Weeks 12 (nothing for Week 11) Flashcards
We should reject a null hypothesis if it is not contained in the confidence interval for a sample.
a) True
b) False
a) True
Explanation … The confidence interval for a sample contains all hypotheses that would be accepted based on that sample. If a null hypothesis (i.e., the hypothesis that the statistic of interest has value zero) is contained in the confidence interval it should be accepted and if it is not contained in the interval it should be rejected.
“CI.95 = 3 ± 2.5” and “CI.95 = [-2.5, +2.5]” are 2 ways of displaying the same confidence interval.
a) true
b) false
b) False
Explanation … You can display a confidence interval by stating where the centre is and how far to go up and down to reach the upper and lower confidence limits, or you can just give the upper and lower limits directly. In the latter style, the centre of the confidence interval uis assumed to be exactly half way between the limits. In the question, the centre of one confidence interval is directly given as “3”, but the centre of the other confidence interval is clearly 0 (half way between -2.5 and 2.5). The confidence intervals are therefore not the same.
A sample with a confidence interval on the mean given by CI.95 = [–12.3, +1.3] could reasonably have come from a population with a mean of 0.
a) true
b) false
a) True
Explanation … “Reasonable”, in this context, means that the sample mean (which must be at half way between the upper and lower confidence limits at M = 6.8) is not significantly different from 0. But 0 is inside this confidence interval. Hence it is one of the values for the population mean that is, indeed, not significantly different from the observed sample mean. Thus the sample could reasonably have come from a population with a mean of 0.
The smaller the confidence interval the greater the power.
a) True
b) False
a) True
Explanation … Power is your ability to reject a false hypothesis. But the smaller a confidence interval is, the more hypotheses you end up rejecting. Thus a small confidence interval is associated with greater power than a large confidence interval.
A sample’s 99% confidence interval is larger than its 95% confidence interval.
a) true
b) false
Explanation … The upper and lower limits for the confidence interval on the mean are given by the formula. We can use this formula to calculate the 95% confidence limits or the 99% limits for a given sample. If we deal with the same sample all the time then the values of M and sM will not change and the only thing that changes is the value of the critical point, t alpha/2. This critical point tells you how many standard errors you have to travel out into the tails of a t-distribution before encountering the desired rejection region (the notation “alpha/2” indicates that we are dealing with a 2-tailed rejection region). Of course this critical point will depend on your choice of alpha and since t. 01/2 is larger than t. 05/2, the 99% confidence interval is wider than the 95% confidence interval. An intuitive way of approaching this same question is that since a 99% confidence interval based on some sample has to capture the true population mean more often that the corresponding 95% confidence interval, the 99% interval must be a vaguer, wider, estimate.
6) The smaller the sample size, the smaller the confidence interval.
a) true
b) false
b) False
Explanation … The confidence interval of the mean tells where we think the true population mean lies. When we use a large sample to calculate the confidence interval we have a lot of information to work with and so our prediction of the location of the mean should be quite good. In that case we would expect the confidence interval to be small. If we use a small sample, the information regarding the true location of the mean will not be as good and so the confidence interval will be large (i.e., the true mean is in there somewhere but we don’t know exactly where).
What is the 95% confidence interval on the mean for a single sample with sX = 5 and N = 100?
a) M ± 0.98
b) M ± 1.96
c) M ± 10
d) M ± 1.28
a) M ± 0.98
Explanation … First, note that in a single sample design the number of observations (n) per sample is the same as the number of observations (N) in the whole design. In other words, n = N. This is not true of a 2-sample design. From the information given in the problem, then, we know not only that N = 200, but also that n = 200.
For a large (n > 30) sample,
CI.05 = M +- 1.06 (Sx/square root of n).
Plugging in values from the question yields
CI.05 = M+-1.96 (5/square root of 100) = M+- 1.96 (5/10) = M+- 0.98
What is the 95% confidence interval for a paired-samples design with MD, = 1, sD = 9, and N = 162?
a) [-0.96, 2.96]
b) [0.96, 2.96]
c) [1.96, 1.96]
d) [-6.00, 8.00]
a) [-0.96, 2.96]
Explanation … For a paired-samples design,
CI. 05= mD +- t.05/2 (Sd square root (1/n)).
.
You can see from this that the CI is centred on the mean of pair differences, MD. This already means that answers b) and c) are incorrect since we know from the question that MD = 1 and neither of these confidence intervals are centred on 1. Answers a) and d) however, are still in consideration. The sample size in this paired samples design is n = N/2 = 162/2 = 81 and, since this means that n > 30, we can assume that the value of the critical point to use is t.05/2 = 1.96. Putting all the values in the formula for the confidence interval gives …
CI.05 = 1 +- 1.96(9(square root of 1/81)) = 1 +- 1.96 (9/square root of 81) = [-0.96, 2.96]
For a single-sample design, what is the sample mean corresponding to an 80% confidence interval of [1.45, 2.55]?
a) 0.56
b) 1.10
c) 1.96
d) 2.00
d) 2.00
Explanation … In this question, you can’t construct the entire confidence interval unless you know the value of the required critical point t.80/2. If you really wanted to know this critical point you would look it up in a table or use some sort of software. However, you actually don’t need to construct the entire confidence interval to answer the question. All we are asked for is the value of the sample mean, and therefore all we have to do is find the centre of the interval. Since the limits of the confidence interval are at 1.45 and 2.55 it is obvious that 2.00 is the right answer.
