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Quiz 9: t-tables; Z-tests and t-tests for a mean

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Question 1

 
 
Use tables to find an interval containing P(t6 > 1.5).
a) (0, 0.05)   b) (0.05, 0.10)
c) (0.10, 1)   d) (0.25, 1)

 

Not correct. Choice (a) is false.
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.
Your answer is correct.
Not correct. Choice (c) is false.
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.
Not correct. Choice (d) is false.
Try again.
 

Question 2

 
 
Let τ = ¯ XS∕-√μn, where ¯X 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
a) c = 1.771   b) c = 1.645
c) c = 0.05   d) c = 1.782

 

Not correct. Choice (a) is false.
Try again. τ does not have the t13 distribution.
Not correct. Choice (b) is false.
Try again. τ = -¯X-√μ- S∕ 13 is not normally distributed.
Not correct. Choice (c) is false.
What is the distribution of τ = -¯X-μ- S∕√13?
Your answer is correct.
τ = -¯X--μ S∕√13 ~ t12 and P(t12 > 1.782) = 0.05.
 

Question 3

 
 
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, ¯x, to be 5.9 units.
What are the null and alternative hypotheses being tested?
a) H0 : μ = 5; H1 : μ > 5   b) H0 : μ = 5; H1 : μ = 5.9
c) H0 : ¯x = 5; H1 : ¯x > 5.9   d) H0 : μ = 5; H1 : μ⁄=5

 

Your answer is correct.
This an upper-tail test, as the anticipated alternative hypothesis is a mean greater than 5 units.
Not correct. Choice (b) is false.
Try again. What is the anticipated alternative?
Not correct. Choice (c) is false.
Try again. ¯x is the observed mean.
Not correct. Choice (d) is false.
Try again. This is not a two-tailed test.
 

Question 4

 
 
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, ¯x, to be 5.9 units.
Assume that the measurement of fineness is normally distributed with a variance of σ2 = 4.0. We can state:
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.

 

Your answer is correct.
Since σ = 2, and this is an upper-tail test, the observed value of the Z-statistic is 52.9∕√-156- = 1.8 with P-value = P(Z ≥ 1.8) = 0.0359.
Not correct. Choice (b) is false.
Try again. The observed value of the Z-statistic is not 0.45.
Not correct. Choice (c) is false.
Try again. The observed value of the Z-statistic is not 3.6.
Not correct. Choice (d) is false.
Try again. This is not a two-tailed test.
 

Question 5

 
 
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, ¯x, 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.
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.

 

Not correct. Choice (a) is false.
Try again. The test statistic is not normal when σ is unknown.
Not correct. Choice (b) is false.
Try again. You have the wrong t distribution.
Not correct. Choice (c) is false.
Try again. You may have misread the t-tables.
Your answer is correct.
τobs = 25.4.9∕-√516 = 1.5 and P(t15 > 1.5) is in the interval (0.05, 0.10).
 

Question 6

 
 
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:
60.3 59.8 62.5 60.8 61.6 59.9 61.2 59.4
61.0 58.9 62.1 60.7 59.1 60.2 63.1 
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?
a) H : μ = 60 0 ; H : μ > 60 1   b) H : μ = 60 0 ; H : μ ⁄= 60 1
c) H : μ = 0 0 ; H : μ > 0 1   d) H : μ = 0 0 ; H : μ ⁄= 0 1

 

Not correct. Choice (a) is false.
Try again. There is no supporting information to anticipate an increase in weight.
Your answer is correct.
We perform a two-sided test as there is no anticipated direction for the alternative hypothesis.
Not correct. Choice (c) is false.
Try again. Think about the value of μ and what sort of test we need to perform.
Not correct. Choice (d) is false.
Try again. Think about the value of μ .
 

Question 7

 
 
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:
60.3 59.8 62.5 60.8 61.6 59.9 61.2 59.4
61.0 58.9 62.1 60.7 59.1 60.2 63.1 
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?
a) τ = X ~ N (60,1.22)   b) -- τ = 1X.2-∕6√015 ~ N (0,1)
c) X--60- τ = 1.2 ~ N (0,1)   d) -- τ =-X-√μ ~ t14 S∕ n

 

Not correct. Choice (a) is false.
Try again, we have a sample of 15 measurements.
Your answer is correct.
Since σ is known, we know that X-~ N (μ, σ2) ~ N (60, 1.22) n 15 . Standardising X gives the test statistic, τ = -X-60- 1.2∕√15 ~N(0,1)
Not correct. Choice (c) is false.
You have not standardised X correctly.
Not correct. Choice (d) is false.
Try again, σ is known.
 

Question 8

 
 
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:
60.3 59.8 62.5 60.8 61.6 59.9 61.2 59.4
61.0 58.9 62.1 60.7 59.1 60.2 63.1 
The management wishes to know if the new process results in a different mean weight. State the P-value of the Z-test.
a) P-value = 0.011.   b) P-value = 0.989.
c) P-value = 0.022.   d) P-value = 0.10

 

Not correct. Choice (a) is false.
Try again, remember we are performing a two-sided test.
Not correct. Choice (b) is false.
Try again.
Your answer is correct.
Large and small values of -- X argue against H0 .
The observed value of -- X is 60.71.
60.71-√60- P (|Z| > 1.2∕ 15 ) = P (|Z | > 2.29) = 2(1- Φ (2.29)) = 2(1- 0.9890) = 0.022.
Not correct. Choice (d) is false.
Check that the sample mean is 60.71 and recalculate the observed value of the Z-statistic.
 

Question 9

 
 
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:
60.3 59.8 62.5 60.8 61.6 59.9 61.2 59.4
61.0 58.9 62.1 60.7 59.1 60.2 63.1 
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.
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.

 

Not correct. Choice (a) is false.
Try again. You need to calculate both ¯x and s and substitute into τobs = s¯x-∕√6015.
Not correct. Choice (b) is false.
Try again. Check your calculation of the sample standard deviation, s = 1.24, and substitute ¯x and s into τobs = s¯x∕-√6105.
Not correct. Choice (c) is false.
Try again. This is not an upper-tail test.
Your answer is correct.
Since σ is unknown, τobs = -x-√μ s∕ n = 60.71-√60 1.24∕ 15 = 2.22
and P-value = P(|t14| > 2.22) = 2P(t14 > 2.22) < 0.05.
 

Question 10

 
 
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:
a) t27 , P-value ≈ 0.05.   b) t27 , P-value ≈ 0.025.
c) t28 , P-value ≈ 0.05.   d) t8 , P-value ≈ 0.025.

 

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