Hypothesis TestsHypothesis Tests

One Sample Means

Example 2: A government agency has received numerous complaints that a particular restaurant has been selling underweight hamburgers. The restaurant advertises that it’s patties are “a quarter pound” (4 ounces).

How can I tell if they really are underweight?

Take a sample & find x.

But how do I know if this x is one that I expectexpect to happen or

is it one that is unlikelyunlikely to happen?

A hypothesis test will help me decide!

What are hypothesis What are hypothesis tests?tests?

Calculations that tell us if a value, x, occurs by random chance or not – if it is statistically significantIs it . . .

–a random occurrence due to natural variation?

–a biased occurrence due to some other reason?

Statistically significant means that it is NOTNOT a random

chance occurrence!

Is it one of the sample means

that are likely to occur?

Is it one that isn’t likely to

occur?

Nature of hypothesis tests Nature of hypothesis tests --•First begin by supposing the

“effect” is NOT present•Next, see if data provides

evidence against the supposition

Example: murder trial

How does a murder trial work?

First - assume that the person is innocentThen – mustmust have

sufficient evidence to prove guilty

Hmmmmm …Hypothesis tests use the same process!

Steps:Steps:

1) Assumptions2) Hypothesis statements &

define parameters3) Calculations4) Conclusion, in context

Notice the steps are the same except we add

hypothesis statements – which you will learn

today

Assumptions for z-test (t-Assumptions for z-test (t-test):test):• Have an SRS of context• Distribution is (approximately)

normal– Given– Large sample size– Graph data

• is known (unknown)

YEA YEA –These are the same

assumptions as confidence intervals!!

Example 1: Bottles of a popular cola are supposed to contain 300 mL of cola. There is some variation from bottle to bottle. An inspector, who suspects that the bottler is under-filling, measures the contents of six randomly selected bottles. Are the assumptions met?

299.4 297.7 298.9 300.2 297 301

•Have an SRS of bottles•Sampling distribution is approximately normal because the boxplot is symmetrical• is unknown

Writing Hypothesis Writing Hypothesis statements:statements:

• Null hypothesis – is the statement being tested; this is a statement of “no effect” or “no difference”

• Alternative hypothesis – is the statement that we suspect is true

H0:

Ha:

The form:The form:Null hypothesis H0: parameter = hypothesized value

Alternative hypothesis Ha: parameter > hypothesized value Ha: parameter < hypothesized value Ha: parameter = hypothesized value

Example 2: A government agency has received numerous complaints that a particular restaurant has been selling underweight hamburgers. The restaurant advertises that it’s patties are “a quarter pound” (4 ounces).State the hypotheses :

Where is the true mean weight of hamburger patties

H0: = 4

Ha: < 4

Example 3: A car dealer advertises that is new subcompact models get 47 mpg. You suspect the mileage might be overrated. State the hypotheses :

Where is the true mean mpg

H0: = 47

Ha: < 47

Example 4: Many older homes have electrical systems that use fuses rather than circuit breakers. A manufacturer of 40-A fuses wants to make sure that the mean amperage at which its fuses burn out is in fact 40. If the mean amperage is lower than 40, customers will complain because the fuses require replacement too often. If the amperage is higher than 40, the manufacturer might be liable for damage to an electrical system due to fuse malfunction. State the hypotheses :Where is the

true mean amperage of the fuses

H0: = 40

Ha: = 40

Facts to remember about Facts to remember about hypotheses:hypotheses:• ALWAYS refer to populations

(parameters)• The null hypothesis for the

“difference” between populations is usually equal to zero

• The null hypothesis for the correlation (rho) of two events is usually equal to zero.

H0: x-y= 0

H0: = 0

Activity: For each pair of hypotheses, indicate which are not legitimate & explain why

0:H;0:H e)6.:H;4.:H d)1.:H;1.:H c)123:H;123:H b)

15:H;15:H a)

a0

a0

a0

a0

a0

xx

Must use parameter (population) x is a statistics

(sample)

is the population proportion!Must use same

number as H0! is parameter for population correlation coefficient – but H0

MUST be “=“ !

