· Q 1. Where can I obtain a general overview of confidence intervals?
A. You can consider the contents list on the left of the page for the HyperStat Chapter on Confidence Intervals. Also, the Students 4 Best Evidence page Confidence intervals should be reported is a good motivational tool for calculating confidence intervals.
· Q 2. I have calculated the mean for a sample of body weights (recorded as percentage of ideal body weight) and I wish to obtain a 95% confidence interval for the population mean. What sample size do I require and what calculations do I need to perform?
A. If you do not know the population standard deviation, you will need to use the sample standard deviation to calculate your confidence interval. All relevant calculations (including that for sample size), together with instructions for using SPSS are provided in the resource
Inference About a Mean: Estimation.
· Q 2. Seventy-five people attended a picnic. Of these, 45 contracted food poisoning. I am interested in comparing incidence between those who did and did not eat the vanilla ice-cream. How may I obtain a 95% confidence interval for the difference in risk between these two groups?
A. Have a look at the resource Cross-Tabulated Counts and Independent Proportions , where you will also find information on an appropriate hypothesis test for testing for a significant difference in risks.
· Q 4. I am conducting a study into the outcomes of pregnancies for women who have increased nuchal translucency recorded on their antenatal booking scan. For these women, I would like to obtain a confidence interval (CI) for the proportion who go on to have a fetal abnormality. Could you please advise on a convenient way to obtain the CI and on its interpretation.
A. You should find the statistical package Minitab very helpful in this respect, as it allows you to enter summary data and get your answer very quickly and conveniently. The confidence interval you are looking for is also presented in the output in a suitable way for writing up in your report, although you may well wish to provide the lower and upper limits to less decimal places! Relevant instructions are provided below.
- Go the menu Stat and select the sequence of commands Basic Statistics –> 1 Proportion…
- Within the resultant dialogue box, choose the menu item Summarized Data under the first drop-down menu.
- Suppose that 34/96 women with recorded increased nuchal translucency go on to have a fetal abnormality. The number of trials in each case is the group size (96). The number of events is the number of cases for which the event you are considering (here, fetal abnormality) occurred (34). Enter the data accordingly and check using the button Options that:
* the level of confidence selected is 95.0%,
* ‘Proportion is not equal to hypothesized proportion’ is selected under ‘Alternative hypothesis’.
Under ‘Method’,
* the appropriate method has been ticked.
To assist you in choosing the appropriate method (normal approximation or Exact) from the drop-down Menu, please refer to the resource Binomial Confidence Intervals, where you can obtain a simple rule of thumb and compare the conditions under which the calculations for each method are possible.
In case you also wish to perform the binomial test (see below), in the first of the two dialogues boxes you have opened, you can also tick the box ‘Perform hypothesis test’ and choose 0.5 for the Hypothesized proportion. This choice of proportion is consistent with the null hypothesis for women in the study population with nuchal translucency, that the proportion who go on to experience a fetal abormality is equal to the proportion who do not go on to experience a fetal abnormality.
- Your output should look like the content which has been highlighted in red below:
Test and CI for One Proportion
Sample X N Sample p 95% CI Z-Value P-Value 1 34 96 0.354167 (0.258496, 0.449837) -2.86 0.004 Using the normal approximation.
- When presenting the results in your report, you would note that 34/96 (35.4%) of women with increased nuchal translucency recorded went on to have a fetal abnormality (95% CI (0.258, 0.450)). Note that the CI tells us in this case that we are 95% certain that the true (or, population) proportion lies between 25.8% and 45.0% (to 1 d.p.).
- Notice that the above confidence interval is quite wide. You should also mention therefore that the accuracy of the sample estimate for the proportion could be improved by increasing the sample size. Increasing the sample size will not only change the value of the sample estimate of the proportion but also decrease the width of the 95% CI for that proportion. In this sense, increasing the sample size, decreases the margin of error which you can associate with your sample proportion as an estimate of the true (or, population) proportion.
You may also wish to calculate separate CIs for chromosomal and structural abnormalities separately using a similar procedure.
Incidentally, the p-value in your output corresponds with that obtained for the binomial test (also known within Minitab as the one-sample proportion test). This test can prove useful where you wish to compare a study proportion (e.g. proportion of infections) against a fixed target proportion or a population proportion from a prior study.
Here, the null hypothesis to be tested is typically of the sort:
current population proportion = target population proportion.
Whilst the test is not particularly useful in the above clinical scenario, suppose that you wished to know whether of those patients with low bone mineral density, there was a significant difference between those who were and were not prescribed anti-resorptive agents. Here, our null hypothesis is that the proportion of patients with low bone mineral density who were prescribed anti-resorptive agents is equal to those with low bone mineral density who were not prescribed anti-resorptive agents and is therefore 0.5 (the target proportion).
