Test of a hypothesis about µ when value of sigma is unknown I



PSY 201 Lecture Notes

Test of a hypothesis about µ when the value of sigma is unknown

Module 17

Problem: Suppose a car dealership wishes to discover ways to improve customer satisfaction with its service department. Over the past five years, surveys have been distributed to randomly selected service department customers who rated their “overall satisfaction” with the service department. The mean of those five years worth of ratings is 70.4 on a 0 – 100 scale. Unfortunately, a fire at the dealership destroyed records of the previous past years, so no information on the five-year population characteristics remains except that the manager of service remembers that the mean was 70.4. The dealership wishes to devise ways to increase the mean rating. It hires a psychologist who implements a strategy to improve satisfaction. The strategy includes training the “service writers”, the persons who interact with the customers, in interpersonal skills, training the persons who accept payment in such skills, improving the physical appearance of the service waiting area, and streamlining and clarifying the invoice for service given to customers. After implementation of the plan, the dealership sent out 25 surveys to randomly selected customers. The question here is: Is the mean satisfaction of the population of service customers equal to 70.4, the value of the population mean before the plan was implemented, or is it not? The mean of the sample of 25 customers was 79.52. The sample standard deviation was 15.382. (Note – for pedagogical reasons, the sample size for this example problem has been set at a value that is too small.

This problem starts out like a typical Z-test problem.

Here are the Hypothesis testing steps

Step 1: Null hypothesis______Population mean = 70.4__________µ=70.4____________

Alternative Hypothesis______Population mean ≠ 70.4____ µ≠70.4________

Step 2 Test Statistic:

Sample mean – Hypothesized value of population mean

Hmm. The Z statistic is Z = ---------------------------------------------------------------------

Population standard deviation

------------------------------------

Square root of sample size

Symbolically, that’s

X-bar - µH X-bar– 70.4

Z = --------------------------- = ------------------------------------------

σ ?????

--- -----

N 5

The problem is: We’re missing a quantity. Specifically, we’re missing the value of sigma (σ), the population standard deviation. For the statistic to be a real Z, the population standard deviation must be plugged into the formula.

What should we do?

The test statistic problem

This problem was faced by statisticians of old (who would now be very old statisticians.)

They wanted to compute a Z statistic but did not know the value of the population standard deviation.

In fact, this is the situation we’ll most likely face.

That is, if we’re so ignorant of the characteristics of a population that we don’t know its mean, how in earth could we know its standard deviation?

The answer is: We couldn’t. We need to find a way of testing such hypotheses without having to know the value of the population standard deviation.

Why do we even care?

We care because we can’t determine the p-value necessary for making a decision. The determination of the p-value depends on the use of the Normal Distribution tables. But if we don’t have a Z statistic, we can’t use the Normal Distribution tables.

Two possible solutions

Solution 1: Take a large sample and use the sample standard deviation as a substitute for the population standard deviation.

This is what statisticians of old did in the early 1900s. They took samples of 100-200, used the sample standard deviation in place of the population standard deviation in the formula, and pretended that they had a Z statistic. They then used the Normal distribution tables to determine p-values.

This procedure was technically incorrect, since the number they put in the Z-statistic formula was NOT the actual population standard deviation but a sample quantity. But since the sample sizes they used were very large, the value they substituted was probably quite close to the true value, and they probably computed p-values that were very close to the correct ones.

So why not just keep on doing this – taking huge samples and pretending the statistic is a Z.

a. Huge samples cost a lot in terms of $ and time.

b. It just isn’t right.

Solution 2. Figure out the correct sampling distribution of the statistic

X-bar - µH

? = ----------------------------------------

S

---

N

Note that the formula has S, rather than σ in the denominator. Note also that it’s not called Z, it’s called ? because we don’t know what it is.

The computations necessary for Solution 2 were provided by a mathematician named William Gosset in the early 1900s. He figured out what the sampling distribution of the above quantity was.

Gosset called the distribution the T distribution. He called the statistic the t statistic.

The Z and the T distributions are pictured below. Note first that they’re pretty similar. But careful inspection shows a difference: The T distribution is more variable than the Z. The probability of a Z close to 5, for example is essentially 0. But the probability of a t of 5 is substantial, as can be seen from the height of the curve near 5.

Degrees of freedom and variability of the T distribution

The T distribution is different from the Normal distribution in one other respect. Its variability decreases as sample size gets larger.

Part of the formula for T distribution is the quantity, N-1. This value is called degrees of freedom. It is a parameter in the T distribution, just as п and e are parameters of the Normal distribution.

In general, degrees of freedom, symbolized as df, equals N-1 for a single sample.

Getting back to the T distribution, as degrees of freedom increases, the variability of the T distribution gets smaller. When degrees of freedom is very large, the T distribution looks more and more like, you guessed it, the Normal Distribution. This validates the practice of the statisticians of old in using large samples to estimate σ.

From Steinberg, p. 201

Gosset’s publication of his result.

The article describing the results concerning the t statistic was published under the pseudonym, Student. The statistic has since been known as Student’s t.

Critical t values

Since the T distribution is more variable than Z, we would expect t values to be farther from 0 than Z values when the null hypothesis is true. Recall that a Z of 1.96 or larger would be unusual if the null were true. But a t of 1.96 would not necessarily be so unusual.

