Tutorial 5

Learning outcomes

After this tutorial the student should be able to:

explain the concept of the sample distribution for the mean;
mention and apply the formula (given simple situations) of the expected value, the variance, and the standard deviation of the sample distribution of the mean;
explain and apply the concept of standard error;
distinguish the standard error from the standard deviation;
explain the concept of the sample distribution for the sum;
mention and apply the formula (given simple situations) of the expected value, the variance, and the standard deviation of the sample distribution of the sum;
mention and apply the Central Limit Theorem;
use the Student t-distributions;
apply the concept of degrees of freedom (denoted by df or \(\nu\));
interpret a ‘confidence interval’;
determine the confidence interval for \(\mu\) of a one sample situation.

Pre-class activity

Watch:

‘Confidence level and Margin of Error’ (Duration: 5:30 min.).

The clip is linked on Brightspace.

Sampling Distribution

Read:

O&L 7^th Edition:
- paragraph 4.12 pp.190-200, or
O&L 6^th Edition:
- paragraph 4.12 pp.181-190,

where the sampling distribution of the sample mean \(\bar{y}\) is discussed as well as the role of the Central Limit Theorem. Additionally the book presents the sampling distribution of the sample sum \(\sum y\).

Rules for the sample mean \(\bar{y}\)

Let \(\bar{y}\) denote the sample mean computed from a random sample of \(n\) independent measurements from a population having an expected value (mean) \(\mu\) and variance \(\sigma^2\).

It follows that:

\(\mbox{E}(\bar{y}) = \mu_{\bar{y}} = \mu\) and \(\mbox{var}(\bar{y}) = \sigma^2(\bar{y}) = \sigma^2/n\)
The standard deviation \(\sigma(\bar{y})\) is also a measure of precision for the estimate of \(\bar{y}\) and for that reason is referred to as standard error of \(\bar{y}\) or \(\mbox{SE}(\bar{y})\).
\(\sigma(\bar{y}) = \mbox{SE}(\bar{y}) = \sigma/\sqrt{n}\)
When the population distribution is normal, the distribution of \(\bar{y}\) is exactly normal.
When the population distribution is not normal, the Central Limit Theorem says that when \(n\) goes to \(\infty\),
- the distribution of \(\bar{y}\) will (in some sense) converge to a normal distribution, and
- the distribution of the standardized \(\bar{y}\) (i.e., \(\frac{\bar{y} - \mu}{\sigma/\sqrt{n}}\)) will converge to the standard normal distribution.

Remark about approximation of a standardized sample mean.

In practice it is often assumed that the distribution of a standardized sample mean is well approximated by a standard normal distribution, when the sample size \(n\) is large enough, see:

O&L 7^th Edition pp.197-198, or
O&L 6^th Edition pp.188–189.

Rules for the sample sum \(\sum y\)

Let \(\sum y\) denote the sample sum computed from a random sample of \(n\) independent measurements from a population having an expected value (mean) \(\mu\) and variance \(\sigma^2\).

It follows that:

\(\mbox{E}(\sum y) = \mu_{\sum y} = n \times \mu\) and \(\mbox{var}(\sum y) = \sigma^2(\sum y) = n \times \sigma^ 2\rightarrow \sigma(\sum y) = \sqrt{n} \times \sigma\).
When the population distribution is normal, the distribution of \(\sum y\) is exactly normal.
When the population distribution is not normal, the Central Limit Theorem says that when \(n\) goes to \(\infty\),
- the distribution of \(\sum y\) will (in some sense) converge to a normal distribution and
- the distribution of the standardised \(\sum y\) (i.e., \(\frac{\sum{y} - n \times \mu}{\sqrt{n} \times \sigma}\) ) will converge to the standard normal distribution.

Confidence interval for \(\mu\) (one sample: quantitative continuous random variable \(y\))

Read:

O&L 7^th Edition:
- paragraphs 5.1 and 5.2 pp.232-240, or
O&L 6^th Edition:
- paragraph 5.1 and 5.2 pp.222-230.

For a random sample with observations \(y_1, y_2, \ldots, y_n\) with \(\mbox{E}(y) = \mu\) and \(\mbox{var}(y) = \sigma^2\).

