The average population of a process was 50. However, part of the process was cha

Question

The average population of a process was 50. However, part of the process was changed with the introduction of a new machine. When the data from the new process were as follows, The deviation is not known and the significance level is then 0.05 53, 52, 47, 49, 54, 50 The average population of a process was 50. However, part of the process was changed with the introduction of a new machine. When the data from the new process were as follows, The deviation is not known and the significance level is then 0.05 53, 52, 47, 49, 54, 50

Explanation / Answer

Ho: Mean of the population =50

H1: Mean of the population =/= 50

observations : 53, 52, 47, 49, 54, 50

sd of the obsevation = 2.639444

here n=6

& X'= 50.83333 i.e. mean(53, 52, 47, 49, 54, 50)

it's a t test as the deviation of the population is unknown so,

t= (X'-mu)/(Sd/sqrt(n))

t= (50.83333-50)/(2.639444/sqrt(6)) = 0.7733573

while the critical t value at 5% significance is from the t table at 5 degrees of freedom is

t= (-2.570582, 2.570582)

and so 0.7733573 is folling between (-2.570582, 2.570582) and so here we can say that we cannot reject the null hypothesis Ho

and so here we conclude that the mean of a process is 50

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.