample Size(n)=49
Confidence Interval = [ 173.45 ± t a/2 ( 7.47/ Sqrt ( 49) ) ]
= [ 173.45 - 2.011 * (1.067) , 173.45 + 2.011 * (1.067) ]
= [ 171.304,175.596 ]
c.
p-value: 0.1286
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