Question

An article in the U.S. News & Works Report (September 28, 1981) states that approximately 21.3 million workers, more than a fifth of the workforce in the United States, have unorthodox working hours. More than 9.3 million work on a flexible schedule (the worker plans his own schedule) or on a weekly "compressed" schedule. A company planning to install flexible hours estimated that an average of 7 hours a day per assembly worker was needed to operate efficiently. Each of the company's 80 assemblers was asked to submit a tentative flexible schedule. If the average number of hours per day for Monday was 6.7 hours, and the standard deviation 2.7 hours. Using a 95% confidence interval, do the data provide evidence that the average number of hours worked every Monday, for all fitters in the company, will be less than 7 hours?

Por medio de una prueba de hipótesis, contestar usando un nivel de significancia del 5%

Answer 1

a.
TRADITIONAL METHOD
given that,
sample mean, x =6.7
standard deviation, s =2.7
sample size, n =80
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 2.7/ sqrt ( 80) )
= 0.302
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 79 d.f is 1.99
margin of error = 1.99 * 0.302
= 0.601
III.
CI = x ± margin of error
confidence interval = [ 6.7 ± 0.601 ]
= [ 6.099 , 7.301 ]
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DIRECT METHOD
given that,
sample mean, x =6.7
standard deviation, s =2.7
sample size, n =80
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 79 d.f is 1.99
we use 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
confidence interval = [ 6.7 ± t a/2 ( 2.7/ Sqrt ( 80) ]
= [ 6.7-(1.99 * 0.302) , 6.7+(1.99 * 0.302) ]
= [ 6.099 , 7.301 ]
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interpretations:
1) we are 95% sure that the interval [ 6.099 , 7.301 ] 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
b.
Given that,
population mean(u)=7
sample mean, x =6.7
standard deviation, s =2.7
number (n)=80
null, Ho: μ=7
alternate, H1: μ<7
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.664
since our test is left-tailed
reject Ho, if to < -1.664
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =6.7-7/(2.7/sqrt(80))
to =-0.9938
| to | =0.9938
critical value
the value of |t α| with n-1 = 79 d.f is 1.664
we got |to| =0.9938 & | t α | =1.664
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :left tail - Ha : ( p < -0.9938 ) = 0.16168
hence value of p0.05 < 0.16168,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=7
alternate, H1: μ<7
test statistic: -0.9938
critical value: -1.664
decision: do not reject Ho
p-value: 0.16168
we do not have enough evidence to support the claim that the average number of hours worked every Monday, for all fitters in the company, will be less than 7 hours.