We assume that blood pressure, X, has N (mu, sigma^2) distribution. To estimate

Question

We assume that blood pressure, X, has N (mu, sigma^2) distribution. To estimate mu and sigma, we randomly sampled 9 people and measured their blood pressure. The sample mean and sample standard deviation are X = 110 and S = 6. Knowing that based on the t(8) distribution, answer the following questions. (a) Write down the point estimate for mu along with its standard error. (b) Write down the margin of error at 0.8 confidence level. (c) Write down the 80% confidence interval estimation for mu based on this sample. (d) Assume that we want to test the null hypothesis H_0: mu = 120 using a one sample t test, what is the t score?

Explanation / Answer

a.
Standard Error= sd/ Sqrt(n)
Where,
sd = Standard Deviation
n = Sample Size

Standard deviation( sd )=6
Sample Size(n)=9
Standard Error = ( 6/ Sqrt ( 9) )
= 2
b.
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)=110
Standard deviation( sd )=6
Sample Size(n)=9
Margin of Error = t a/2 * 6/ Sqrt ( 9)
= 1.397 * (2)
= 2.794
c.
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)=110
Standard deviation( sd )=6
Sample Size(n)=9
Confidence Interval = [ 110 ± t a/2 ( 6/ Sqrt ( 9) ) ]
= [ 110 - 1.397 * (2) , 110 + 1.397 * (2) ]
= [ 107.206,112.794 ]

d.
Given that,
population mean(u)=120
sample mean, x =110
standard deviation, s =6
number (n)=9
null, Ho: µ=120
alternate, H1: µ!=120
level of significance, a = 0.2
from standard normal table, two tailed t a/2 =1.397
since our test is two-tailed
reject Ho, if to < -1.397 OR if to > 1.397
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =110-120/(6/sqrt(9))
to =-5
| to | =5
critical value
the value of |t a| with n-1 = 8 d.f is 1.397
we got |to| =5 & | t a | =1.397
make decision
hence value of | to | > | t a| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -5 ) = 0.0011
hence value of p0.2 > 0.0011,here we reject Ho

ANSWERS
---------------
null, Ho: µ=120
alternate, H1: µ!=120
test statistic: -5
critical value: -1.397 , 1.397
decision: reject Ho
p-value: 0.0011

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