In: Statistics and Probability
Ten randomly selected people took an IQ test A, and next day they took a very similar IQ test B. Their scores are shown in the table below.
Person | A | B | C | D | E | F | G | H | I | J |
Test A | 85 | 104 | 129 | 112 | 88 | 93 | 116 | 79 | 96 | 104 |
Test B | 85 | 104 | 130 | 117 | 88 | 96 | 119 | 78 | 95 | 104 |
Consider (Test A - Test B). Use a 0.01 significance level to test
the claim that people do better on the second test than they do on
the first. Assume differences are normal. (Note: You may wish to
use software.)
(a) The test statistic is (at least 2 digits after the decimal)
(b) The critical value is (at least 3 digits after the decimal)
(c) Is there sufficient evidence to support the claim that
people do better on the second test?
A. No
B. Yes
(d) Construct a 99% confidence interval for the mean of the
differences. Again, use (Test A - Test B) Use 3 digits after the
decimal.
a)
test statistic is =-1.58
b)
critical value is =-2.821
c)
A) NO (since test statistic is not less than critical value)
d)
for 99% CI; and 9 degree of freedom, value of t= | 3.250 | ||
therefore confidence interval=sample mean -/+ t*std error | |||
margin of errror =t*std error= | 2.055 | ||
lower confidence limit = | -3.055 | ||
upper confidence limit = | 1.055 | ||
from above 99% confidence interval for population mean =(-3.055,1.055) |