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MATH1005 Quizzes

Quiz 9: t-tables; Z-tests and t-tests for a mean
Question 1 Questions
Use tables to find an interval containing P(t6 > 1.5). Exactly one option must be correct)
a)
(0, 0.05)
b)
(0.05, 0.10)
c)
(0.10, 1)
d)
(0.25, 1)

Choice (a) is incorrect
Draw a diagram and mark in 1.5 on the x-axis. From the tables, P(t6 > 1.943) = 0.05. Now mark in 1.943 on the x-axis, and it will be obvious that P(t6 > 1.5) must be greater than 0.05.
Choice (b) is correct!
Choice (c) is incorrect
Draw a diagram and mark in 1.5 on the x-axis. From the tables, P(t6 > 1.440) = 0.10. Now mark in 1.44 on the x-axis, and it will be obvious that P(t6 > 1.5) must be smaller than 0.10.
Choice (d) is incorrect
Try again.
Let τ = X̄μ Sn , where X̄ and S are variables representing the mean and standard deviation of a sample of size n = 13 from the normal N(μ,σ2) population. From the tables, the value of c such that P(τ > c) = 0.05 is Exactly one option must be correct)
a)
c = 1.771
b)
c = 1.645
c)
c = 0.05
d)
c = 1.782

Choice (a) is incorrect
Try again. τ does not have the t13 distribution.
Choice (b) is incorrect
Try again. τ = X̄μ S13 is not normally distributed.
Choice (c) is incorrect
What is the distribution of τ = X̄μ S13 ?
Choice (d) is correct!
τ = X̄μ S13 t12 and P(t12 > 1.782) = 0.05.
Questions 3-5 use the following information.
The mean fineness μ, of yarn is expected to be greater than the standard value of 5 units. To test this claim, the factory graded 16 specimens of the yarn and found the sample average, x̄, to be 5.9 units.
What are the null and alternative hypotheses being tested? Exactly one option must be correct)
a)
H0 : μ = 5;H1 : μ > 5
b)
H0 : μ = 5;H1 : μ = 5.9
c)
H0 : x̄ = 5;H1 : x̄ > 5.9
d)
H0 : μ = 5;H1 : μ5

Choice (a) is correct!
This an upper-tail test, as the anticipated alternative hypothesis is a mean greater than 5 units.
Choice (b) is incorrect
Try again. What is the anticipated alternative?
Choice (c) is incorrect
Try again. x̄ is the observed mean.
Choice (d) is incorrect
Try again. This is not a two-tailed test.
Questions 3-5 use the following information.
The mean fineness μ, of yarn is expected to be greater than the standard value of 5 units. To test this claim, the factory graded 16 specimens of the yarn and found the sample average, x̄, to be 5.9 units.
Assume that the measurement of fineness is normally distributed with a variance of σ2 = 4.0. We can state: Exactly one option must be correct)
a)
The P-value is P(Z 1.8) = 0.0359 and there is strong evidence that the mean exceeds 5 units.
b)
The P-value is P(Z 0.45) = 0.3264 which indicates the data are consistent with a mean of 5 units.
c)
The P-value is P(Z 3.6) = 0.0002 and there is strong evidence that the mean exceeds 5 units.
d)
The P-value is P(|Z| 1.8) = 0.0718 which does not constitute strong evidence against a mean of 5 units.

Choice (a) is correct!
Since σ = 2, and this is an upper-tail test, the observed value of the Z-statistic is 5.95 216 = 1.8 with P-value = P(Z 1.8) = 0.0359.
Choice (b) is incorrect
Try again. The observed value of the Z-statistic is not 0.45.
Choice (c) is incorrect
Try again. The observed value of the Z-statistic is not 3.6.
Choice (d) is incorrect
Try again. This is not a two-tailed test.
Questions 3-5 use the following information.
The mean fineness μ, of yarn is expected to be greater than the standard value of 5 units. To test this claim, the factory graded 16 specimens of the yarn and found the sample average, x̄, to be 5.9 units. Assume that the measurement of fineness is normally distributed with unknown variance. State the distribution of the appropriate test statistic, τ, and find the P-value if the standard deviation of the sample is s = 2.4. Exactly one option must be correct)
a)
τ N(0,1) and P-value < 0.05.
b)
τ t16 and P-value < 0.05.
c)
τ t15 and P-value < 0.05.
d)
τ t15 and P-value > 0.05.

