Probleml: Ten new devices were tested under two weather conditions. High wind sp

Question

Probleml: Ten new devices were tested under two weather conditions. High wind speeds with low visibility and Low wind speeds with high visibility. Their accuracies are measured, the measurements of their accuracies are as follows. 140 139 148 136 129 145 155 130 123 192 188 190 186 179 194 183 191 187 10 At the 3% significance level, test if the accuracy of the device is higher at Low wind speed with high visibility than at High wind speed with low visibility condition. Find a 90% confidence interval for the difference for the means of the device accuracies. What are the assumptions needed for the test in part (b). a) b) c)

Explanation / Answer

PART A.
Given that,
mean(x)=138.3333
standard deviation , s.d1=10.0995
number(n1)=9
y(mean)=187.7778
standard deviation, s.d2 =4.6845
number(n2)=9
null, Ho: u1 > u2
alternate, H1: u1 < u2
level of significance, = 0.03
from standard normal table,left tailed t /2 =2.189
since our test is left-tailed
reject Ho, if to < -2.189
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =138.3333-187.7778/sqrt((101.9999/9)+(21.94454/9))
to =-13.3237
| to | =13.3237
critical value
the value of |t | with min (n1-1, n2-1) i.e 8 d.f is 2.189
we got |to| = 13.32373 & | t | = 2.189
make decision
hence value of | to | > | t | and here we reject Ho
p-value:left tail - Ha : ( p < -13.3237 ) = 0
hence value of p0.03 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 > u2
alternate, H1: u1 < u2
test statistic: -13.3237
critical value: -2.189
decision: reject Ho
p-value: 0

PART B.
given that,
mean(x)=138.3333
standard deviation , s.d1=10.0995
sample size, n1=9
y(mean)=187.7778
standard deviation, s.d2 =4.6845
sample size,n2 =9
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 = [( 138.3333-187.7778) ± t a/2 * sqrt((102/9)+(21.945/9)]
= [ (-49.445) ± t a/2 * 3.711]
= [-56.347 , -42.542]
PART C.
interpretations:
1. we are 90% sure that the interval [-56.347 , -42.542] 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

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