Must be NOT equal!

P-values -P-values -

•Assuming H0 is true, the probability that the test statistic would have a value as extreme or more than what is actually observedIn other words . . . is it

far out in the tails of the distribution?

Level of significance -Level of significance - • Is the amount of evidence

necessary before we begin to doubt that the null hypothesis is true

• Is the probability that we will reject the null hypothesis, assuming that it is true

• Denoted by – Can be any value– Usual values: 0.1, 0.05, 0.01– Most common is 0.05

Statistically significant –• The p-value is as smallas small or

smaller smaller than the level of significance ()

• If p > , “fail to rejectfail to reject” the null hypothesis at the level.

• If p < , “rejectreject” the null hypothesis at the level.

Facts about p-values:• ALWAYS make decision about the

null hypothesis!• Large p-values show support for

the null hypothesis, but never that it is true!

• Small p-values show support that the null is not true.

• Double the p-value for two-tail (=) tests

• Never acceptNever accept the null hypothesis!

Never “accept” the null hypothesis!Never “accept” the null

hypothesis!

Never “accept” the null hypothesis!

At an level of .05, would you reject or fail to reject H0

for the given p-values?a) .03b) .15c) .45d) .023

Reject

Reject

Fail to rejectFail to reject

Draw & shade a curve & calculate the p-value:1) right-tail test t = 1.6; n = 20

2) left-tail test z = -2.4; n = 15

3) two-tail test t = 2.3; n = 25

P-value = .0630

P-value = .0082

P-value = (.0152)2 = .0304

Writing Conclusions:1) A statement of the decision

being made (reject or fail to reject H0) & why (linkage)

2) A statement of the results in context. (state in terms of Ha)

AND

“Since the p-value < (>) , I reject (fail to reject) the H0. There is (is not) sufficient evidence to suggest that Ha.” Be sure to write Ha

in context (words)!

Example 5: Drinking water is considered unsafe if the mean concentration of lead is greater than 15 ppb (parts per billion). Suppose a community randomly selects of 25 water samples and computes a t-test statistic of 2.1. Assume that lead concentrations are normally distributed. Write the hypotheses, calculate the p-value & write the appropriate conclusion for = 0.05.

H0: = 15Ha: > 15Where is the true mean concentration of lead in drinking waterP-value = tcdf(2.1,10^99,24) =.0232

t=2.1

Since the p-value < , I reject H0. There is sufficient evidence to suggest that the mean concentration of lead in drinking water is greater than 15 ppb.

Example 6: A certain type of frozen dinners states that the dinner contains 240 calories. A random sample of 12 of these frozen dinners was selected from production to see if the caloric content was greater than stated on the box. The t-test statistic was calculated to be 1.9. Assume calories vary normally. Write the hypotheses, calculate the p-value & write the appropriate conclusion for = 0.05.

H0: = 240Ha: > 240Where is the true mean caloric content of the frozen dinners

P-value = tcdf(1.9,10^99,11) =.0420

t=1.9Since the p-value < , I reject H0. There is sufficient evidence to suggest that the true mean caloric content of these frozen dinners is greater than 240 calories.

Formulas:known:

statistic of deviation standardparameter - statisticstatistic test

z =

x

n

Formulas: unknown:

statistic of deviation standardparameter - statisticstatistic test

t =

x

ns

Example 7: The Fritzi Cheese Company buys milk from several suppliers as the essential raw material for its cheese. Fritzi suspects that some producers are adding water to their milk to increase their profits. Excess water can be detected by determining the freezing point of milk. The freezing temperature of natural milk varies normally, with a mean of -0.545 degrees and a standard deviation of 0.008. Added water raises the freezing temperature toward 0 degrees, the freezing point of water (in Celsius). The laboratory manager measures the freezing temperature of five randomly selected lots of milk from one producer with a mean of -0.538 degrees. Is there sufficient evidence to suggest that this producer is adding water to his milk?

