Download Math 144 tutorial 9

Question 1
Question 2
Question 3
Question 4
Math 144 tutorial 9
26 April, 2010
Question 5
Question 6
Question 1
Question 2
Question 3
Question 4
Question 5
1. A random sample of 64 bags weighted, on average, 5.23 ounces.
Assume that the standard deviation is 0.24 ounces. Test the
hypothesis that µ = 5.5 ounces against the alternative hypothesis,
µ < 5.5 ounces at the 0.05 level of significance.
Question 6
Question 1
Question 2
Question 3
Question 4
Question 5
1. A random sample of 64 bags weighted, on average, 5.23 ounces.
Assume that the standard deviation is 0.24 ounces. Test the
hypothesis that µ = 5.5 ounces against the alternative hypothesis,
µ < 5.5 ounces at the 0.05 level of significance.
Solution: 1. H0 : µ = 5.5 ounces.
2. H1 : µ < 5.5 ounces.
3. α = 0.05.
4. Critical region: z < −1.645.
5. Computations: x̄ = 5.23 ounces, σ = 0.24 ounces, and
5.23−5.5
√
z = 0.24/
= −9.
64
6. Decision: Reject H0 , accept H1 .
Question 6
Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
2. Past experience indicated that the time for high school seniors to
complete a standardized test is a normal random variable with a mean
of 35 minutes. If a random sample of 20 high school seniors took an
average of 33.1 minutes to complete this test. Now, suppose the
standard deviation of the population is of 4.3 minutes, test the
hypothesis at the 0.05 level of significance that µ = 35 minutes
against the alternative that µ < 35 minutes.
Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
2. Past experience indicated that the time for high school seniors to
complete a standardized test is a normal random variable with a mean
of 35 minutes. If a random sample of 20 high school seniors took an
average of 33.1 minutes to complete this test. Now, suppose the
standard deviation of the population is of 4.3 minutes, test the
hypothesis at the 0.05 level of significance that µ = 35 minutes
against the alternative that µ < 35 minutes.
Solution: 1. H0 : µ = 35 ounces.
2. H1 : µ < 35 ounces.
3. α = 0.05.
4. Critical region: z < −1.645.
5. Computations: x̄ = 33.1 ounces, σ = 4.3 ounces, and
33.1−35
√
z = 4.3/
= −1.976
20
6. Decision: Reject H0 , accept H1 .
Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
3. A manufactures of sports equipment has developed a new synthetic
fishing line that he claims has a mean breaking strength of 8 kilograms
with a standard deviation of 0.5 kilogram. Test the hypothesis that
µ = 8 kilograms against the alternatives that µ 6= 8 kilograms if a
random sample of 50 lines is tested and found to have a mean
breaking strength of 7.8 kilograms. Use a 0.01 level of significance.
Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
3. A manufactures of sports equipment has developed a new synthetic
fishing line that he claims has a mean breaking strength of 8 kilograms
with a standard deviation of 0.5 kilogram. Test the hypothesis that
µ = 8 kilograms against the alternatives that µ 6= 8 kilograms if a
random sample of 50 lines is tested and found to have a mean
breaking strength of 7.8 kilograms. Use a 0.01 level of significance.
Solution: 1. H0 : µ = 8 kilograms.
2. H1 : µ 6= 8 kilograms.
3. α = 0.01.
x̄−µ
√0 .
4. Critical region: z < −2.575 and z > 2.575, where z = σ/
n
5. Computations: x̄ = 7.8 kilograms, n = 50, and hence
7.8−8
√
z = 0.5/
= −2.83
50
6. Decision: Reject H0 . Conclude that the average breaking strength
is not equal to 8 but is, in fact, less than 8 kilograms.
Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
4. A builder claims that heat pumps are installed in 70% of all homes
being constructed today in the city of Richmond, Virginia. Would you
agree with this claim if a random survey of new homes in this city
shows that 8 out of 15 had heat pumps installed? Use a 0.10 level of
significance.
Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
4. A builder claims that heat pumps are installed in 70% of all homes
being constructed today in the city of Richmond, Virginia. Would you
agree with this claim if a random survey of new homes in this city
shows that 8 out of 15 had heat pumps installed? Use a 0.10 level of
significance.
Solution: 1. H0 : p = 0.7.
2. H1 : p 6= 0.7.
3. α = 0.10.
4. Test statistic: Binomial variable X with p = 0.7 and n = 15.
5. Computations: x = 8 and np0 = 15 × 0.7 = 10.5. Therefore, by
the binomial probability table, the computed P-value is
P = 2P(X ≤ 8 when p = 0.7) = 2
8
X
b(x; 15, 0.7) = 0.2622 > 0.10.
x=0
6. Decision: Do not reject H0 . Conclude that there is insufficient
reason to doubt the builder’s claim.
Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
5. A manufacturer of car batteries claims that his batteries will last, on
average, 3 years with a variance of 1 year. If 5 of this batteries have
lifetimes of 1.9, 2.4, 3.0, 3.5 and 4.2 years, construct a 95%
confidence interval for σ 2 and decide if the manufacturer’s claim that
σ 2 = 1 is valid. Assume the population of battery lives to be
approximately normally distributed.
Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
5. A manufacturer of car batteries claims that his batteries will last, on
average, 3 years with a variance of 1 year. If 5 of this batteries have
lifetimes of 1.9, 2.4, 3.0, 3.5 and 4.2 years, construct a 95%
confidence interval for σ 2 and decide if the manufacturer’s claim that
σ 2 = 1 is valid. Assume the population of battery lives to be
approximately normally distributed.
Solution:2 Because
2
P( (n−1)S
< σ 2 < (n−1)S
) = 1 − α, n = 5, α = 0.05, then we have
χ2
χ2
α/2
1−α/2
n
S2 =
5 × 48026 − 225
1 X
(xi − x̄)2 =
= 0.815,
n−1
20
i=1
and degree of freedom ν = 4, χ20.025 = 11.143, χ21−0.025 = 0.484,
then the 95% confidence interval for σ 2 is:
(n − 1)S2 (n − 1)S2
[
, 2
] = [0.29256, 6.73554].
χ2α/2
χ1−α/2
Since 1 lies the interval, so the claim that σ = 1 is valid.
Question 1
Question 2
Question 3
Question 4
Question 5
6. A random sample of 20 students obtained a mean of x̄ = 72and a
variance of S2 = 16 on a college placement test in mathematics.
Assuming the scores to be normally distributed, construct a 98%
confidence interval for σ 2 .
Question 6
Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
6. A random sample of 20 students obtained a mean of x̄ = 72and a
variance of S2 = 16 on a college placement test in mathematics.
Assuming the scores to be normally distributed, construct a 98%
confidence interval for σ 2 .
Solution: Similarly to Question 5, we have ν = 19, S2 = 16,
χ20.01 = 36.191, χ21−0.01 = 7.633. So the 98% confidence interval for
σ 2 is:
(n − 1)S2 (n − 1)S2
[[
, 2
] = [8.39988, 39.8271].
χ2α/2
χ1−α/2