Question

The mean income per year of employees who produce internal and external newsletters and magazines for corporations was reported to be $51,500. These editors and designers work on corporate publications but not on marketing materials. As a result of poor economic conditions and oversupply, these corporate communications workers may be experiencing a decrease in salary. A random sample of 38 corporate communications workers revealed that x̄ = $49,762. Conduct a hypothesis test to determine whether there is any evidence to suggest that the mean income per year of corporate communications workers has decreased. Assume that the population standard deviation is $3750. Please use 4 decimal places for all critical values.

(0.5 pt.) a) Should a z or t distribution be used for statistical procedures regarding the mean? Remember to explain your answer.

(6 pts.) b) Is there any evidence to suggest the true mean income per year of corporate communications workers has decreased? Perform the hypothesis test at a 1% significance level. Include the complete four step hypothesis test with all of the work.

(2 pts.) c) Calculate and interpret the appropriate 99% bound that is consistent with what you did in part b).

(1 pt.) d) If the employer also found a cheaper medical insurance that reduced the amount of money that the employees have to pay for health insurance by $300 per year, in practical terms, are the employees are really taking home less money than before? Please explain your answer.

Answer 1

Given that,
population mean(u)=51500
standard deviation, σ =3750
sample mean, x =49762
number (n)=38
null, Ho: μ=51500
alternate, H1: μ<51500
level of significance, α = 0.01
from standard normal table,left tailed z α/2 =2.326
since our test is left-tailed
reject Ho, if zo < -2.326
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 49762-51500/(3750/sqrt(38)
zo = -2.86
| zo | = 2.86
critical value
the value of |z α| at los 1% is 2.326
we got |zo| =2.86 & | z α | = 2.326
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value : left tail - ha : ( p < -2.86 ) = 0
hence value of p0.01 > 0, here we reject Ho
ANSWERS
---------------
null, Ho: μ=51500
alternate, H1: μ<51500
test statistic: -2.86
critical value: -2.326
decision: reject Ho
p-value: 0

a.
we use z distribution as the reason the value of the population standard deviation is provided

b.
have evidence that the true mean income per year of corporate communications workers has decreased

c.
Upper Bound Confidence Interval
given that,
standard deviation, σ =3750
sample mean, x =49762
population size (n)=38
level of significance, α = 0.01
from standard normal table,left tailed z α/2 =2.3263
since our test is left-tailed
value of z table is 2.3263
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 = [ 49762 ± Z a/2 ( 3750/ Sqrt ( 38) ) ]
= [ 49762 - 2.3263 * (608.33) , 49762 + 2.3263 * (608.33) ]
= [ 48346.84,51177.16 ]

d.
true, the employee is very much lessar than the reported since the value of mean
mean stated is $49,762+ $300 is also lessar than initial value $51,500