Confidence interval for \(\mu\)

A confidence interval gives, based on the data of a random sample, all likely values for the unknown value of \(\mu\).
An unbiased estimator for \(\mu\) is \(\hat{\mu} = \bar{y}\) with \(\mbox{SE}(\bar{y}) = \sigma\ /\ \sqrt{n}\).
In general \(\sigma\) will be unknown and have to be estimated using the sample standard deviation \(s\) (see also Tutorial 4); in that case the standard error for \(\bar{y}\) becomes: \(\mbox{SE}(\bar{y})=s/\sqrt{n}\).
When sample sizes are large (say \(n > 120\)) we may assume, based on the Central Limit Theorem, that the sample mean \(\bar{y}\) is normally distributed: \(\bar{y} \sim \mbox{N}(\mu,\sigma)\).
In that case the limits of a \((1- \alpha) \times 100\%\) Confidence Interval for \(\mu\) are: \(\bar{y} \pm z_{\alpha/2} \times s/\sqrt{n}\), with \(z_{\alpha/2}\) coming from the standard normal distribution \(\mbox{N}(\mu = 0,\ \sigma = 1)\).
When sample sizes are small (say \(n \leq 120\)) we may not assume that the sample mean \(\bar{y}\) is normally distributed. In that case we have to use a \(t\)-distribution and the limits of a \((1 - \alpha) \times 100\%\) Confidence Interval for \(\mu\) are: \(\bar{y} \pm t_{\alpha/2} \times s/\sqrt{n}\), with \(t_{\alpha/2}\) coming from the \(t\)-distribution with \(\nu = n - 1\) degrees of freedom.

The value \(z_{\alpha}\) is the upper \(\alpha\)-point of a standard normal distribution, so \(z_{0.05} = 1.645\) and \(z_{0.05/2} = z_{0.025} = 1.960\). The value \(t_{\alpha}\) is the upper \(\alpha\)-point of a \(t\)-distribution with the appropriate degrees of freedom (\(= \nu\)): e.g., when \(\nu = 10:\ t_{0.05} = 1.812\) and \(t_{0.05/2} = t_{0.025} = 2.228\) (See O&L: Table 1, with the Standard Normal Distribution, and Table 2 with the Inverse Student’s \(t\) Distributions).

Exercises to be done during the tutorial

Exercise 5.1 and Exercise 5.2 are in the presentation handouts of Tutorial 5. For answers/feedback check Brightspace.

Post-class activity

Watch:

‘Confidence level and Margin of Error’ (Duration: 5:30 min.), when not done as pre-class activity,
‘Sampling distribution and estimation of population mean’ (Duration: 4:56 min.),
The first 3:30 min. of ‘Understanding Confidence Intervals’ (Duration: 4:02 min.).

All of the clips are linked on Brightspace.

Exercises to be done after the tutorial

For answers/feedback check Brightspace.

Exercise 5.3

A factory delivers packages of sugar. A shop owner suspects that the weight of these packages is systematically less than 1000 g.

To prove this, the shop owner takes a random sample of 30 packages and weighs each package individually. The observed weights are denoted by \(y_1, y_2,\ldots,y_{30}\). The 30 observed weights can be considered as independent and normally distributed with \(\mbox{E}(y) = \mu\) and \(\sqrt{\mbox{var}(y)} = \sqrt{\sigma^2_y} = \sigma\).

Computational results: sample mean \(\bar{y} = 998.62\) g and sample standard deviation \(s = 5\) g.

a. Determine the expected value \(\mbox{E}(\bar{y})\) and the standard deviation \(\sigma_{\bar{y}}\) for the sampling distribution of the mean.

b. Determine the 90% Confidence Interval for the population mean \(\mu\).

c. Give two case specific interpretations for the Confidence Interval from b.

Exercise 5.4

Do either

Exercise 4.85 O&L 7^th Edition p.225, or
Exercise 4.85 O&L 6^th Edition p.215.

Exercise 5.5

Do either

Exercise 5.5 O&L 7^th Edition p.286 [Note: Typographical error in the text on line 4, “\(\hat{\sigma} = 7.1\ \mbox{n}\)” should be “\(\sigma = 7.1\ \mbox{mg}\)”.], or
Exercise 5.5 O&L 6^th Edition p.276 [Note: Before answering a., and b., answer the following additional question: “Identify the population about which inferences can be made from the sample data.”. This is question a. in the O&L \(7^{\mbox{th}}\) Edition.].

Exercise 5.6

Do either

Exercise 5.6 O&L 7^th Edition p.286, or
Exercise 5.7 O&L 6^th Edition p.276.

Exercise 5.7

Do either

Exercise 5.7 O&L 7^th Edition p.286, or
Exercise 5.6 O&L 6^th Edition p.276 [Note: Typographical error in the text of Exercise 5.6c on line 2, “how many of the 95% confidence intervals\(\ldots\)” should be “how many of the 99% confidence intervals\(\ldots\)”].