Chapter 10:Introduction to Inference

“If you believe in miracles, head for the Keno lounge.”

Jimmy the Greek

10.1 Estimating with Confidence(pp. 506-526)

Confidence intervals are important in statistics.

This chapter provides information on: How such intervals can be constructed

from samples. How to interpret such intervals.

C% confidence interval for a parameter

An interval computed from sample data by a method that has probability C% of producing an interval containing the true value of the parameter.

A confidence interval for an unknown population, μ,

calculated from a sample size n with mean , has

the form

where z* is obtained from the normal distribution

table and σ is the standard deviation of the

population.

x

*x zn

Example: Suppose that the following scores represent a random sample from a population with a

known standard deviation σ = 3.88 Find a 95 % confidence interval

for the mean of the population. {85, 83, 91, 88, 88, 92, 81, 83,

85, 83, 86, 84} Calculate the mean.

Mean = 85.75 Remember that the Central Limit

Theorem states that the means of sample means of a specific size are normally distributed.

The 95% C.I. for the mean of the population from which the sample was chosen is (83.555, 87.945).

3.88* 85.75 1.96

12x z

n

INTERPRETATION OF 95% CONFIDENCE INTERVAL:

If we took 100 random samples from the population and computed 95% confidence intervals for each sample, we would expect 95 of the 100 samples to contain the mean of the population.

Results of polls produced by polling organizations (Gallup) are often provided with a margin of error.

Example: “64% of those polled favored Proportion

A with a margin of error of 3%.” Interpretation:

This determines a 95% confidence interval (0.61, 0.67).

Margin of Error Formula

Margin of error determines the length of the confidence level.

Therefore, the interval for the previous example would be (0.61, 0.67).

. . *M E zn

. .

.64 .03

x M E

Find sample size using margin of error

Suppose we want a 95% C.I. that is 3 units long and the M.E. to be no more than 1.5 units.

We can do this by taking a larger sample, but how large?

Therefore, we need a sample size of at least 26 to obtain an M.E. of 1.5 units.

Rule: *z mn

3.88Solve: 1.96 1.5

3.881.96

1.5

25.7, round to 26

n

n

n

Using the TI-84:

z* values for:99% confidence

interval: 2.57695% confidence

interval: 1.96

90% confidence interval: 1.645

80% confidence interval: 1.28

Some Cautions:

The data must be an SRS from the population.

The formula is NOT CORRECT for probability sampling designs more complex than an SRS.

There is no correct method for inference from data haphazardly collected with bias of unknown size.

Because is strongly influenced by a few extreme observations, outliers can have a LARGE EFFECT on the confidence interval.

x

Some Cautions: (continued)

If the sample size is small and the population is not normal, the true confidence interval will be different from the C value used in computing the interval.

You must know the standard deviation σ of the population.

If the sample size is large, the sample standard deviation s will be close to the unknown σ. Then,

*s

x zn

Margin of Error

The M.E. of a confidence interval gets

smaller as:the confidence level C decreases.the population standard deviation σ

decreases.The sample size n increases.

10.2 Test of Significance(pp. 531-556)

THIS IS ONE OF THE MOST IMPORTANT SECTIONS IN THE BOOK!!!!

IT CONTAINS A GREAT DEAL OF IMPORTANT MATERIAL.

READ CAREFULLY!!

Example #1: Tire Manufacturing

A tire manufacturer advertised that a new brand of tire has a mean life of 40,000 miles with a standard deviation of 1,500 miles. A research team tested a random sample of 100 of these tires and obtained a mean life of 39,500 miles.

Is the manufacturer’s claim reasonable? How likely is it that one would obtain a random

sample of 100 tires with a mean life of 39,500 miles from a population with a mean life of 40,000 miles and a standard deviation of 1,500 miles?

Example #1: Tire Manufacturing Consider the set of

means of ALL samples of size 100:

The Central Limit Theorem says that the set has a mean of 40,000 and a standard deviation of 150.

The normal distribution table shows that it is HIGHLY UNLIKELY that one would obtain such a sample if the population is as give.

