Lab 5: Confidence Intervals for Normal Means with an Illustration using the Cavendish Data

• [OBJECTIVES:] We first consider the problem of estimating the parameters of an underlying (normal) population when all we have to work with is a sample of data. This leads us to consider what our best ``point estimate'' of the parameter might be and how ``confident'' we can be that some interval estimate contains the parameter of interest. Once we have worked some with data which SYSTAT has generated for us, we will then move on to a real data set.
• [DATA:] The first data we will use for this lab will be simulated (constructed) by Systat. Once we have worked with this simulated data for a while, we will then move on to the Cavendish data set which we have previously examined.
• [DIRECTIONS:] Be sure to answer all questions. Write out all answers to the questions neatly and turn them in to the lab instructors.

1.

Open up a new data file in Systat. Title 7 columns NORM1, NORM2, NORM3, NORM4, NORMAVG, LOWBND and UPBND. Use the Fill Worksheet option under the Data menu to fill the worksheet to 200 rows. Then, use the Math option under the Data menu to set NORM1 to ZRN (select NORM1 on the left, ZRN on the right). This fills the column with 200 values taken at random from a standard Normal distribution. Repeat this procedure for NORM2 through NORM4.

The population mean and variance of a standard normal ($\mu$ and $\sigma^2$) are and 1, repectively.

2.

Using Data/Math, set NORMAVG to (NORM1 + NORM2 + NORM3 + NORM4)/4. You will see in class that the probability that a standard normal random variable Z is between -1.645 and 1.645 is 0.90. You will also see (or may already have seen) that if $X_{1}, X_{2}, \ldots, X_{n}$ are n are iid $N(\mu,\sigma^2)$ random variables then $\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}$ will be N(0,1). Thus,

$$P(-1.645 < \frac{\bar{X}-\mu}{\sigma/\sqrt{n}} < 1.645) = 0.90.$$

Let's rearrange this. Multiply all terms in the inequality by $\sigma/\sqrt{n}$, subtract $\bar{X}$ from all terms, and multiply all terms by -1. This gives

$$P(\bar{X}-1.645\frac{\sigma}{\sqrt{n}} < \mu < \bar{X}+1.645\frac{\sigma}{\sqrt{n}}) = 0.90.$$

Thus, for each sample we can construct an interval which we are ``90% confident'' contains the true mean $\mu$. Here, we know that $\mu=0$ and $\sigma=1$. Thus, $1.645\sigma/\sqrt{n}= 1.645/2 \approx 0.823$. Using Data/Math, set LOWBND to NORMAVG - 0.823, and UPBND to NORMAVG + 0.823. We expect the true population mean, , to lie in about $90\%$ of the (LOWBND,UPBND) intervals. As we filled the data tables to 200 rows, this means that we expect about 20 cases where we don't catch the true mean.

• Going down the LOWBND column, in how many cases is LOWBND > 0? Going down the UPBND column, in how many cases is UPBND <0? In how many cases is the true mean not in the confidence interval you constructed (in other words, how many cases have either LOWBND > 0 or UPBDN < 0)? How well does this match the theoretical prediction?
• The interval

  $$P(\mu < \bar{X} + 1.645\frac{\sigma}{\sqrt{n}}) = 0.95$$

  is said to be a one-sided confidence interval. This would correspond to $\mu$ being less than UPBND. How well does the observed value match the predicted value?
• If we used 1.96 and -1.96 instead of 1.645 and -1.645, respectively, then we would have a $95\%$ confidence interval (check the tables). Reset LOWBND to NORMAVG - 0.98 and UPBND similarly. In how many cases is the true mean not in the confidence interval? How well does this match the theoretical prediction? Note that we are now ``$95\%$ confident'' that our new interval contains the true mean, so the predicted value here is 0.05 * 200 = 10. Also note that to achieve higher confidence, we had to make our intervals wider.

The reason we call this a ``confidence interval'' and not a probability interval is that the true mean has a specific value. For a given sample, the probability that the true mean is in the confidence interval associated with that sample is either 0 or 1. What we know is that if we construct intervals in this way for a lot of samples, we will catch the true mean some specified fraction of the time.

3.

Now, before we used a value of 1 for the standard deviation, this being the true population value. What if we didn't know this value? Then we would have to substitute an estimate of $\sigma$. The most obvious estimate is the sample standard deviation s. Create a new column, STDEV. Using Data/Math, set STDEV to STD(NORM1,NORM2,NORM3,NORM4). Then, set UPBND to NORMAVG + 0.98*STDEV, and LOWBND to NORMAVG-0.98*STDEV.

