Question

A well-designed and safe workplace can contribute greatly to increasing productivity. It is especially important that workers not be asked to perform tasks, such as lifting, that exceed their capabilities. The following data on maximum weight of lift (MWOL, in kilograms) for a frequency of 4 lifts per minute were reported in a certain article: 25.3 36.1 25.8 21.3 26.7 Suppose that it is reasonable to regard the sample as a random sample from the population of healthy males, age 18-30. Do the data suggest that the population mean MWOL exceeds 24.5? Answer the following questions using the following text box. Please label each answer according to question number. 1. State null and alternative hypothesis. 2. Report the normality test p-value and conclude the normality assumption. 3. Report the p-value for the t-test and conclude the analysis using p-value, at 5% level of significance.

Answer 1

Given that,
population mean(u)=24.5
sample mean, x =27.04
standard deviation, s =5.4706
number (n)=5
null, Ho: μ=24.5
alternate, H1: μ>24.5
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =2.132
since our test is right-tailed
reject Ho, if to > 2.132
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =27.04-24.5/(5.4706/sqrt(5))
to =1.0382
| to | =1.0382
critical value
the value of |t α| with n-1 = 4 d.f is 2.132
we got |to| =1.0382 & | t α | =2.132
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :right tail - Ha : ( p > 1.0382 ) = 0.1789
hence value of p0.05 < 0.1789,here we do not reject Ho
ANSWERS
---------------
1.
null, Ho: μ=24.5
alternate, H1: μ>24.5
test statistic: 1.0382
critical value: 2.132
2.
the normality test p-value and conclude the normality assumption from calculator
p value is 0.2994
decision: do not reject Ho
3.
p-value: 0.1789
we do not have enough evidence to support the claim that the population mean MWOL exceeds 24.5.