0.043% is the P-VALUE of the test.

Since the probability is SO SMALL, we would likely conclude that the manufacturer’s claim is incorrect and that the mean life of the tires is something less than 40,000 miles.

1500150

100s

n

39,500

39,500 40,0003.33

150z

Example #1: Tire ManufacturingFormalized as: Null hypothesis: Alternate Hypothesis: 1-tail test Level of significance: 1%

Usually determined before the test

Critical region: from invNorm = 2.33

Calculated sample mean: 39,500

Mean of sample means: 40,000 Standard deviation of sample

means: 150 Calculated z for sample: -3.33 P=value for test: .04%

: 40,000OH : 40,000AH

.01

Example #1: Tire Manufacturing

Conclusion: Since the calculated z

is in the critical region, we reject the null hypothesis.

Since the P-value is less than 1% (the level of significance), this supports the rejection of the null hypothesis.

Example #1: Tire Manufacturing

A 1-tail test was involved because we were interested in a deviation in one direction only.

It wasn’t a concern to consumers if the mean life of the tires is more than the advertised value of 40,000 miles.

Example #2: Parachutes

An automatic opening device for parachutes has a stated mean release time of 10 seconds with a standard deviation of 3 seconds. To test the claim, a parachute club tested a random sample of 36 of these devices and found the mean release time to be 10.6 seconds. Is the result significant at the 5% level of significance?

Example #2: Parachutes

Null Hypothesis:

Alternate Hypothesis:

Type of test: 2-tail

Level of significance:

Critical Region:

Calculated sample

mean: 10.6

: 10OH

: 10AH

.05

1.96 or 1.96z z

Example #2: Parachutes

Mean of the sample means: 10

Standard deviation of sample means: 0.5

Calculated z for the sample: 1.2

Since calculated z is NOT in the critical region we DO NOT reject the null hypothesis.

P-Value for the test: .8414

3

36s

n

10.6 10

.5z

Example #2: Parachutes (TI-84)

Since the P-value is GREATER than 5%, this supports the decision to not reject the null hypothesis.

There IS NOT strong evidence to suggest that this sample did not come from a population with mean = 10.

Note well:

A null hypothesis is basically a hypothesis of no change. If one does not reject a null hypothesis, then the

test results are not statistically significant. Accepting a null hypothesis at a LOW

LEVEL of significance is NOT STRONG evidence that it is true. Acceptance of a null hypothesis simply means

that it is not unreasonable to assume that the population mean μ is the stated value.

For all you know, it might be some other number even closer to the stated value.

Note well:

A P-value is the probability that one would obtain a statistic as “extreme” as that which was calculated from the sample. A small P-value, such as .01, means that the

statistic is NOT LIKELY the result of pure chance. A P-value, such as .35, means that the statistics is

not an unusual or unexpected result. The level of significance for a test is usually

set beforehand. If a calculated P-value is smaller than the level of

significance, then the test statistic is statistically significant.

Note well:

A statistical test could be 2-tailed or 1-tailed. The one used is dependent on the

purpose and nature of the test. If both positive and negative deviations

from a parameter are important, use a 2-tailed test.

If only positive (or negative) deviations are important, use a 1-tailed test.

Finding P-values

contains Right tail

P-value *AH

P z z

contains Left tail

P-value *AH

P z z

contains ? Two-tailed

P-value * *

AH

P z z P z z

Note well:

Rejecting a null hypothesis is equivalent to saying that the test statistic is statistically significant. i.e. the calculated test statistic is not a

likely result of pure chance. The null and alternate hypothesis are

both stated in terms of population parameters, not sample statistics. You are attempting to use sample

statistics to come to reasonable conclusions about population parameters.

10.3 Using Significance Tests(pp. 560-567)

This “short” section points out things that need to be considered when attempting to determine if test results are significant.