Gosset’s calculations allowed statisticians to compute the true p-values associated with various values of t. The formula he derived or one like it is used by all computer programs to compute p-values when t is used as the test statistic.

Statistics textbooks publish tables of critical t values – the values of t whose p-values were exactly equal to .05, .01 and other common significance levels.

A portion of the t Table from Steinberg, p. 201

1 12.706

This is all nice to know, but in the modern age, it’s only useful if you’re stranded on a deserted island.

In all other instances, our computer program will compute the p-value associated with the t.

Worked Out Example

Data (The 25 survey responses from the problem described above were as follows . . .)

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Menu Sequence

Analyze -> Compare Means -> One-Sample T Test...

Talking to the t-test procedure

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The Output

T-Test

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SPSS computes a confidence interval for the population mean.

However, the interval is in deviation from the hypothesized mean. To make it more useable, add the Test Value to each limit. (Yes – the test value.)

Actual lower limit = “Lower” + Test Value = 2.77 + 70.4 = 73.17

Actual upper limit = “Upper” + Test Value = 15.47 + 70.4 = 85.87

Conclusions

Hypothesis test conclusion

The sample mean is larger than the hypothesized mean. The p-value is less than .05, So we reject the null hypothesis that the mean of the population of ratings of the new service procedure is 70.4 and conclude that the mean is larger than 70.4.

Confidence interval conclusion

Lower limit = Test value + Lower = 70.4 + 2.77 = 73.17

Upper limit = Test value + Upper = 70.4 + 15.47 = 85.87

The probability is .95 that the population mean falls between 73.17 and 85.87.

Example problem II

Suppose a manufacturer of radial tires claims that the average mileage of its tires is 40,000 miles. To test the claim, you purchase 9 of the tires and run them until their tread reaches the minimum legal depth. The number of miles for each of the tires is 36900, 33300, 35500, 32000, 37800, 40000, 43200, 39500, and 34900. Set up and test the appropriate hypothesis.

The Hypothesis Testing Answer sheet. This is the sheet you’ll be expected to fill out for each hypothesis testing problem.

Describe the population or populations whose characteristics are being investigated.

Population of mileages of tires made by a particular manufacturer.

Null Hypothesis:_____________________________________________________________________

Formally state the

Alternative Hypothesis:______________________________________________________________

Give the name and the formula of the test statistic that will be employed to test the null hypothesis.

One sample t statistic because we don’t know the value of the population standard deviation. If we did know the value of the population SD, we’d use the Z statistic.

One sample t = (X-bar – 40000)/(S/sqrt(9))

What significance level will you use to separate "likely" value from "unlikely" values of the test statistic?

Hint: .05 is a popular choice.________________________________________________________________________________

What is the value of the test statistic

computed from your data? ___________________________________________________________________

What is the probability of a value as extreme as the

above value if the null hypothesis were true, i.e., the p-value?______________________________________________________

What is your conclusion?

Do you reject or not reject the null hypothesis? _____________________________________________________________

What are the upper and lower limits of a 95% confidence interval appropriate for the problem? Present them in a sentence, with standard interpretive language.

The probability is .95 that the interval, 34,298 to 39, 725 contains the population mean.

State the implications of your conclusion for the problem you were asked to solve. That is, relate your statistical conclusion to the problem.

It appears that the manufacturer’s claim is incorrect. Our evidence suggests that the population mean is less than 40,000, somewhere between 34,298 and 39,725.

SPSS Computations for Tire Mileage Example Problem

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T-Test

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To form the 95% confidence interval, add the test value to the value under Lower and to the value under Upper. So the 95% confidence interval is

34,297.69 to 39,724.53.

Two-tailed vs. One-tailed alternative hypotheses. Module 17

If the null is false, there are actually three possible alternative hypotheses . . .

1: Population mean does not equal the hypothesized value (e.g. Pop mean ≠ 5000)

2: Population mean is only less than the hypothesized value (e.g., Pop mean < 5000)

3: Population mean is only larger than the hypothesized value (e.g., Pop mean > 5000)

We usually have no preconceived notions concerning the direction of difference if the null is not true. In those cases, we employ #1 above. It is called a two-tailed alternative hypothesis, and we compute the p-value as the probability of an outcome as extreme as the obtained outcome in the positive or in the negative direction.

On the other hand, if we know that there is no way that one of the two directions of difference could be observed if the null were false, then we don't consider that outcome as a possibility, and we employ a one-tailed alternative hypothesis. In doing so, we compute the p-value as the probability of an outcome as extreme as the obtained outcome in only one direction.

Specifically, if the alternative hypothesis is represented only by outcomes which are more positive than that represented by the null, then the p-value is the probability of an outcome as extreme as the obtained outcome in the positive direction only.

And, if the alternative hypothesis is represented only by outcomes which are more negative than that represented by the null, the p-value is the probability of an outcome as extreme as the obtained outcome in the negative direction only.

Example: An experiment is designed to test the effect of consumption of a newly formulated alcohol on time to react to complex traffic situations. Since there is a tremendous accumulation of evidence that alcohol only slows decision times in complex situations, the two hypotheses would like be:

Null: Mean new alcohol reaction time = Mean No Alcohol reaction time

Alternative: Mean new Alcohol reaction time > Mean No Alcohol reaction time.

In this case, the null would be rejected only if the test statistic were positive and p ................
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