Question

1. At the alpha equals 5% level, a researcher was testing whether the average annual salary for nurses in a hospital setting in City Y is below from the national average salary in a hospital setting of $45,000. Assume that the salaries are approximately normal. a) What is the null hypothesis? What is the alternative hypothesis? Describe any symbols that you used. Now you go out and collect the data. Assume that the following salaries are from a normal distribution. $42,000; $40,000; $38,000; $32,000; $46,000, $41,000. b) What test will you use to analyze the data? What is your test statistic? p-value? c) Assume that alpha-5%. What conclusion did you make - Reject the null or not? State the conclusion in the context of the problem d) Find the corresponding 95% confidence interval for mu. Interpret the confidence interval in the context of the problem

Answer 1

1.

Given that,
population mean(u)=45000
sample mean, x =39833.33
standard deviation, s =4665.476
number (n)=6
null, Ho: μ=45000
alternate, H1: μ<45000
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =2.015
since our test is left-tailed
reject Ho, if to < -2.015
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =39833.33-45000/(4665.476/sqrt(6))
to =-2.7126
| to | =2.7126
critical value
the value of |t α| with n-1 = 5 d.f is 2.015
we got |to| =2.7126 & | t α | =2.015
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :left tail - Ha : ( p < -2.7126 ) = 0.02107
hence value of p0.05 > 0.02107,here we reject Ho
ANSWERS
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a.
null, Ho: μ=45000
alternate, H1: μ<45000
b.
test statistic: -2.7126
critical value: -2.015
c.
decision: reject Ho
p-value: 0.02107
we have enough evidence to support the claim that average salary of nurse is below than
national average salary in a hospital is 45000$
d.
TRADITIONAL METHOD
given that,
standard deviation, σ =4665.476
sample mean, x =39833.33
population size (n)=6
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 4665.476/ sqrt ( 6) )
= 1904.67
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 1904.67
= 3733.16
III.
CI = x ± margin of error
confidence interval = [ 39833.33 ± 3733.16 ]
= [ 36100.17,43566.49 ]
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DIRECT METHOD
given that,
standard deviation, σ =4665.476
sample mean, x =39833.33
population size (n)=6
level of significance, α = 0.05
from standard normal table, two tailed z α/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 = [ 39833.33 ± Z a/2 ( 4665.476/ Sqrt ( 6) ) ]
= [ 39833.33 - 1.96 * (1904.67) , 39833.33 + 1.96 * (1904.67) ]
= [ 36100.17,43566.49 ]
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interpretations:
1. we are 95% sure that the interval [36100.17 , 43566.49 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean