Download Formula for one-sample z-test

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Formula for one-sample z-test
To test the null hypothesis of H0 : µ = µ0 against one of the alternatives from below:
HA : µ > µ0
(Upper-tailed alternative)
HA : µ < µ0
(Lower-tailed alternative)
HA : µ 6= µ0
(Two-tailed alternative)
We use the test statistic of
z =
X − µ0
√ ,
σ/ n
and with significance level α,
HA
rejection region
p-value
P ower
upper-tailed
z > zα
1 − Φ(z)
1 − Φ(zα − µA )
lower-tailed
z < −zα
Φ(z)
Φ(−zα − µA )
z < −zα/2 or z > zα/2
2 1 − Φ(|z|)
1 − Φ(z α2 − µA ) + Φ(−z α2 − µA )
two-tailed
µA =
µ−µ
√0
σ/ n
2
HW 7 on One-sample Inference
Name:
Questions
1. A random sample of 100 lightning flashes in a certain region resulted in a sample
average radar echo duration of .91 sec. Assume that the observations are random
sample from the normal distribution with (true) standard deviation of .34 sec.
(a) with α = 5%, test the null hypothesis of µ = 1 sec against lower-tail alternative.
Picture below represents the behavior of test statistic z under H0 and HA .
i. mean of curve under H0 is
0
ii. mean of curve under HA is
(formula)
iii. SD of curve under H0 is
1
iv. SD of curve under HA is
1
µ−µ
√0
σ/ n
−zα = −1.65
v. critical value is
vi. We reject H0 if test statistic is greater/less than the CV. Less
vii. Label α, β, and Power on the picture below:
(b) Calculate the test statistic and report your conclusion with signifiance level of
5%.
Test statistic z is
z=
x̄ − µ0
.91 − 1
√ =
√
= −2.64
σ/ n
.34/ 100
Since the test statistic is less than CV, we reject H0 . There is an evidence that
true mean duration of the lightning µ is less than 1 second.
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(c) Calculate P-value from above test statistic.
Since this is lower-tailed test, P-value is Φ(−2.64) = .004.
(d) Calculate probability of false negative of above test, when µ = .9.
P( false negative) = β. Using the formula from the first page,
µ − µ0 .9 − 1 √
√
1 − Φ − zα −
= 1 − Φ − 2.33 −
σ/ n
.34/ 100
= 1 − Φ(.61) = .73
So if true µ is .9, then this test will reject H0 with probability .27.
Since α=5%, we have zα = 1.65, and above solution is wrong. It should be
.9 − 1 √
β = 1 − Φ − 1.65 −
= 1 − Φ(1.29) = 0.0985.
.34/ 100
I gave everybody +1 point on this.
2. Suppose we are interested in modeling the distribution of tread life for a premium
brand of radial tire. Assuming that the life of the tires are normaly distributed with
unkown mean µ and standard deviation σ = 5120, the company has decided to test
randomly choose n tires and test their lifetime.
(a) With α = 5%, we wish to test µ = 40000 miles against upper-tail alternative.
Picture below represents the behavior of test statistic z under H0 and HA .
i. mean of curve under H0 is
0
ii. mean of curve under HA is
(formula)
iii. SD of curve under H0 is
1
iv. SD of curve under HA is
1
v. critical value is
µ−µ
√0
σ/ n
zα = 1.65
vi. We reject H0 if test statistic is greater/less than the CV. Greater
vii. Label α, β, and Power on the picture below:
(b) Suppose in the actual experiment, random sample of size n = 100 were tested,
and resulted in sample mean of 43563 miles. Perform the test, and report P-value
and conclusion.
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Test statistic z is
z=
x̄ − µ0
43564 − 40000
√ =
√
= 6.96
σ/ n
5120/ 100
Since test stat is greater than the critical value, we can reject H0 , and say there is
strong evidence that true mean lifetime of the tire µ is greater than 40000 miles.
(c) What is the p-value of the above test?
Since this is upper-tailed test, P-value is 1 − Φ(6.96) ≈ 0. In the z−table, 6.96
is out of chart. So that means Φ(6.96) is very close to 1. So the p-value is very
close to 0.
(d) Calculate the sample size n we need in order to have probability of false negative
of above test equal to .1 when µ = 41000.
This is upper-tailed test, so our β is
µ − µ0 41000 − 4000 √
√
β = Φ zα −
= Φ 1.65 −
= .1
σ/ n
5120/ n
Since from the z-table we know that Φ(−1.28) = .1, we have equation
1.65 −
41000 − 40000
√
= −1.28
5120/ n
Solving the equation for n, we get n = 225.05.
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3. An article in the Journal of Heat Transfer (Trans. ASME, Sec. C, 96. 1974. p.
59) described a new method of measuring the thermal conductivity of Armco iron.
Using a temperature of 100F and a power input of 550 watts, the 10 measurements of
thermal conductivity (in Btu/hr-ft-F) resulted in sample mean of 41.92. Assuming the
measurements were random sample from Normal distribution with mean µ and SD of
.3.
(a) Test the null hypothesis that µ = 42 against two-sided alternative with significance level of 5%.
i. Critical Values are
and
ii. Rejection regions are
and
iii. Test Statistic is
iv. P-value is
. ±1.96
. z < −1.96 and z > 1.96
-.8432
. .4
v. State your conclusion:
Test statistic z is
z=
41.92 − 42
x̄ − µ0
√ =
√
= −.8432
σ/ n
.3/ 10
Since this is two-tailed test, P-value is 2(1 − Φ(| − .8432|)) = 2(.20) = .4. This is
very high p-value. If you use α = 5%, (or any other reasonable α) you wouldn’t
be able to reject H0 . There’s no evidence that µ is different from 42. The test is
inconclusive.