In this particular case, the p-value is 0.004 and thus there is a significant difference in the proportions of patients with low bone mineral density who are and are not prescribed anti-resorptive agents. In fact, since our sample proportion is 35.4%, we can now say that for a sample size of 96, at the 5% significance level, there is sufficient evidence that the proportion of persons with low bone mineral density who are prescribed anti-resorptive agents is likely to be less than those who are not (p = 0.004). This is quite an important result; not only is it surprising to find that of those patients indicated for treatment, less patients are prescribed the treatment than those who are not but we are able to see that this difference is statistically significant. As in the second of our two clinical scenarios, it is useful to take into consideration both the clinical and statistical significance of results when drawing our conclusions.
Note, however, that if your target proportion is not 0.5, you will need to change the ‘Hypothesized proportion’ in the corresponding Minitab dialogue box ‘One-Sample proportion’ to your chosen value (e.g. 0.64) before running the test.
· Q 5. I am conducting a study into the sexual health needs of young people (aged 16 to 35) who have a learning disability. One of my objectives is to compare the proportions of males and females who obtained the correct answer for a question on sexual risk. How may I obtain a confidence interval for the difference between these proportions and how should I interpret it in my write-up?
- You should find the statistical package Minitab very helpful in this respect, as it allows you to enter summary data and get your answer very quickly and conveniently. The confidence interval you are looking for is also presented in the output in a suitable way for writing up in your report, although you may well wish to provide the lower and upper limits to less decimal places! Relevant instructions are provided below.
- Go the menu Stat and select the sequence of commands Basic Statistics –> 2 Proportions
- Within the resultant dialogue box click the button Summarized Data.
- Suppose that 34/40 females 42/45 males gave the correct answer. The number of trials in each case is the group size (40 for females and 45 for males). The number of events is the number of cases for which the event you are considering (here, getting the correct answer) occurred (34 for females and 42 for males). Enter the data accordingly and check using the button Options that the level of confidence selected is 95.0% and that ‘not equal’ has been selected under ‘Alternative:’.
- Your output should look like this:
Difference = p (1) – p (2)
Estimate for difference: -0.0833333
95% CI for difference: (-0.215834, 0.0491668)
Test for difference = 0 (vs not = 0): Z = -1.23 P-Value = 0.218
* NOTE * The normal approximation may be inaccurate for small samples.
Fisher’s exact test: P-Value = 0.295
- Just focus on the first three lines, as we are not discussing hypothesis tests at present.
- When presenting the results in your report, you would note that 34/40 females got the answer correct compared with 40/42 males with a corresponding difference in proportions for females versus males of – 0.083 (95% CI (-0.216, 0.0492)). Note that this particular confidence interval contains zero. This show us that zero is included in the range of possible values which we are 95% certain contain the true or population difference. Thus, we have insufficient evidence to refute the null hypothesis of no difference in proportions.
- Hence, in this example, you may say that whilst the sample proportion (-0.083) indicates that females performed less well than males, assuming a level of confidence of 95%, there is insufficient evidence to conclude that there is a true difference in performance between females and males.
- Moreover, notice that the above confidence interval is quite wide. You should also mention therefore that the accuracy of the sample estimate for difference in proportions could be improved by increasing the sample size. Increasing the sample size might in turn ensure that the direction of the difference represented by the confidence interval is consistent across the two limits, thus enabling you to be 95% certain as to whether females performed less well or better than males.
· Q 6. I have worked my way through Q. 5 and the corresponding solution, as I wish to compare proportions. However, I am considering a single cohort of 25 malignant patients and I wish to compare the proportions who have been referred to two different types of clinic (‘multidiscipinary’ and ‘single clinician’) . As my patients are derived from a single group (the malignant cohort), I am puzzled as to how to identify two separate numbers for each of the two trial boxes in the corresponding Minitab dialogue box. Also, I did not obtain a result for Fisher’s Exact test in my output. How should I proceed?
A. As the patients referred to the multidisciplinary and single patient clincics originate from the same cohort – the malignant patients – just enter the same figure (25) into the two boxes for number of trials and proceed as before. Think about it; given the study design, to calculate your two proportions of referrals, you would automatically use the same denominator to represent the number of presenting patients, as they come from the same source in each case. Based on the output provided, the most commonly recommended test is the pooled version of the Z-test for comparing two proportions.
· Q 7. Where can I find illustrations of how to quote confidence intervals in the text of my report?
A. Different journals have different formatting styles in the presentation of confidence intervals. You can find some illustrations of these styles by searching for all instances where ’95’ occurs in the following articles:
Confidence Intervals by Margaret MacDougall is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License.
