Please do all work manually (not with excel), showing all steps clearly. Thank y
ID: 3235496 • Letter: P
Question
Please do all work manually (not with excel), showing all steps clearly. Thank you so much for your help!
It is to be determined whether car ownership affects student's academic achievement. A random sample of male students was drawn and their grade point average was measured and is tabulated below. Do the data present sufficient evidence to indicate a difference in the mean achievements between car owners and nonowners? Test using a 0.05 Car nonowners Car owners 100 100 Sample size Mean 2.7 2.54 Standard deviation 0.40 0.36Explanation / Answer
standard deviation, s = sqrt( (0.36^2+0.40^2)/2) =0.3805 (Since both groups are equal in number)
t = (2.7-2.54) /[ (0.3805/sqrt(2/100))] = 2.9732
Intermediate values used in calculations:
t = 2.9732
df = 100+100-2 = 198
P value and statistical significance:
The two-tailed P value equals 0.0033 from tables.
By conventional criteria, this difference is considered to be very statistically significant.
The mean of Group One minus Group Two equals 0.1600