Question

Clearly label your responses (Question 1.a, Question 1.b., etc.), show all your work, results, figures and interpretation in the Word document. At the end of the Word document, create a section entitled “Appendix: R Syntax”, and copy and paste all the R codes you used for each question part.

The target activation force of the buttons on a clicker is 1.967 newtons. Variation exists in activation force due to the nature of the manufacturing process. A sample of 9 clickers showed a mean activation force of 1.88 newtons. The population standard deviation is known to be 0.145 newton. Too much force makes the keys hard to click, while too little force means the keys might be clicked accidentally. We want to use an appropriate hypothesis test to detect excessive deviations in either direction. What is the appropriate hypothesis test? What is the test statistic value? At α = .05, does the sample indicate a significant deviation from the target?

Answer 1

Given that,
population mean(u)=1.967
standard deviation, σ =0.145
sample mean, x =1.88
number (n)=9
null, Ho: μ=1.967
alternate, H1: μ!=1.967
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 1.88-1.967/(0.145/sqrt(9)
zo = -1.8
| zo | = 1.8
critical value
the value of |z α| at los 5% is 1.96
we got |zo| =1.8 & | z α | = 1.96
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -1.8 ) = 0.072
hence value of p0.05 < 0.072, here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=1.967
alternate, H1: μ!=1.967
test statistic: -1.8
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.072
we do not have enough evidence to support the claim that the sample indicate a significant deviation from the target