HOW CONFIDENT ARE YOU?

Confidence Intervals and the Missing Link

What Do Students Need to Know?

You’ve taught students the normal curve and the central limit theorem, but they just don’t get confidence intervals.

Z-Scores – the missing link

What are z-scores? Sue got her math test back with a mark

of 82%. The teacher said the marks were normally distributed with a mean of 72% and a standard deviation of 10. How many standard deviations was Sue’s mark above the mean mark?

deviationsstdofdevstd

meanmark.#

.

Z-Scores are the number of standard deviations some x-value is above or below the mean. If z is

positive, x is above the mean; negative, x is below the mean.

x

z

In Sue’s case, her z-score is 1. That is, her mark is one standard deviation above the mean her mark is being compared to. Using the 68-95-99.7 rule, we can conclude that Sue had a higher mark than 84% of the class.

But what if the z-score does not work out to be a whole number? How can we estimate percentages if all we have is the 68-95-99.7 rule?

Z-Scores use a chart that let’s us easily look-up the values we want. Convenient tables allow us to find the numbers needed

http://www.epatric.com/documentation/statistics/z-score_table.html

Example Standard Normal Distribution:  On

Human Gestation Studies have determined that the average gestation period (time between conception and giving birth) for humans to be approximately 266 days, while the standard deviation (σ) is about 16 days.  If the length of gestation is considered to follow a normal distribution, then find the percent of human pregnancies which would be expected to last more than 39 weeks?

Using the z-score chart we find the % below this z value is 67%. Since we want the % above 39 weeks, we subtract this from 100%. 33% of pregnancies would be expected to last longer than 39 weeks.

44.016

266)39*7(

z

What percent would be expected to last between 266 days and 286 days?

For this example, we need to calculate 2 z-score values and find the % in between.

Again, using the table, there is 89.44% below a z-value of 1.25. Subtracting 50% from this, we have 39.44% of pregnancies are expected to last between 266 and 286 days.

016

266266

z

25.116

266286

z

Example Coach recorded all the practice times for

the Horton track team. These times were N(5.6, 0.76) minutes – meaning a normal distribution with mean of 5.6 and standard deviation of 0.76. The coach says if you are in the lowest 5% you can compete in provincials. What time do you need?

5% matches a z-value of -1.64 Using the formula, we can solve for x You must have a time of < 4.35 mins.

x

x

35.476.0

6.564.1

Making the Connection

Central Limit Theorem

Example

The average height of a NS grade 11 student is 63” with a standard deviation of 4”; heights are normally distributed.

If a single grade 11 student is chosen at random what is the probability their height will be >66”?

2266.07734.01

75.04

6366

tableFrom

z

Sample vs. individual…

If 16 grade 11 students are chosen at random, what is the probability that their average height ( ) is greater than 66”?

x

0013.09987.01

316/4

6366

tableFrom

z

Example – not normal…

The average price of a house sold in NS in 2007 was $150000 with a standard deviation of $15000. Housing prices are right skewed. If you randomly selected one house from those sold, what is the probability it sold for < $125000?

Sol’n - We can’t use z-scoresbecause the distribution is not normal. Given our background, This cannot be found.

Non –normal continued…

You randomly choose 40 houses from those sold in 2007. what is the probability that the average selling price of these 40 houses is < $125000?

Sol’n – Since the sample size is ≥ 30, the sampling distribution of all possible

sample means is normal.

This value is “off the chart”. Basically, the probability is zero!

54.1040/15000

150000125000

z

Confidence Intervals

The recent scores of literacy tests for Grade 9 NS students was N(400, 36). If we choose one student at random , what is the probability their score is between 375 & 425?

69.036

400375

z

69.036

400425

z

5098.02451.07549.0 tableFrom

Taking a sample from the population…

If we take one sample of size 25 from all these scores, what is the probability the mean of the sample will be between 375 & 425?

From table -

47.325/36

400375

z

47.325/36

400425

z

9994.00003.09997.0

Creating an interval…

What if we take another sample of 25? How can we create an interval, using the mean, that will include µ?

If the mean of our sample was 407, we could add/subtractone std. dev. And get an intervalthat contains µ=400. The problemis we could get almost any value forthe mean of our sample AND we usually

don’t know the value of µ.

What if we want to be 90% sure?

What if we want to be 90% sure that the interval we calculate from our sample mean will contain µ? How many standard deviations do we need to add and subtract to create this interval?

Using z-scores, we need to look-up a z-value with 95% below it. Z = 1.645We have to ± 1.645 std. dev.to the sample mean to get our interval.

n

zxCI

Applications …

The average weight of infants born in Canada in 1947 was N(3, 0.6) kg. We think babies are getting heavier so we took a SRS of 36 babies born in 2007 and got a sample mean of 3.4 kg. Are babies getting heavier?

Sol’n – we need a CI for the population of newborns in 2007. We decide to use a 90% confidence level. We can…

)56.3,24.3(

36

6.0645.14.3

OR using the calculator…

Conclusion…

Since our population mean of 3 kg from 1947 is not contained in the interval estimate for the 2007 population , we can conclude with 90% confidence that newborns are getting heavier.