- Problem 5.6.
- Sample mean will be mid point of confidence interval or 71.
- Margin of error will be 71-65 or 6.
- Standard deviation will be margin of error or ME (6) divided by t with 24 (n-1) DF at 0.90 (1.3178) multiplied by squire root of n (25) or 5. The answer is 22.7646.
t24_90<-qt(0.90, df = c(24))
(6/t24_90)*5
## [1] 22.76459
- Problem 5.48
- H0: Means for all five groups is the same or Mean1=Mean2=Mean3=Mean4=Mean5 H1: At least one Mean is not equal
- We assume independence of samples. We assume normality of group distribution. We can see from Box plots that some samples are skewed. However, our sample sizes are pretty big so some skewedness is acceptable.
- We have 4 DF. Residuals DF is 1,167. Total DF is 1,171. Degree Sum Squires are 2,004.101. Total SS equals to 269,386.1. Mean Sq Residuals equals 229.1191. F value is 2.188992.
- We will accept H0 at 5% level of significance, but we would reject it at 10% level of significance.
DSS<-(38.67-40.45)**2*121+(39.6-40.45)**2*546+(41.39-40.45)**2*97+(42.55-40.45)**2*253+(40.85-40.45)**2*155
DSS+267382
## [1] 269386.1
MSR<-267382/1167
501.54/MSR
## [1] 2.188992