Question

Using the same information and data as was given in Problem 6, provide an 90% confidence interval for mean time advantage of right-hand over left-hand threads. Do you think that the time saved would be of practical importance if the task were performed many times - for example, by an assembly line worker? To help answer this question, find the mean time for right-hand threads as a percent of the mean time for left-hand threads. Be descriptive.

by the same information, the question is referring to the information/question below:

The design of controls and instruments has a large effect on how easily people can use them. A student project investigated this effect by asking 25 right-handed students to turn a knob (with their right hands) that moved an indicator by screw action. There were two identical instruments, one with a right-hand thread (the knob turns clockwise) and the other with a left-hand thread (the knob turns counter-clockwise). The table below gives the times required (in seconds) to move the indicator a fixed distance.

Subject ID	Right-hand Thread	Left-hand Thread
1	113	137
2	105	105
3	130	133
4	101	108
5	138	115
6	118	170
7	87	103
8	116	145
9	75	78
10	96	107
11	122	84
12	103	148
13	116	147
14	107	87
15	118	166
16	103	146
17	111	123
18	104	135
19	111	112
20	89	93
21	78	76
22	100	116
23	89	78
24	85	101
25	88	123

1. The project hoped to show that right-handed people find right-hand threads easier to use. State the appropriate H_o and H_a about the mean time required to complete the task.
2. Carry out a test of your hypotheses. Give the P-value, report and explain your conclusion in a brief paragraph.

Answer 1

i.
TRADITIONAL METHOD
given that,
mean(x)=104.12
standard deviation , s.d1=15.796
number(n1)=25
y(mean)=117.44
standard deviation, s.d2 =27.262
number(n2)=25
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((249.514/25)+(743.217/25))
= 6.302
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.1
from standard normal table,right tailed and
value of |t α| with min (n1-1, n2-1) i.e 24 d.f is 1.318
margin of error = 1.318 * 6.302
= 8.305
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (104.12-117.44) ± 8.305 ]
= [-21.625 , -5.015]
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DIRECT METHOD
given that,
mean(x)=104.12
standard deviation , s.d1=15.796
sample size, n1=25
y(mean)=117.44
standard deviation, s.d2 =27.262
sample size,n2 =25
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 104.12-117.44) ± t a/2 * sqrt((249.514/25)+(743.217/25)]
= [ (-13.32) ± t a/2 * 6.302]
= [-21.625 , -5.015]
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interpretations:
1. we are 90% sure that the interval [-21.625 , -5.015] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion
90% confidence interval for mean time advantage of right-hand over left-hand threads
[-21.625 , -5.015]

ii.
Given that,
mean(x)=104.12
standard deviation , s.d1=15.796
number(n1)=25
y(mean)=117.44
standard deviation, s.d2 =27.262
number(n2)=25
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.1
from standard normal table, two tailed t α/2 =1.711
since our test is two-tailed
reject Ho, if to < -1.711 OR if to > 1.711
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =104.12-117.44/sqrt((249.51362/25)+(743.21664/25))
to =-2.1138
| to | =2.1138
critical value
the value of |t α| with min (n1-1, n2-1) i.e 24 d.f is 1.711
we got |to| = 2.11377 & | t α | = 1.711
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -2.1138 ) = 0.045
hence value of p0.1 > 0.045,here we reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -2.1138
critical value: -1.711 , 1.711
decision: reject Ho
b.
p-value: 0.045
we have enough evidence to support the claim that There were two identical instruments, one with a right-hand thread (the knob turns clockwise) and the other with a left-hand thread (the knob turns counter-clockwise)