10.4 Inference as Decision(pp. 567-577)

This section introduces three concepts that were added to the AP Statistics syllabus for the 1998-1999 academic year. Type I error Type II error Power of a test

Type I and Type II Errors

Let’s try an example:Quality control tests at the 5% level of significance:

Smith’s produces a machine-produced product that weighs 1500 lbs. The population of produced items has an allowable standard deviation of 40 lbs. Samples of size 100 are periodically examined to see if production standards are being maintained.

Consider the set, M, that consists of mean weights of all samples of size 100. The CLT states that M will have a normal distribution with mean = 1500 lbs. and s= 4 lbs.

Quality control tests at the 5% level of significance:

Null Hypothesis:

Alternate Hypothesis: Type of test: 2-tailed Level of significance:

5% 2.5% in each tail

Critical values of z:

: 1500OH

: 1500AH

1.96 or 1.96z z

Quality control tests at the 5% level of significance:

Smith’s will reject the null hypothesis if a sample of 100 yields a mean that is not within 1.96 standard deviations of 1500 lbs. The null will be rejected if a mean weight is

less than 1500 – 1.96(4) which equals 1492.16 lbs or if a mean weight is more than 1500 + 1.96(4) which equals 1507.84 lbs.

In other words, Smith’s will accept the null if a mean weight is in the range 1492.16 < x < 1507.84.

Quality control tests at the 5% level of significance:

Related calculator computations:

Suppose a random sample produces a mean of 1509 lbs.:

The null would be rejected. There is a “suggestion” that production standards are

not being met. The probability of obtaining a mean as large as 1509 is

normalcdf(1509,1E99,1500,4) which equals .012224 or about 1.2%.

Therefore, if the null is true, there is a 1.2% chance the Smith’s will incorrectly reject it.

This is a Type I error. By setting the level of confidence at 5% PRIOR to

doing any testing, Smith’s is allowing for a 5% chance of making a Type I error. In real life, a Type I error might result in stopping

production to try to find a problem that doesn’t exist.

NOW assume that it is extremely undesirable to have an item produced that weighs 1515 or more lbs.

Smith realizes that things can go wrong in a mass production process. Some products may be too heavy.

She is interested in knowing the probability that her quality control test will incorrectly accept the null if the mean weight somehow shifts to 1515 lbs.

If the null is incorrectly accepted, this is a Type II error.

Type II Error (continued) The null will be accepted if a sample mean weight is less

than 1507.84 lbs. and greater than 1492.16 lbs. If the population mean has shifted to 1515 lbs., the z-

value for 1507.84 is -1.79. Using the normal distribution table, the probability that

z < -1.79 is .0367. THIS IS THE PROBABILITY OF MAKING A TYPE II ERROR. In other words, if the mean has shifted to 1515 lbs.,

Smith’s will incorrectly accept the null about 3.67% of the time.

1507.84 1515

4z

The Power of a Test

THE POWER OF A TEST IS THE PROBABILITY THAT THE NULL WILL BE REJECTED FOR A PARTICULAR VALUE OF A POPULATION PARAMETER.

In this case, the population parameter is μ = 1500 lbs., and the particular alternative value is μ = 1515 lbs.

The power of the test for μ = 1515 lbs. is 0.9633.(1 – 0.0367)

That is, if μ = 1515 lbs., Smith’s can expect to correctly reject the null about 96.33% of the time.

Real-life situations frequently involve Type I and Type II errors.

Consider the legal world and a null hypothesis “The accused man is innocent.” Type I error occurs when the man is found guilty

when, in fact, he is innocent. Type II error occurs when the man is found not

guilty, but he is guilty. Decreasing the chance of one error type

frequently increases the chance of the other error type.

In real-world situations, one must often decide which error type is more important to minimize.

Things to Remember:1. A Type I error can occur only when a null

hypothesis is true. You incorrectly reject a TRUE null hypothesis.

2. A Type II error can only occur when a null hypothesis is false. You incorrectly accept a FALSE null hypothesis.

3. The Power of a Test is 1 - probability (Type II error). This is the probability that you correctly reject a false null hypothesis.

4. One needs an alternative to the null hypothesis in order to calculate a Type II error. Without an alternative hypothesis, the question “What is the probability of a Type II error?” is meaningless.