Confidence Intervals


Confidence Intervals

Suppose you have a large i.i.d. sample. The CLT implies that with chance about 95%, the sample mean is within 2 SDs of the population mean.

This can be expressed in a different way:

  • In about 95% of all samples, the sample mean is in the range population mean $\pm ~ 2 \sigma/\sqrt{n}$.

That is the same as saying:

  • In about 95% of all samples, the population mean is in the range sample mean $\pm ~ 2 \sigma/\sqrt{n}$.

That is why the interval sample mean $\pm ~ 2 \sigma/\sqrt{n}$ is used as an interval of estimates of $\mu$.

Confidence Intervals for $\mu$

The interval sample mean $\pm ~ 2 \sigma/\sqrt{n}$ is called an approximate 95% confidence interval for the parameter $\mu$. The interval has a confidence level of 95%.

You could choose a different confidence level, say 80%. With that choice, you would expect the interval to be narrower. To find out exactly how many SDs you have to go on either side of the center to pick up a central area of about 80%, you have to find the corresponding $z$ on the standard normal curve, as shown below.

As you know from Data 8 and can see in the figure, the interval runs from the 10th to the 90th percentile of the distribution. So $z$ is the 90th percentile of the standard normal curve, also known as the "90 percent point" of the curve. The scipy method is therefore called ppf and takes a decimal value as its argument.


Therefore an approximate 80% confidence interval for the population mean $\mu$ is given by "sample mean $\pm ~ 1.28\sigma/\sqrt{n}$".

Let's double check that 2 is a good choice of $z$ for a 95% interval. The $z$ we need is the 97.5 percent point:


That's $z = 1.96$, which we have been calling 2. It's good enough, but $z = 1.96$ is also commonly used for constructing 95% confidence intervals.

General Definition

Let $\lambda$ be any confidence level. Let $z_\lambda$ be the point such that the interval $(-z_\lambda, ~ z_\lambda)$ contains $\lambda$% of the area under the standard normal curve. In our example above, $\lambda$ was 80 and $z_\lambda$ was 1.28.

Then if $n$ is large,

$$ \frac{\lambda}{100} ~ \approx ~ P(\bar{X}_n \in \mu ~ \pm ~ z_\lambda \sigma/\sqrt{n}) ~ = ~ P(\mu \in \bar{X}_n ~ \pm ~ z_\lambda \sigma/\sqrt{n}) $$

The random interval $\bar{X}_n ~ \pm ~ z_\lambda \sigma/\sqrt{n}$ is called an approximate $\lambda$% confidence interval for the population mean $\mu$. There is about a $\lambda$% chance that this random interval contains the parameter $\mu$.

The only difference between confidence intervals of different levels is the choice of $z$ which depends on the level. The other two components are the sample mean and its SD.

A Data 8 Example Revisited

Let's return to an example very familiar from Data 8: a random sample of 1,174 pairs of mothers and their newborns.

baby = Table.read_table('baby.csv')
Birth Weight Gestational Days Maternal Age Maternal Height Maternal Pregnancy Weight Maternal Smoker
120 284 27 62 100 False
113 282 33 64 135 False
128 279 28 64 115 True
108 282 23 67 125 True
136 286 25 62 93 False
138 244 33 62 178 False
132 245 23 65 140 False
120 289 25 62 125 False
143 299 30 66 136 True
140 351 27 68 120 False

... (1164 rows omitted)

The third column consists of the ages of the mothers. Let's construct an approximate 95% confidence interval for the mean age of mothers in the population. We did this in Data 8 using the bootstrap, so we will be able to compare results.

We can apply the methods of this section because our data come from a large random sample.

n = 1174
ages = baby.column('Maternal Age')

samp_mean = np.mean(ages)

The observed value of $\bar{X}_n$ in the sample is 27.23 years. We know that $n = 1174$, so all we need is the population SD $\sigma$ and then we can complete our calculation.

But of course we don't know the population SD $\sigma$.

So we estimate $\sigma$ by the SD of the data. There is some error in this estimate, of course, but it gets divided by $\sqrt{n}$ and therefore doesn't have much effect. Remember that our methods rely on the CLT and are only valid when $n$ is large.

The estimate of $\sigma$ is about 5.82 years.

sigma_estimate = np.std(ages)

So an approximate 95% confidence interval for the mean birth weight of babies in the population is $(26.89, 27.57)$ years.

samp_mean - 2*sigma_estimate/(n**0.5), samp_mean + 2*sigma_estimate/(n**0.5)
(26.888831911866099, 27.567726861558093)

No bootstrapping required! But let's compare our interval to the interval we got in Data 8 by using the bootstrap percentile method. Here is the function bootstrap_mean from Data 8.

def bootstrap_mean(original_sample, label, replications):

    """Displays approximate 95% confidence interval for population mean.
    original_sample: table containing the original sample
    label: label of column containing the variable
    replications: number of bootstrap samples
    just_one_column = original_sample.column(label)
    n = len(just_one_column)
    means = make_array()
    for i in np.arange(replications):
        bootstrap_sample = np.random.choice(just_one_column, size=n)
        resampled_mean = np.mean(bootstrap_sample)
        means = np.append(means, resampled_mean)

    left = percentile(2.5, means)
    right = percentile(97.5, means)

    resampled_means = Table().with_column(
    'Bootstrap Sample Mean', means
    print('Approximate 95% confidence interval for population mean:')
    print(np.round(left, 2), 'to', np.round(right, 2))
    plt.plot(make_array(left, right), make_array(0, 0), color='yellow', lw=8);

Let's construct a bootstrap 95% confidence interval for the population mean. We will use 5000 bootstrap samples as we did in Data 8.

bootstrap_mean(baby, 'Maternal Age', 5000)
Approximate 95% confidence interval for population mean:
26.9 to 27.56

The bootstrap confidence interval is essentially identical to the interval (26.89, 27.57) that we got by using the normal approximation.

As we did in Data 8, let's observe that the distribution of maternal ages in the sample is far from normal:'Maternal Age').hist()

But the probability distribution of the mean of the large sample is approximately normal, by the Central Limit Theorem.


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