On Thursday, we talked about confidence intervals.

A

*confidence interval* gives an estimated range of values which is likely to include an unknown population parameter, the estimated range being calculated from a given set of sample data.

We had this problem…

*Some Senior 4 students in a large high school want to change a tradition at graduation. Instead of wearing the usual cap and gown, they want to wear formal clothes. A quick survey of 96 randomly selected students shows that 41 prefer formal wear. *To answer this question, construct a “confidence interval”, most often a “95% confidence interval”.

Find the probability (P), the mean (µ), and the standard deviation (σ) first.

**p** = 41 / 96

=

0.43**µ** = n ∙ p

= 96 (0.43)

=

41.28**σ** = √ n ∙ p ∙ (1 - p)

= √ 96 (0.43) (1 – 0.43)

= √ 23.5296

=

4.851For data that have a normal distribution with mean and standard deviation, a 95% confidence interval is:

**µ ± 1.96σ**

This is the range of values that lie within 1.96 standard deviations of the mean. The probability that a particular data value lies in that range is 0.95. This is shown in the graph.

*95% confidence interval* = µ ± 1.96σ

= 41.28 ± 1.96 (4.851)

= 41.28 ± 9.508

= 41.28 – 9.508

= 31.772

or

= 41.28 + 9.508

=

50.788We write it this way:

(31.772, 50.788)

= (31.772 / 96) × 100

=

33%= (50.788 / 96) × 100

=

53%And we write it this way:

(33%, 53%)With 95% confidence, we know that between 33% and 53% of 96 students prefer formal wear.

Now, find the Margin of Error and Percent Margin of Error.

The

*margin of error* is the proportion that we add to, and subtract from, the mean to construct the confidence interval.

For a 95% confidence interval:

**Margin of Error = ± 1.96σ**1.96 × 4.851 or

9.508 represents the half-width of the interval. This is the margin of error. To express this as a percent, we divide it by the sample size, which is 96, and multiply by 100.

*% Margin of Error* = ± (9.508 / 96) × 100

= ± 9.904%

Is it possible that a majority of students prefer formal wear?

*** Yes, but not likely, because the high end of the 95% confidence interval is just 3% over the half of the sample size and the Margin of Error is a little bit high. When you conduct a survey, the smaller the Margin of Error, the better the results of the survey. *The next scribe is **Donna (",)**