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

we shall answer this using the open source statistical package R , the complete R snippet is as follows

HighWind <- c(140,139,148,136,129,145,155,130,123)
LowWind <- c(192,188,190,186,179,194,183,191,187)

# perform the test
t.test(HighWind,LowWind,paired = TRUE,conf.level = 0.97,alternative = "less")

## 90% confidence interval

t.test(HighWind,LowWind,paired = TRUE,conf.level = 0.90,alternative = "less")

The results are

t.test(HighWind,LowWind,paired = TRUE,conf.level = 0.97,alternative = "less")

   Paired t-test

data: HighWind and LowWind
t = -14.239, df = 8, p-value = 2.883e-07 ## as the p value is less than 0.05 , hence we reject the null hypothesis in favor of alternate hypothesis to conclude that low wind has higher accuracy
alternative hypothesis: true difference in means is less than 0
97 percent confidence interval:
-Inf -41.84248
sample estimates:
mean of the differences
-49.44444

>
>
>
> ## 90% confidence interval
>

> t.test(HighWind,LowWind,paired = TRUE,conf.level = 0.90)

   Paired t-test

data: HighWind and LowWind
t = -14.239, df = 8, p-value = 5.767e-07
alternative hypothesis: true difference in means is not equal to 0
90 percent confidence interval:
-55.90183 -42.98706
sample estimates:
mean of the differences
-49.44444

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