A governmental survey taken from 1999-2002 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
p-value, 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 1999-2002. 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 p-value
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 one-sided)
using table B:
The closest degrees of freedom are either 100 or infinity. Let's use 100: