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Quiz 10: Paired and two-sample tests for means

Last unanswered question  Question  Next unanswered question
 

Question 1

 
 
Questions 1 - 3 use the following information.
Eleven patients were weighed before and after a diet which was designed to help patients reduce weight. Eight of these patients did lose weight but three patients had a weight increase.
To test whether the diet is successful, state the appropriate null and alternative hypotheses in terms of p+ (the probability that the diet increases weight).
a) H0 : p+ = 0; H1 : p+ < 0.   b) H0 : p+ = 0.5; H1 : p+ < 0.5.
c) H0 : p+ = 0.5; H1 : p+⁄=0.5.   d) H0 : p+ = 0.5; H1 : p+ > 0.5.

 

Not correct. Choice (a) is false.
A zero change in weight is not the same as p+ = 0.
Your answer is correct.
We are looking for evidence of a decrease in weight on average, so we expect the proportion with a positive weight change to be below 0.5.
Not correct. Choice (c) is false.
This is not a two-sided test. What is the anticipated alternative?
Not correct. Choice (d) is false.
What is the anticipated alternative in terms of p+?
 

Question 2

 
 
Questions 1 - 3 use the following information.
Eleven patients were weighed before and after a diet which was designed to help patients reduce weight. Eight of these patients did lose weight but three patients had a weight increase.
Is there evidence that the diet is sucessful? Provide a P-value and state your findings.
a) P-value = 0.1133. There is insufficient evidence to claim that the diet is successful.
b) P-value = 0.9673. The data do not provide evidence that the diet is successful.
c) P-value = 0.0302. There is very strong evidence that the diet is successful.
d) P-value = 0.1133. The probability that the diet is successful is 0.1133.

 

Your answer is correct.
Write τ to represent the number of weight increases. Under H0, τ ~B(11,0.5). Since this is a lower tail test, and since the observed value of the test statistic is τobs = 3, the P-value is P(τ ≤ 3) = 0.1133.
Not correct. Choice (b) is false.
The P-value calculates the probability of a result at least as extreme as the observed result of 3 positive weight gains.
Not correct. Choice (c) is false.
The P-value calculates the probability of a result at least as extreme as the observed result of 3 positive weight gains.
Not correct. Choice (d) is false.
Revise the interpretation of a P-value.
 

Question 3

 
 
Questions 1 - 3 use the following information.
Eleven patients were weighed before and after a diet which was designed to help patients reduce weight. Eight of these patients did lose weight but three patients had a weight increase.
The observed sample mean (in kg) and sample variance (in kg2) of the eleven changes in weight (after - before) were ¯x = -0.6 and s2 = 1.44. Writing μd to represent the mean change in weight (after - before), provide a P-value for testing H0 : μd = 0 vs H1 : μd < 0 assuming that the weight changes follow a normal distribution. (Round τobs to 2dp.)
a) P-value = P(Z ≤-1.66) ≈ 0.0485   b) P-value= P(t11 < -1.66) > 0.05.
c) P-value = P(t10 < -1.38) > 0.05   d) P-value = P(t10 < -1.66) > 0.05

 

Not correct. Choice (a) is false.
The population standard deviation of the differences is unknown, so you should not use the Z-test.
Not correct. Choice (b) is false.
You have used the wrong t-distribution.
Not correct. Choice (c) is false.
Have you used s2 by mistake, instead of s in the calculation of the observed test statistic?
Your answer is correct.
τobs = 1-.2∕0.√611 = -1.66, and P-value = P(t10 ≤-1.66) which is in the interval (0.05, 0.1).
 

Question 4

 
 
Twelve athletes were each given two different exercise routines, and their heartbeats were monitored closely. The aim of the study was to test for a difference in the effect of the routines on heartbeat rate. The twelve differences were (in appropriate units):
di: 1.5 2.3  4.7  6.1  -1.2  2.6  1.5  -0.4  -2.7  1.9  1.1  -1.5.
(Elementary calculations give: ∑ idi = 15.9 and ∑ idi2 = 91.81.) Assuming the differences are normally distributed, provide bounds for the P-value from the appropriate t-test.
a) (0.025, 0.05)   b) (0.05, 0.10)
c) (0.10, 0.20)   d) (0.01, 0.025)

 

Not correct. Choice (a) is false.
This is a two-sided test.
Your answer is correct.
The sample mean difference is 1.325 and the sample standard deviation of the differences is 2.536, so τobs = --1.325√-- 2.536∕ 12 = 1.809. The test is two-sided, so the P-value = P(∣t11∣ > 1.809) = 2P(t11 > 1.809), which is in the interval (0.05, 0.10).
Not correct. Choice (c) is false.
Check your calculation of the sample mean and standard deviation.
Not correct. Choice (d) is false.
Check your calculation of the sample mean and standard deviation.
 

Question 5

 
 
A panel of psychiatrists is asked to compare the condition of 65 patients before and after intermittent psychotherapy, which is expected to improve the condition in general. 40 patients showed an improvement, 24 were worse and one was considered unchanged. Use the sign test with the normal approximation to find the P-value.
a) P(X ≥ 40) = 0   b) P(Z ≥ 40) = 0
c) P(Z ≥ 2) = 0.0228   d) P(Z > 1.875) ≈ 0.03

 

Not correct. Choice (a) is false.
You need the approximating normal variable to find P(X ≥ 40).
Not correct. Choice (b) is false.
You need to use the approximating normal variable and standardise it.
Not correct. Choice (c) is false.
Perhaps you have forgotten the continuity correction.
Your answer is correct.
Ignore the one unchanged patient, and let X represent the number improved out of 64. Under the hypothesis of no improvement, X ~B(64,0.5) and the P-value is P(X ≥ 40), since this is an upper-tail test. Using a normal approximation with correction for continuity, P-value ≈ P(Y ≥ 39.5), where Y ~N(32,16) is the approximating normal variable. Thus, P-value ≈ P(Z ≥39.√5-32 16) = P(Z ≥ 1.875) ≈ 0.03.
 

Question 6

 
 
A test of H0 : μx = μy vs H1 : μx⁄=μy based on independent samples of sizes nx = 12 and ny = 10, yields a two-sample t-statistic with observed value, τobs = 1.95. The P-value is in the interval:
a) (0, 0.01)   b) (0.01,0.025)
c) (0.025,0.05)   d) (0.05,0.10)

 

Not correct. Choice (a) is false.
Check the tables again.
Not correct. Choice (b) is false.
Check the tables again.
Not correct. Choice (c) is false.
Recall that this is a two-tail test.
Your answer is correct.
P-value = P(∣t20∣≥ 1.95) = 2P(t20 ≥ 1.95). Since P(t20 ≥ 1.95) is in the interval (0.025, 0.05), the P-value is in the interval (0.05, 0.1).
 

Question 7

 
 
Questions 7 - 9 use the following information.
In an attempt to compare the durability of two different materials (X and Y), 10 pieces of type X and 14 pieces of type Y were used. From these samples we calculate the statistics: ¯x = 129.44, ¯y = 122.65, sx = 9.15, sy = 11.02.
The pooled estimate of the standard deviation, sp, in the two-sample test statistic is:
a) 10.30   b) 105.72
c) 106.01   d) 10.28

 

Your answer is correct.
sp = ∘-9×9.152+13×11.022- 10+14-2 = 10.30.
Not correct. Choice (b) is false.
Use the formula for sp2 and then find its square root.
Not correct. Choice (c) is false.
This is the pooled estimate of the variance, not of the standard deviation.
Not correct. Choice (d) is false.
Check the formula for sp2 and then square root.
 

Question 8

 
 
Questions 7 - 9 use the following information.
In an attempt to compare the durability of two different materials (X and Y), 10 pieces of type X and 14 pieces of type Y were used. From these samples we calculate the statistics: ¯x = 129.44, ¯y = 122.65, sx = 9.15, sy = 11.02.
To test for equal means, the appropriate t-distribution is
a) t24   b) t22
c) t10 - t14.   d) t23

 

Not correct. Choice (a) is false.
You do not simply add the sample sizes 10+14 to give the df of the t-distribution.
Your answer is correct.
The appropriate t-variable has df = 10 + 14 - 2 = 22.
Not correct. Choice (c) is false.
Try again.
Not correct. Choice (d) is false.
With two independent samples of sizes m and n, df = m + n - 2.
 

Question 9

 
 
Questions 7 - 9 use the following information.
In an attempt to compare the durability of two different materials (X and Y), 10 pieces of type X and 14 pieces of type Y were used. From these samples we calculate the statistics: ¯x = 129.44, ¯y = 122.65, sx = 9.15, sy = 11.02.
Assuming the two populations sampled are normal with equal variance, find the P-value.
a) (0.05, 0.10)   b) (0.1, 0.2)
c) (0.01, 0.025)   d) (0.02, 0.05).

 

Not correct. Choice (a) is false.
This is a two-sided test.
Your answer is correct.
sp = ∘ -------------- 9×9.152+13×11.022 10+14-2 = 10.30, so τobs = 129.44√-122.65 10.30 110+ 114- = 1.52.
P-value = P(∣t22∣≥ 1.52) = 2P(t22 ≥ 1.52), with bounds (0.1, 0.2), since P(t22 ≥ 1.52) is in the range (0.05, 0.10).
Not correct. Choice (c) is false.
Try again. Check that sp = 10.30 and that τobs = 1.52. Use a two-sided test.
Not correct. Choice (d) is false.
Try again. Check that sp = 10.30 and that τobs = 1.52.
 

Question 10

 
 
Two methods A and B, were used to find the latent heat of fusion of ice. 13 measurements were made with method A and 8 measurements were made with method B with statistics: ¯xA = 80.02, ¯xB = 79.98, sA = 0.024 and sB = 0.031. To test equal means assuming normality and equal variances, the observed test statistic (to 1dp) and the P-value are:
a) τobs = 1.2; P-value = P(∣t19∣ > 1.2) > 0.2
b) τobs = 3.3; P-value = P(t21 > 3.3) < 0.005
c) τobs = 3.3; P-value = P(t19 > 3.3) < 0.005
d) τobs = 3.3; P-value = P(∣t19∣ > 3.3) < 0.01

 

Not correct. Choice (a) is false.
Recalculate your test statistic.
Not correct. Choice (b) is false.
You have used a one-tail test and have also used the wrong t-distribution.
Not correct. Choice (c) is false.
This should be a two-tail test
Your answer is correct.
sp = ∘ --------------- 12×0.0242+7×0.0312 13+8-2 = 0.02679, so τobs = --80.02-79.98-- 0.02679√ 113+18 = 3.3,
with P-value = P(∣t19∣ > 3.3) = 2P(t19 > 3.3), which has bounds (0.002, 0.01).