(confidence interval for means) The average number of sick days taken annually b

Question

(confidence interval for means) The average number of sick days taken annually by Canadian government employees is approximately normally distributed. We know the population standard deviation is Sigma-2.5 2.5 days, since it stays constant year to year. However, the population mean often changes (due to, e.g., differences in the yearly flu virus). A 2016 survey with n-100 finds a sample mean of x 7.9x-bar=7.9 days. We are interested in a confidence interval for the number of sick days taken by the entire population at the 95% confidence level. (a) Do we use the z-score or t-score in this case? (b) Find the appropriate (z or t) score (c) Find the confidence interval at the 95% confidence level

Explanation / Answer

a.
we use Z test score because they given population standard deviation so that we go for Z test

b.
Given that,
population mean(u)=6
standard deviation, =2.5
sample mean, x =7.9
number (n)=100
null, Ho: =6
alternate, H1: !=6
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 = 7.9-6/(2.5/sqrt(100)
zo = 7.6
| zo | = 7.6
critical value
the value of |z | at los 5% is 1.96
we got |zo| =7.6 & | 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 != 7.6 ) = 0
hence value of p0.05 > 0, here we reject Ho
ANSWERS
---------------
null, Ho: =6
alternate, H1: !=6
test statistic: 7.6
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0

c.
TRADITIONAL METHOD
given that,
standard deviation, =2.5
sample mean, x =7.9
population size (n)=100
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 2.5/ sqrt ( 100) )
= 0.25
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
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
margin of error = 1.96 * 0.25
= 0.49
III.
CI = x ± margin of error
confidence interval = [ 7.9 ± 0.49 ]
= [ 7.41,8.39 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =2.5
sample mean, x =7.9
population size (n)=100
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 = [ 7.9 ± Z a/2 ( 2.5/ Sqrt ( 100) ) ]
= [ 7.9 - 1.96 * (0.25) , 7.9 + 1.96 * (0.25) ]
= [ 7.41,8.39 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [7.41 , 8.39 ] 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
[ANSWERS]
best point of estimate = mean = 7.9
standard error =0.25
z table value = 1.96
margin of error = 0.49
confidence interval = [ 7.41 , 8.39 ]

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