Question

A pudding manufacturer packages its product into bags weighing 1 kilogram, on average. The manufacturer’s statistician has discovered that the setting of the machine is causing the fill weights to drift. The statistician needs to detect shifts in the mean weight as quickly as possible and reset the machine when appropriate. In order to detect shifts in the mean weight, he collects a random sample of 50 bags periodically, weighs them, and calculates the mean and standard deviation. The data from this afternoon’s sample yields a sample mean of 1.03 kg and a sample standard deviation of 0.08 kg. Determine the p-value for the hypothesis test H0: μ = 1 versus H1: μ not equal to 1.

Given the Minitab hypothesis testing information and p-value you posted above, do you reject the null hypothesis and why?

researcher plans to conduct a study to determine how three different types of treatment affect depression. However, before the researcher carries out this study, he wants to make sure that the 40 participants taking part have depression scores that are considered to be 'normal'. Let's imagine that a score of 4.0 is considered to reflect 'normal' depression. Lower scores indicate less depression and higher scores indicate greater depression. Therefore, the depression of all 40 participants is measured and a one-sample t-test is used to determine whether this sample is representative of a normal population (i.e., whether the sample-estimated population mean score is significantly different from 4.0). Sample mean is 3.7243 and sample standard deviation is .737.

What can you conclude from the hypothesis testing T-test of the study?

Answer 1

a.
Given that,
population mean(u)=1
sample mean, x =1.03
standard deviation, s =0.08
number (n)=50
null, Ho: μ=1
alternate, H1: μ!=1
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.01
since our test is two-tailed
reject Ho, if to < -2.01 OR if to > 2.01
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =1.03-1/(0.08/sqrt(50))
to =2.6517
| to | =2.6517
critical value
the value of |t α| with n-1 = 49 d.f is 2.01
we got |to| =2.6517 & | t α | =2.01
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 2.6517 ) = 0.0108
hence value of p0.05 > 0.0108,here we reject Ho
ANSWERS
---------------
null, Ho: μ=1
alternate, H1: μ!=1
test statistic: 2.6517
critical value: -2.01 , 2.01
decision: reject Ho
p-value: 0.0108
we have enough evidence to support the claim that A pudding manufacturer packages its product into bags
weighing 1 kilogram, on average.
b.
Given that,
population mean(u)=4
sample mean, x =3.7243
standard deviation, s =0.737
number (n)=40
null, Ho: μ=4
alternate, H1: μ!=4
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.023
since our test is two-tailed
reject Ho, if to < -2.023 OR if to > 2.023
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =3.7243-4/(0.737/sqrt(40))
to =-2.3659
| to | =2.3659
critical value
the value of |t α| with n-1 = 39 d.f is 2.023
we got |to| =2.3659 & | t α | =2.023
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -2.3659 ) = 0.0231
hence value of p0.05 > 0.0231,here we reject Ho
ANSWERS
---------------
null, Ho: μ=4
alternate, H1: μ!=4
test statistic: -2.3659
critical value: -2.023 , 2.023
decision: reject Ho
p-value: 0.0231
we have enough evidence to support the claim that whether the sample-estimated population mean score is significantly different from 4.0