A governmental survey taken from 19992002 found that the average 7 year old
boy weighed 59.8 lbs and 15% of those boys were overweight. Some
researchers want to see if, in the past decade, 7 year old boys have gotten
heavier. They take a random sample of 150 boys and Fnd, in this random
sample, that 17% are overweight. The average weight of a 7 year old boy in
their sample was 61.2 lbs and the standard deviation for the weight was 13.6
lbs.
A) Is there evidence that 7 year old boys, on average, are getting heavier?
Remember to state your hypotheses, Fnd your test statistic, calculate your
pvalue, and draw your conclusions.
There's a lot of data mentioned in this problem. So the best place to start with this problem is with the
question: "Is there evidence that 7 year old boys, on average, are getting heavier?"
•
Whenever you see the phrase: "Is there evidence.
.." you are being asked to perform a
hypothesis test. You will need to do all 4 steps.
•
Any time you see the word "average", you know you're dealing with a quantitative variable, and
your hypotheses need to be about
µ
.
THE FOUR STEPS
1.
Hypotheses:
The problem asks: are 7 year old boys getting heavier? Heavier than what? The problem
mentions how heavy they were, on average, in 19992002. So the established value, or null
value, must be what they weighed a decade ago; i.e. the null value must be
µ
0
= 59.8 lbs. As for
the alternative, the problem does specifically state direction, which is we want to see if boys are
getting heavier, or greater than, 59.8 lbs. Therefore, the Hypotheses are:
Ho :
µ
= 59.8 lbs.
Ha :
µ
> 59.8 lbs.
2.
Test statistic:
I always write down all the data from the sample before I start calculating.
x
=
61.2
lbs
s
=
13.6
lbs
n
=
150
Then I find the standard error (se) :
se
=
s
n
=
13.6
lbs
150
=
1.11
lbs
Finally, the test statistic. Since we're dealing with averages, it is best to use t. Of course, since
our n=150, you can use z if you'd like, but I'll proceed with t for consistency:
t
=
x
−
0
se
=
61.2
lbs
−
59.8
lbs
1.11
lbs
=
1.26 and df = 150  1 = 149
3.
The pvalue
using R:
> pt(1.26, 149)
[1] 0.8951804
But the alternative is >, so the p value = 1 0.895 = 0.105 (no need to double b/c onesided)
using table B:
The closest degrees of freedom are either 100 or infinity. Let's use 100:
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 Summer '11
 aaa
 Null hypothesis, Statistical hypothesis testing

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