How many of these confidence intervals contain the true mean? Is this consistent with $95\%$ confidence? Do the intervals need to be made wider (to catch more often) or shorter?

4.

We now want to take our intuitions from simulating confidence intervals to a real data set. The Cavendish data is as follows:

5.50 5.55 5.57 5.34 5.42 5.30

5.61 5.36 5.53 5.79 5.47 5.75

4.88 5.29 5.62 5.10 5.63 5.68

5.07 5.58 5.29 5.27 5.34 5.85

5.26 5.65 5.44 5.39 5.46

5.

Open a new data file, title a column CAVEND and enter the above data in that column.

6.

The best modern measurements of the density of the earth correspond to a measurement of 5.517 in Cavendish's experiment. One way to evaluate Cavendish's results is to use them to come up with a range of values for the true density of the earth which are plausible at some degree of confidence - in short, to construct a confidence interval for the the true value of the earth's density. We will say that any value within this interval is plausible given this data. The problem now is to choose how wide to make this interval. If we make it too wide, the interval will contain the true value, but it will not let us make precise statements about it. If it is too short, we can make precise statements, but our statements are likely to be wrong. Thus, we must compromise. We will try to construct an interval that is as short as possible subject to the constraint that the probability of our making a mistake is quite small. We need to set a probability of excluding the true value which we are willing to live with. This probability of saying that the true value is not plausible is said to be the level of significance associated with our test and is denoted $\alpha$. The most common values chosen for $\alpha$ in practical work are .10 and .05, corresponding to making a mistake one time in 10 or making a mistake one time in 20. In this case, we will let $\alpha = .05$. Thus, we want to choose the upper and lower bounds on our interval in such a way that

$$P(lowbnd < \mu < upbnd) = 1 - \alpha = 0.95.$$

This probability is difficult to evaluate. So, we look for an interval involving $\mu$ whose probabilities we can evaluate. By the Central Limit Theorem, the sample mean is normally distributed about the true mean of the population,

$$\bar{y} \sim N(\mu,\sigma/\sqrt{n}).$$

Now, if we knew the true standard deviation $\sigma$ associated with these measurements, we could compute an interval of the following form

$$P(lowbnd2 < \frac{\bar{y} - \mu}{\sigma / \sqrt{n}} < upbnd2) = 0.95$$

using the fact that the quantity in the middle has a standard Normal distribution. It turns out that the shortest intervals that we can construct are those which divide the probability of making a mistake evenly between values which are too large and those which are too small. In other words, half of the time we make a mistake it will be by saying the true value is too large to be plausible, and the other half it will be by saying the true value is too small to be plausible. In the context of the standard Normal shown above, this corresponds to choosing a value for upbnd2 such that a standard Normal will exceed that value only .025 of the time and a value of lowbnd2 such that a standard Normal will be less than that value only .025 of the time. These values are 1.96 and -1.96, respectively. Since we do not know $\sigma$, we use the sample standard deviation s in its place. This means that we have to replace the normal values with those from a Student t distribution with n-1 degrees of freedom. (You will learn about this procedure in a couple of weeks in class; for now, just accept that it is valid.) For a t-distribution with 28 degrees of freedom, the upper and lower bound values can be found from Table E of your book to be 2.048 and -2.048. Thus,

$$P(-2.048 < \frac{\bar{y} - \mu}{s/\sqrt{n}} < 2.048) = 0.95$$

and

$$(\bar{y} - 2.048\frac{s}{\sqrt{n}}, \bar{y} + 2.048\frac{s}{\sqrt{n}})$$

is a 95% confidence interval for the true mean.

• Is the true value of 5.517 in this interval? In order to figure this out, you will need to find the sample mean $\bar{y}$ and standard deviation s. To do this (just in case you have forgotten how), go to stats/stats/statistics.
• Are Cavendish's measurements and the modern value of the density of the earth significantly different? Give possible explanations for a discrepancy, if one exists. If you were presenting your results to modern scientists, what would you tell them about Cavendish's experiment?
• If we were willing to tolerate a higher probability of error (such as $\alpha=0.10$) could we make more or less precise statements about the plausible values? Is a 90% confidence interval longer or shorter than a 95% interval?