Instructions: Please write neatly. Be sure to use correct notation! 1. You are p
ID: 3340883 • Letter: I
Question
Instructions: Please write neatly. Be sure to use correct notation! 1. You are planning on applying to the Nezog Institute for a well-regarded program in eastern healing methods. The average salary for a new graduate trained in these eastern healing methods is normally distributed with an average of $50,000 per year and a standard deviation of $3,900. Somewhat skeptical (but hopeful), you decide to take a random sample of salaries from 15 new graduates of Nezog and you get an average of $53,100. a. Test the claim that first-year Nezog employees make more than $50,000 a year. b. Find a 95% confidence interval to capture the average first-year salary with NEZOG. c. Explain how your hypothesis test and confidence interval are consistent.
Explanation / Answer
PART A.
Given that,
population mean(u)=50000
standard deviation, =3900
sample mean, x =53100
number (n)=15
null, Ho: =50000
alternate, H1: >50000
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 53100-50000/(3900/sqrt(15)
zo = 3.07853
| zo | = 3.07853
critical value
the value of |z | at los 5% is 1.645
we got |zo| =3.07853 & | z | = 1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : right tail - ha : ( p > 3.07853 ) = 0.00104
hence value of p0.05 > 0.00104, here we reject Ho
ANSWERS
---------------
null, Ho: =50000
alternate, H1: >50000
test statistic: 3.07853
critical value: 1.645
decision: reject Ho
p-value: 0.00104
ezog employees make more than $50,000 a year
PART B.
given that,
standard deviation, =3900
sample mean, x =53100
population size (n)=15
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
we use 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
confidence interval = [ 53100 ± Z a/2 ( 3900/ Sqrt ( 15) ) ]
= [ 53100 - 1.96 * (1006.9757) , 53100 + 1.96 * (1006.9757) ]
= [ 51126.3277,55073.6723 ]
employees make more than $50,000 a year
c.
both hypothesis, CI are arrived at similar reason. We have evidence to support
employees make more than $50,000 a year