Suppose you and a friend are arguing about the average height of students at a p
ID: 3203102 • Letter: S
Question
Suppose you and a friend are arguing about the average height of students at a particular college. Your friend says the average height at this colllege is 5'10", or 70 inches. You think that is not true. Suppose we have a sample of n = 11 with a mean height of 65 inches.
a. Set up a null and alternative hypothesis
b. Assume the population standard deviation is known to be 5 inches. Set up the test statistic (that is, find the T or Z stastistic)
c. Find the probability associated with this statistic, given your hypothesis test and conclude.
d. Construct the 95% confidence interval if the population standard deviation is known to be 5 inches. Interpret meaning.
e. Construct the 95% confidence interval if the sample standard deviation is found to be 5 inches. Interpret meaning.
Explanation / Answer
Given that,
population mean(u)=70
standard deviation, =5
sample mean, x =65
number (n)=11
null, Ho: =70
alternate, H1: !=70
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 65-70/(5/sqrt(11)
zo = -3.31662
| zo | = 3.31662
critical value
the value of |z | at los 5% is 1.96
we got |zo| =3.31662 & | z | = 1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -3.31662 ) = 0.00091
hence value of p0.05 > 0.00091, here we reject Ho
ANSWERS
---------------
null, Ho: =70
alternate, H1: !=70
test statistic: -3.31662
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.00091
the average height of students at a particular college is not at 5'10
d.
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=65
Standard deviation( sd )=5
Sample Size(n)=11
Confidence Interval = [ 65 ± Z a/2 ( 5/ Sqrt ( 11) ) ]
= [ 65 - 1.96 * (1.51) , 65 + 1.96 * (1.51) ]
= [ 62.05,67.95 ]
e.
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)=65
Standard deviation( sd )=5
Sample Size(n)=11
Confidence Interval = [ 65 ± t a/2 ( 5/ Sqrt ( 11) ) ]
= [ 65 - 2.228 * (1.508) , 65 + 2.228 * (1.508) ]
= [ 61.641,68.359 ]