Please use the following info to answer question B. Means = Curling N = 41 Mean
ID: 3217247 • Letter: P
Question
Please use the following info to answer question B.
Means =
Curling N = 41 Mean = 3.8049 StDev = 4.3944 95% CI = (2.2787, 5.3311)
Figure Skating N = 178 Mean = 4.1601 StDev = 5.9770 95% CI = (3.4276, 4.8859)
Ice Hockey N = 300 Mean = 4.3217 StDev = 4.3555 95% CI = (3.7575, 4.8859)
Pooled StDev = 4.97432
B. Use the five-step hypothesis testing procedure to determine if there is evidence that the mean number of hours exercised per week is not equal for students who prefer these three winter sports.
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
Explanation / Answer
Answer:
Means =Curling N = 41 Mean = 3.8049 StDev = 4.3944 95% CI = (2.2787, 5.3311)
Figure Skating N = 178 Mean = 4.1601 StDev = 5.9770 95% CI = (3.4276, 4.8859)
Ice Hockey N = 300 Mean = 4.3217 StDev = 4.3555 95% CI = (3.7575, 4.8859)
Pooled StDev = 4.97432
B. Use the five-step hypothesis testing procedure to determine if there is evidence that the mean number of hours exercised per week is not equal for students who prefer these three winter sports.
R code
mydata.mean <- c(3.8049,4.1601,4.3217)
mydata.sd <- c(4.3944,5.9770,4.3555)
mydata.n <- c(41,178,300)
library(rpsychi)
library(gtools)
ind.oneway.second(mydata.mean,mydata.sd,mydata.n)
R output:
$anova.table
SS df MS F
Between (A) 10.8 2 5.395 0.218
Within 12767.8 516 24.744
Total 12778.6 518
Table value of F with (DF1=2, DF2=516 ) at 5% level =3.013
Calculated F=0.218 < 3.013 the table value
The null hypothesis is not rejected.
The data indicate there is no significant difference among the three groups
There is no evidence that the mean number of hours exercised per week is not equal for students who prefer these three winter.