Download Hypothesis Testing Exercise 1 We reject H0 if and only if pvalue ≤ α

Hypothesis Testing
Exercise 1 We reject H0 if and only if pvalue ≤ α = 0.05. a) Reject H0 b) Reject H0
c) Do not reject H0 d) Reject H0 e) Do not reject H0
Exercise 2 This is an upper-tailed test. Let µ be the mean fat content of the hotdogs.
Then we are interested in the hypothesis
H0 : µ = 0.25 vs H1 : µ > 0.25
Exercise 3 This is a Two-tailed Test. Let µ be the mean lifetime of light-bulbs.
H0 : µ = 1000 vs H1 : µ 6= 1000
Exercise 4 a) This is an upper-tailed test. µ is the mean IQ score in the general
population and we want to show that the students in the gifted program have a
higher IQ than µ = 100. So, we are interested in the hypothesis
H0 : µ = 100 vs H1 : µ > 100
b) The p-value is 0.0287.
c) The p-value is 0.0287 which is less than the specified value of α = 0.05 so we reject
the null hypothesis.
Exercise 5 a) If let X be the amount of acetaminophen in a randomly selected tablet.
We are told that X ∼ N (µ, σ 2 ) but µ and σ are unknown. We are interested in
showing that the tablets have a mean value different from µ0 = 500. Therefore,
H0 : µ = 500 versus H1 : µ 6= 500
b) Using the rejection method to perform the test of hypothesis,
tobs =
499.6 − 500
x̄ − µ0
√ =
√
≈ −3.142
s/ n
0.9/ 50
For a two-tailed test, the critical value is t49,α/2 = t49,0.01/2 = 2.68. Since the quantile
of the Student distribution with 49 degrees of freedom is not in your statistical table,
you can consider the distribution whose the degrees of freedom is the closest to 49
(here, it is 40 and the associated value is t40,0.01/2 = 2.704)
Therefore, since
|tobs | = | − 3.142| > 2.68 = t49,0.01/2
we can reject the null hypothesis.
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c) The summary statistics are: x̄ = 499.6, s = 0.9 and n = 50, therefore a (1 − α)%
confidence interval is
s
s
= (499.26, 499.94)
x̄ + tn−1,α/2 √ , x̄ + tn−1,1−α/2 √
n
n
Exercise 6 pvalue=0.0006. With an error probability 0.0006, we can claim that the
bottles are underfilled
Exercise 7
H0 : µ = 6500 vs. H1 : µ 6= 6500
a) In this case, the rejection region is
s
W = {x̄ : |x̄ − 6500| > −tn−1,α/2 √ }
n
Here |x̄ − 6500| = 319 and n − 1 = 35, considering a type I risk of 0.05, tn−1,α/2 =
−2.045 and −tn−1,α/2 √sn = −437.27. Therefore, because the observed value of x̄
does not belong to the rejection region with level 0.05, we do not reject H0 .
b) The p-value is the probability α? with −t35,α? /2 =
Exercise 8
√
319 36
.
1265
Thus, α? = 0.14
a)
H0 : µ = −1 vs. H1 : µ = 0
b) Reject if x ∈ W = {x : x > −0.5065}
c) 0.9543
d) Reject if x ∈ W = {x : x > −0.3021}
e) 0.8430
f ) 0.0359
Exercise 9
a)
H0 : p = 0.063 vs. Ha : p 6= 0.063
0.080 − 0.063
p − p0
=q
= 1.214
z=q
p0 (1−p0 )
n
0.063(1−0.063)
301
The critical value for this test is Zα = Z0 .025 = 1.96. The p-value is 0.2262.
Therefore, because |1.214| < 1.96 and the p-value is greater than α we do not reject
H0 and state that there is not sufficient evidence against H0 which suggests the true
proportion is not different from 0.063, based on this sample.
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b)
H0 : p = 0.063 vs. H1 : p > 0.063
0.080 − 0.063
p − p0
=q
= 1.214
z=q
p0 (1−p0 )
n
0.063(1−0.063)
301
The critical value for this test is Zα = Z0 .05 = 1.645. The p-value is 0.1131.
Therefore, because |1.214| < 1.645 and the p-value is greater than α we do not
reject H0 and state that there is not sufficient evidence against H0 which suggests
the true proportion is not greater then 0.063, based on this sample.
c)
r
p̂q̂
= 0.080 ± 1.96(0.01564) = 0.080 ± 0.0307.
n
Therefore, we can be 95% confident that the true proportion is between (0.0493, 0.1107).
So, comparing to the two-sided hypothesis test we can see that both the interval and
H0 suggest that p is not different from 0.063.
p̂ ± zα/2
Exercise 10 If let X be number of students who drop out of high school, then
X ∼ Binomial(n = 250, p)
Therefore, we will test
H0 : p = 0.24 vs. H1 : p < 0.24
The p-value is given by
P (Z < −1.33) = 1 − P (Z < 1.33) = 1 − 0.908 = 0.092
Therefore, at α = 0.05 we do not reject the null hypothesis.
Exercise 11 a) A type I error consists of judging one of the two companies favored
over the other when in fact there is a 50-50 split in the population. A type II error
involves judging the split to be 50-50 when it is not.
b) We expect 25(0.5) = 12.5 successes when H0 is true. So, any X values less than 6
are at least as contradictory to H0 as x = 6. But since the alternative hypothesis
states p 6= 0.5, the X values that are just as far away on the high side are equally
contradictory. Those are 19 and above. So, values at least as contradictory to H0
as x = 6 are {0, 1, 2, 3, 4, 5, 6, 19, 20, 21, 22, 23, 24, 25}.
c) When H0 is true, X has a binomial distribution with n = 25 and λ = 0.5. From
part (b), pvalue=P (X ≤ 6 or X ≥ 19) = F (6) + (1 − F (18)) = 0.015 where F is
the cdf of the binomial distribution with parameters (25,0.5).
d) A two tailed pvalue of 0.044 (2 × 0.022) occurs when x = 7. Tha is, saying we will
reject H0 if and only if the pvalue is less than 0.044 must be equivalent to saying
we will reject H0 if and only if X ≤ 7 or X ≥ 18 (the same distance from 12.5 but
on the high side). Therefore, for any value of p 6= 0.5, β(p) =P(do not reject H0
when X ∼ B(25, λ)). Thus β(0.4) = 0.845 while β(0.3) = 0.488. By symmetry (or
re-computation) β(0.6) = 0.845 while β(0.7) = 0.488.
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e) From part (c), the pvalue associated with x = 6 is 0.014. Since 0.014 ≤ 0.044, the
procedure in (d) leads us to reject H0.
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