Suppose the 95% confidence interval for a single sample with N = 144 is –1.96 to +1.96. What is the standard deviation of the sample?
a) 2 x 1.96
b) 12
c) 1
d) 2/12
b) 12
Explanation … The confidence interval in this problem can be diagrammed in the following way
line at -1.96 to 1.96, M at 0
The confidence interval is centred on zero (because that is half-way between -1.96 and +1.96) and has 2 arms of equal lengths. It is obvious from the diagram that each arm has length 1.96. This is all obvious just from diagramming the confidence interval so it is always a good idea to draw such a diagram when answering a question like this.
There is a completely separate way of determining the lengths of the arms of this confidence interval. It is by using the formula for a confidence interval. For a single sample situations such as in this problem the formula (for = .05 and a 2-tailed test) is …
CI0.95 = M ± t.05/2 (Sx(square root of 1/n))
This formula shows the centre of the confidence interval (M) and how long the arms are. The part that shows the length of the arms is circled in red. But from diagramming the confidence interval we already know that it is centred on 0 and the arm length is 1.96, i.e., ….
Now, just looking at the part of this equation circled in red you should be able to see that following relationship holds for the length of the arms ….
1.96 = t.05/2 (Sx(square root of 1/n))
We can now start to fill in some of the values in this equation. We know from the problem we know that N = 144. But in a single-sample experimental design the number of observations overall (N) is the same as the number of observations (n) per sample, so n = 144. Moreover, this allows us to determine the value of the 2-tail critical point (t.05/2) because it means we have more than 30 degrees of freedom and for so many degrees of freedom we always have t.05/2 = 1.96.
Plugging n = 144 and t.05/2 = 1.96 into the equation gives ….
1.96 = 1.96 s_x √(1⁄144)
= 1.96 s_x/12
This is now an equation between 2 things. On the left of the equation is what we first understood, from looking at the initial diagram of the confidence interval, to be length of the arms. On the right of the equation is the arm length as derived from the formula for a confidence interval. These 2 sides of the equation need to be equal to each other. Remember that the whole point of this question is to find sx and now we can do that by finding the value of sx that makes the equation true.
Writing out the latest version of the equation one last time …
1.96 = 1.96 s_x/12
… it should now be obvious that we need sx = 12 for the 2 sides to be equal. Even if it isn’t obvious, there are only 4 possible answers offered in this multiple-choice question and trying those values of sx one by one will show you that only sx = 12 works
Suppose the 95% confidence interval for a balanced 2-sample design with n = 288 is –1.96 to +1.96. What is the value of spooled for the 2 samples?
a) 2 x 1.96
b) 12
c) 1
d) 2/12
b) 12
Explanation … We can undertake the same general procedure as for the pervious question but this time adapted for a 2-sample design. First, diagram the confidence interval …
Once again we realize from the diagram that each arm of the confidence interval has length 1.96. Now write out the formula for the confidence interval, this time for a balanced 2-sample design, and realize that it just describes where the centre is and how long the arms are…
CI0.95 = (M1-M2) ± t_(.05⁄2) s_pooled √(2⁄n)
Concentrating on the part circled in red you should be able to see that following relationship holds for the length of the arms ….
` 1.96 = t_(.05⁄2) s_pooled √(2⁄n)
Now fill in some of the values on the right based on what you have been told in the problem. In a balanced 2-sample design “n” is the number of observations per sample and we are told that n = 288. The overall degrees of freedom in a 2-sample design like this is N – 2 where “N” is the overall number of observations in the entire study. Here, N = 2n = 576, so df = 574. That is many more degrees of freedom that 30 so we must have t.05/2 = 1.96.
Putting all the information in the equation gives…
1.96 = 1.96 s_pooled √(2⁄288)
= 1.96 s_pooled/12
Once again the only way to make the 2 sides of this equation equal to each other is if spooled = 12 and this is the answer we are seeking.
Suppose the 95% confidence interval for a balanced 2-sample design with N = 576 is –1.96 to +1.96. What is the value of spooled for the 2 samples?
a) 2 x 1.96
b) 12
c) 1
d) 2/12
Explanation … This is the same question with the same answer as before. The only difference is that the overall number of observations (N) in the design is specified instead of the number of observations (n) per sample. In a balanced 2-sample design, n = N/2 and since we are told here that N = 576, we have n = 576/2 = 288. This makes it plain that the question is the same as before.
