Increases in worker injuries and disability claims have prompted
renewed interest in workplace design and regulation. As one
particular aspect of this, employees required to do regular lifting
should not have to handle unsafe loads. One study reported
information regarding a random sample of 18 postal workers. The
sample mean rating of acceptable load attained with a
work-simulating test was found to be ? "=9.7 kg. Suppose we know
that the standard deviation is ?=4.3 kg. Suppose that in a
population of all male postal workers, the distribution of rating
of acceptable load can be modeled using a normal distribution with
mean ?.
(e) Holding all else constant, how would the width of the
confidence interval change if the standard deviation were 2.5 kg
rather than 4.3 kg? Explain why.
(f) Suppose that you don’t know the population standard deviation
but you do know that the sample standard deviation for the sample
of 18 postal workers is 4.3 kg. Construct a 95% confidence interval
for based on this information.
(g) Why is the confidence interval in (a) different from the
confidence interval in (f)?
since in part e the difference is positive therefore we can conclude that width of C.I is greater for sigma=4.3 as compared to sigma=2.5.
g) hii. you had not provide part a so i am not able to conclude exactly. but as i think in part one population standard deviation will be known. therefore confidence interval will be different because here in part f apply t statistic for critical value for unknown population variance by replacing sample variance but if population variance known than we apply z critical valie rather than t.
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