Unformatted text preview: 7. What would you conclude about the the idea that 15% of the people have Type 0 blood, if your 2 score calculated for the preceding problem was equal to 2.8?
he data you observed contradicted the 15% Type 0 measurement b. The data is in the realm where you cannot make any conclusions.
A c. The data you observed supported the 15% Type 0 measurement. 8. If you know that the percentage of people who have Type 0 blood is 15%, what is the probability that
the 10“1 person that you test is the ﬁrst person to have Type 0 blood? @035
I. .197
c. .803 d. .965
9. If the percentage of people with Type 0 blood 15%, what is the percentage of people who do m have
Type 0 blood?
a. 0%
b. 15%
85%
. 115% 10. A z—score of 1.5 means that
@The observation has a 1.5% chance of occurring.
@The observation is 1 5 standard deviations to the right of the mean.
c. The observation is 1.5 averages away from the standard deviation.
d. The observation 18 1.5 standard deviations to the left of the mean.
e. The observation IS 150% chance of happening. 1 1. If you look at the following residual plot, what would you think about the ﬁt of the equation to the points?
Pick themstanswer. ' ' ' " a. The equation ﬁts the data well.
b. All that is left is random variation.
There is some other variable still yet to be used to explain y. @AandB
e. B andC 12. interpret Rsquared for the following TI 83 printout, where x — static weight and y: weight In LinReg 3+b
giai‘esaasz b=1. #666413'a
quégs 734331 mo '
A, 93% of the variation of weight in motion can be explained by static 55weight.
b. 93% of the variation of static weight can be explained by weight in motion.
c. 97% of the variation of weight in motion can be explained by static weight.
d. 97% of the variation of static weight can be explained by weight in motion.
e. Rsquared has the same meaning as the correlation. ...
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 Spring '08
 Hardy
 Statistics

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