Assumptions:•I have an SRS of milk from one producer•The freezing temperature of milk is a normal distribution. (given)• is known

SRS?Normal?How do

you know?

Do you know

?H0: = -0.545Ha: > -0.545 where is the true mean freezing temperature of milk

What are your hypothesis

statements? Is there a key

word? 9566.1

5008.

545.538.

z Plug values

into formula.

p-value = normalcdf(1.9566,1E99)=.0252 Use normalcdf

to calculate p-value.

= .05

Conclusion: Compare your p-value to a & make

decisionSince p-value < , I reject the null hypothesis.

Write conclusion in context in terms of

Ha.

There is sufficient evidence to suggest that the true mean freezing temperature is greater than -0.545. This suggests that the producer is adding water to the milk.

Example 8: The Degree of Reading Power (DRP) is a test of the reading ability of children. Here are DRP scores for a random sample of 44 third-grade students in a suburban district:

(data on note page)At the = .1, is there sufficient evidence to suggest that this district’s third graders reading ability is different than the national mean of 34?

• I have an SRS of third-graders•Since the sample size is large, the sampling distribution is approximately normally distributedOR•Since the histogram is unimodal with no outliers, the sampling distribution is approximately normally distributed• is unknown

SRS?Normal?How do

you know?Do you know

?What are your

hypothesis statements? Is

there a key word?

6467.

44189.11

34091.35

t Plug values

into formula.

p-value = tcdf(.6467,1E99,43)=.2606(2)=.5212 Use tcdf to

calculate p-value.

= .1

H0: = 34 where is the true mean readingHa: = 34 ability of the district’s third-graders

Conclusion: Compare your p-value to & make

decisionSince p-value > , I fail to reject the null hypothesis.

Write conclusion in context in terms of Ha.

There is not sufficient evidence to suggest that the true mean reading ability of the district’s third-graders is different than the national mean of 34.

Example 9: The Wall Street Journal (January 27, 1994) reported that based on sales in a chain of Midwestern grocery stores, President’s Choice Chocolate Chip Cookies were selling at a mean rate of $1323 per week. Suppose a random sample of 30 weeks in 1995 in the same stores showed that the cookies were selling at the average rate of $1208 with standard deviation of $275. Does this indicate that the sales of the cookies is different from the earlier figure?

Assume: •Have an SRS of weeks•Distribution of sales is approximately normal due to large sample size• s unknownH0: = 1323 where is the true mean cookie salesHa: ≠ 1323 per week

Since p-value < of 0.05, I reject the null hypothesis. There is sufficient to suggest that the sales of cookies are different from the earlier figure.

0295.29.230

27513231208

valuept

Example 9: President’s Choice Chocolate Chip Cookies were selling at a mean rate of $1323 per week. Suppose a random sample of 30 weeks in 1995 in the same stores showed that the cookies were selling at the average rate of $1208 with standard deviation of $275. Compute a 95% confidence interval for the mean weekly sales rate.CI = ($1105.30, $1310.70)Based on this interval, is the mean weekly sales rate statistically different from the reported $1323?

What do you notice about the decision from the confidence interval & the

hypothesis test?What decision would you make on Example 10 if = .01?

What confidence level would be correct to use?

Does that confidence interval provide the same decision?

If Ha: < 1323, what decision would the hypothesis test give at = .02?

Now, what confidence level is appropriate for this alternative hypothesis?

You should use a 99% confidence level for a two-sided hypothesis test at = .01.

You would fail to reject H0 since the p-value > .

CI = ($1068.6 , $1346.40) - Since $1323 is in this interval we would fail to reject H0.

Remember your, p-value = .01475At = .02, we would reject H0.

The 98% CI = ($1084.40, $1331.60) - Since $1323 is in the interval, we would

fail to reject H0.

Why are we getting different answers?

