Transcript Two Sample β

One Sample β Example
1. Over the past five hundred years or so, the amount of time
it has taken Santa Claus to climb down chimneys, deposit
presents and get back to his sleigh has been normally
distributed with a mean of 10 seconds per chimney and a
standard deviation of 3 seconds. This year, since he's
beginning to feel his age, Santa has engaged in a rigorous
physical fitness program which he believes will allow him to
significantly reduce his time. To test his belief, he plans to
have one of the elves time him on a random sample of 36
chimneys and then to conduct the hypothesis test with α ≤
.025. If, in fact, his true mean time is now 8 seconds, what is
the probability that Santa will incorrectly conclude that his
exercise had no effect?
One Sample β Example
µ = 10
µA = 8
σ=3
α = .025
n = 36
Z = 1.96
1. First step is to obtain the value of X that must
be exceeded in order to reject HO.
Z = -1.96 = X – 10
3/√36
X = -1.96 (1/2) + 10 = 9.02
One Sample β Example
2. Second step is to work out the probability that
sample mean does NOT exceed that value,
given that the new population mean is 8.
Z = 9.02 – 8.0 = 2.04
3/√36
P(Z ≤ 2.04) = .4793. Desired P = .4 - .4793) = .0207
Two Sample β Example 1
2. The Christmas exam scores of two sections of a
course are compared to see if a significant
difference exists between their means. It is
assumed that students were originally randomly
assigned to their sections.
Section 1
Section 2
X1 = 66.0
s1 = 10.75
n1 = 50
X2 = 62.0
s2 = 12.46
n2 = 38
Two Sample β Example 1
a) Perform the appropriate test on the data using α ≤
.05:
b) Suppose that, over a period of years, section 1 has
actually outperformed section 2 by 2.8 points, on
average. What is the probability that this year's
sample data would allow you to correctly reject the
null hypothesis of no difference between the
sections (still using α ≤ .05 and a 2-tailed alternative
hypothesis)?
Two Sample β Example 1
HO: µ1 – µ2 = 0
HA: µ1 – µ2 ≠ 0
Zcrit = Z.025 = 1.96
Zobt =
(66 – 62) – 0
10.752 + 12.462
50
38
Do not reject HO
√
= 1.58
Two Sample β Example 1
b. The alternative hypothesis would specify that
(µ1 – µ2) = 2.8. The question asks, with the
actual difference in the populations being 2.8,
and with sample sizes n1 and n2, and given s1
and s2, what is the probability that we exceed
the critical value for X1 – X2 (and thus reject
HO)?
Two Sample β Example 1
1. First we work out the critical value of X1 – X2.
1.96 = (X1 – X2) – 0
2.53
0 + 1.96 (2.53) = (X1 – X2) = 4.96
These are sampling
distributions of the
difference between
the sample means
.025
0
4.96
.3023
.1977
2.8
4.96
Two Sample β Example 2
3. One of the reasons more people buy Florida orange
juice is that Florida oranges have historically had higher
levels of vitamin C (on average 120 mg. per orange
with a standard deviation of 10) than California
oranges (on average 110 mg. per orange with a
standard deviation of 15). This year, California orange
growers are using a new fertilizer in an effort to
increase the level of vitamin C in their oranges. The
alternative hypothesis that you plan to test is that
because of the new fertilizer the difference in vitamin C
concentrations between California and Florida oranges
is now less than 10 mg.
Two Sample β Example 2
To carry out this test you plan to sample 40
oranges from each State.
If, because of the fertilizer, the true difference is
now only 2 mg., what is the probability that you
would be able to reject the null hypothesis that
the true difference has not changed (α ≤ .01)?
Two Sample β Example 2
HO: µF – µC = 10
HA: µF – µC < 10
Zcrit = Z.01 = 2.33
Zobt =
(X1 – X2) – 10
= 2.33
10 2 + 152
40 40
(X1 – X2) = 10 – 2.33 (2.85) = 3.36
√
Two Sample β Example 2
Z = 3.36 – 2.0
2.85
= .48
P (Z < .48) = .1844
Desired P = .5 + .1844 = .6844
These are sampling
distributions of the
difference between
the sample means
3.36
10
.5
.1844
2
3.36