Choice (a) is incorrect
Try again. The test statistic is not normal when σ is unknown.
Choice (b) is incorrect
Try again. You have the wrong t distribution.
Choice (c) is incorrect
Try again. You may have misread the t-tables.
Choice (d) is correct!
τobs = 5.95 2.416 = 1.5 and P(t15 > 1.5) is in the interval (0.05, 0.10).
Questions 6-9 use the following information.
A manufacturing process produces components which have weight normally distributed about 60g with a standard deviation of 1.2g. A new process has been developed to cut costs. Below are the weights of a random sample of 15 components produced by the new process:
60.359.862.560.861.659.961.259.4
61.0 58.962.160.759.160.263.1
The management wishes to know if the new process results in a different mean weight. Which of the following are the most appropriate null and alternative hypotheses? Exactly one option must be correct)
a)
H0 : μ = 60; H1 : μ > 60
b)
H0 : μ = 60; H1 : μ60
c)
H0 : μ = 0; H1 : μ > 0
d)
H0 : μ = 0; H1 : μ0

Choice (a) is incorrect
Try again. There is no supporting information to anticipate an increase in weight.
Choice (b) is correct!
We perform a two-sided test as there is no anticipated direction for the alternative hypothesis.
Choice (c) is incorrect
Try again. Think about the value of μ and what sort of test we need to perform.
Choice (d) is incorrect
Try again. Think about the value of μ.
Questions 6-9 use the following information.
A manufacturing process produces components which have weight normally distributed about 60g with a standard deviation of 1.2g. A new process has been developed to cut costs. Below are the weights of a random sample of 15 components produced by the new process:
60.359.862.560.861.659.961.259.4
61.0 58.962.160.759.160.263.1
The management wishes to know if the new process results in a different mean weight. Assuming the standard deviation is unchanged for the new process, which of the following is the most appropriate test statistic, τ? Exactly one option must be correct)
a)
τ = X N(60,1.22)
b)
τ = X¯60 1.215 N(0,1)
c)
τ = X¯60 1.2 N(0,1)
d)
τ = X¯μ Sn t14

Choice (a) is incorrect
Try again, we have a sample of 15 measurements.
Choice (b) is correct!
Since σ is known, we know that X¯ N(μ, σ2 n ) N(60, 1.22 15 ). Standardising X¯ gives the test statistic, τ = X¯60 1.215 N(0,1)
Choice (c) is incorrect
You have not standardised X¯ correctly.
Choice (d) is incorrect
Try again, σ is known.
Questions 6 - 9 use the following information. A manufacturing process produces components which have weight normally distributed about 60g with a standard deviation of 1.2g. A new process has been developed to cut costs. Below are the weights of a random sample of 15 components produced by the new process:
60.359.862.560.861.659.961.259.4
61.0 58.962.160.759.160.263.1
The management wishes to know if the new process results in a different mean weight. State the P-value of the Z-test. Exactly one option must be correct)
a)
P-value = 0.011.
b)
P-value = 0.989.
c)
P-value = 0.022.
d)
P-value = 0.10

Choice (a) is incorrect
Try again, remember we are performing a two-sided test.
Choice (b) is incorrect
Try again.
Choice (c) is correct!
Large and small values of X¯ argue against H0.
The observed value of X¯ is 60.71.
P(|Z| > 60.7160 1.215 ) = P(|Z| > 2.29) = 2(1 Φ(2.29)) = 2(1 0.9890) = 0.022.
Choice (d) is incorrect
Check that the sample mean is 60.71 and recalculate the observed value of the Z-statistic.
Questions 6-9 use the following information.
A manufacturing process produces components which have weight normally distributed about 60g with a standard deviation of 1.2g. A new process has been developed to cut costs. Below are the weights of a random sample of 15 components produced by the new process:
60.359.862.560.861.659.961.259.4
61.0 58.962.160.759.160.263.1
The management wishes to know if the new process results in a different mean weight. If we are not prepared to assume that the standard deviation is 1.2g for the new process, find τobs, the observed value of the appropriate t-statistic, and provide a P-value. Exactly one option must be correct)
a)
τobs = 60.71, P-value = 0.
b)
τobs = 1.79, P-value =P(|t14| > 1.79) > 0.05.
c)
τobs = 2.22, P-value = P(t14 > 2.22) < 0.025.
d)
τobs = 2.22, P-value = P(|t14| > 2.22) < 0.05.

Choice (a) is incorrect
Try again. You need to calculate both x̄ and s and substitute into τobs = x̄60 s15 .
Choice (b) is incorrect
Try again. Check your calculation of the sample standard deviation, s = 1.24, and substitute x̄ and s into τobs = x̄60 s15 .
Choice (c) is incorrect
Try again. This is not an upper-tail test.
Choice (d) is correct!
Since σ is unknown, τobs = x¯μ sn = 60.7160 1.2415 = 2.22
and P-value = P(|t14| > 2.22) = 2P(t14 > 2.22) < 0.05.
The breaking strengths of a certain brand of marine rope follow a normal distribution, with unknown variance. To test whether the mean breaking strength is 8 units, 28 lengths of the rope were tested. From this sample, the observed value of the test statistic was τobs = 2.05. The appropriate t distribution and approximate P-value are: Exactly one option must be correct)
a)
t27 , P-value 0.05.
b)
t27 , P-value 0.025.
c)
t28 , P-value 0.05.
d)
t8 , P-value 0.025.

Choice (a) is correct!
The P-value is P(|t27| > 2.05) = 2P(t27 > 2.05) 2 × 0.025.
Choice (b) is incorrect
Try again. Notice that we are conducting a two-sided test.
Choice (c) is incorrect
Try again. The test statistic is not t28.
Choice (d) is incorrect
Try again. The test statistic is not t8.