In a one-sided test, all of goes into that tail (lower tail).

= .02

In a CI, the tails have equal area – so there should also be 2% in the upper tailThat leaves 96%96% in the middle & that should be your confidence confidence levellevel

.02.96

A 96% CI = ($1100, $1316).Since $1323 is not in the interval, we

would reject H0.

Tail probabilities between the significant level ()

and the confidence level MUST match!)

Matched Pairs Test

A special type of t-inference

Matched Pairs – two forms

• Pair individuals by certain characteristics

• Randomly select treatment for individual A

• Individual B is assigned to other treatment

• Assignment of B is dependent on assignment of A

• Individual persons or items receive both treatments

• Order of treatments are randomly assigned or before & after measurements are taken

• The two measures are dependent on the individual

Is this an example of matched pairs?

1)A college wants to see if there’s a difference in time it took last year’s class to find a job after graduation and the time it took the class from five years ago to find work after graduation. Researchers take a random sample from both classes and measure the number of days between graduation and first day of employmentNo, there is no pairing of individuals, you have two independent samples

Is this an example of matched pairs?

2) In a taste test, a researcher asks people in a random sample to taste a certain brand of spring water and rate it. Another random sample of people is asked to taste a different brand of water and rate it. The researcher wants to compare these samplesNo, there is no pairing of individuals, you have two independent samples – If you would have the same people taste both brands in random order, then it would bean example of matched pairs.

Is this an example of matched pairs?

3) A pharmaceutical company wants to test its new weight-loss drug. Before giving the drug to a random sample, company researchers take a weight measurement on each person. After a month of using the drug, each person’s weight is measured again.Yes, you have two measurements that are dependent on each individual.

A whale-watching company noticed that many customers wanted to know whether it was better to book an excursion in the morning or the afternoon. To test this question, the company collected the following data on 15 randomly selected days over the past month. (Note: days were not consecutive.)

Day 1 2 3 4 5 6 7 8 9 10 11

12

13

14

15

Morning 8 9 7 9 1

013 10 8 2 5 7 7 6 8 7

After-noon 8 10 9 8 9 1

1 8 10 4 7 8 9 6 6 9First, you must find the differences for

each day.

Since you have two values for each day, they are

dependent on the day – making this data matched

pairs

You may subtract either way – just be careful

when writing Ha

Day 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Morning 8 9 7 9 10 13 10 8 2 5 7 7 6 8 7After-noon 8 10 9 8 9 11 8 10 4 7 8 9 6 6 9

Differences 0 -1 -2 1 1 2 2 -2 -2 -2 -1 -2 0 2 -2

Assumptions:• Have an SRS of days for whale-watching• unknown•Since the normal probability plot is approximately linear, the distribution of difference is approximately normal.

I subtracted:Morning – afternoon

You could subtract the other way!

You need to state assumptions using the differences!

Notice the granularity in this plot, it is still displays a nice linear relationship!

Differences 0 -1 -2 1 1 2 2 -2 -2 -2 -1 -2 0 2 -2

Is there sufficient evidence that more whales are sighted in the afternoon?

Be careful writing your Ha!

Think about how you subtracted: M-A

If afternoon is more should the differences be

+ or -?Don’t look at numbers!!!!

H0: D = 0Ha: D < 0Where D is the true mean difference in whale sightings from morning minus afternoon

Notice we used D for differences

& it equals 0 since the null should be that there is NO

difference.

If you subtract afternoon – morning; then Ha: D>0

finishing the hypothesis test:

Since p-value > , I fail to reject H0. There is insufficient evidence to suggest that more whales are sighted in the afternoon than in the morning.

05.141803.

945.15

639.104.

dfp

nsxt Notice that if

you subtracted A-M, then your test statistic

t = + .945, but p-value would be the same

In your calculator, perform a t-test

using the differences (L3)

Differences 0 -1 -2 1 1 2 2 -2 -2 -2 -1 -2 0 2 -2