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Chapter 10

One-Sample Tests of Hypothesis

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Goals1. Define a hypothesis and hypothesis testing

2. Describe the five step hypothesis testing procedure

3. Distinguish between a one-tailed and a two-tailed test of hypothesis

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Goals4. Conduct a test of hypothesis about a

population mean

5. Conduct a test of hypothesis about a population proportion

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Goals

Chapters 1 – 9

Everything we have learned so far, we can use in this chapter!

Chapters 10

We will test peoples claims by running experiments and then

conclude whether the initial claims are reasonable or not!

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Define A Hypothesis A Hypothesis is a statement about the value of a

population parameter developed for the purpose of testing.

Examples of hypotheses made about a population parameter are: The mean monthly income for systems analysts is

$3,625 Twenty percent of all customers at Bovine’s Chop

House return for another meal within a month The mean yearly salary for a real estate agent is

$85,000

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Define Hypothesis TestingHypothesis testing is a procedure, based on

sample evidence and probability theory, used to determine whether:The hypothesis is a reasonable statement

and should not be rejectedor

The hypothesis is unreasonable (not reasonable) and should be rejected

Parallel Examples: Legal System, Doctors

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Define Hypothesis Testing Illustration (Example 5): The hypothesized mean yearly salary earned by

full-time realtors is $85,000 (sigma = $12,549) The mean salary is not significantly different from

$85,000 If we take a sample and get a sample mean of

$88,595, we must make a decision about the difference of $3,595 Is it a true difference

or Is it sampling error

Five steps to hypothesis testing

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5 Steps In Hypothesis Testing Procedure• Step 1: State Null Hypothesis (H0)and Alternate

Hypothesis (H1).• Step 2: Select a Level of Significance (alpha).• Step 3: Select the Test Statistic (z or t). (Draw picture

and calculate Critical Value).• Step 4: Formulate the Decision Rule based on steps

1-3.• Step 5: Make a decision about the Null Hypothesis

based on sample information. (Take a random sample, compute the test statistic, compare it to critical value, and make decision to reject or not reject null hypotheses). Interpret the results of the test.

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Step 1: State Null Hypothesis (H0)and Alternate Hypothesis (H1)

Null Hypothesis = H0

“H sub zero,” or “H sub not” A statement about the value of a population

parameter This is the value we test After the experiment, we will either:

Reject H0, and accept alternative hypothesis

Fail to reject H0 (Does not prove that H0 is true)

Example: H0 : μ ≤ $85,000

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Alternative Hypothesis = H1

“H sub one” A statement that is accepted if the sample data

from the experiment rejects the null hypothesis

If H1 taken: Reject H0

H1 replaces H0

If H1 rejected: Fail to reject H0

Assumption, H0, holds

Example: H1 : μ > $85,000

Step 1: State Null Hypothesis (H0)and Alternate Hypothesis (H1)

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The trick to hypothesis testing is translating the words into , <, > for H1 It is usually easier to state H1 first and then state

H0 .H0 will always have the equal sign: =, ≥, ≤H1 will never have the equal sign: , >, <Step 1 for Example 5:

If someone claims: “The mean yearly salary earned by full-time realtors is more than $85,000” Write H1 > $85,000 first Then write H0 ≤ $85,000

Step 1: State Null Hypothesis (H0)and Alternate Hypothesis (H1)

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H0 : The mean yearly salary earned by full-time realtors is $85,000 Second we would write H0 : µ ≤ $85,000 Why ≤ ?

In our test, anything equal to or less than $85,000 contradicts µ > $85,000!

H1 : The mean yearly salary earned by full-time realtors is more than $85,000 First we would write H1 : µ > $85,000

Step 1: State Null Hypothesis (H0)and Alternate Hypothesis (H1)

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Step 2: Select a Level of Significance (alpha)

Level of Significance () The probability of rejecting the null hypothesis

when it is actually true “The risk we are willing to take in committing an

error” The lower the number, the lower the risk of

committing an error Standards for Level of Significance:

= .10 (political polls) = .05 (consumer research project) = .01 (quality assurance)

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Type I Error Rejecting the null hypothesis when it is actually

true

H0 was true, but experiment rejected H0

“Innocent, but found guilty”Type II Error β

Accepting the null hypothesis when it is actually false

H0 was false, but experiment failed to reject H0

“Guilty, but found not guilty”

Errors

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Step 3: Select the Test Statistic (z or t). (Draw picture and

calculate Critical Value)1. Is the critical value and the test statistic z or t?

2. Given level of significance, look up critical value in table

The dividing point between the region where the null hypothesis is rejected and the region where it is not rejected

3. The actual test statistic will be calculated in step 5 A value, determined from sample information, used to

determine whether or not to reject the null hypothesis

Before we can draw our picture, we must distinguish between a one-tailed and a two-tailed test of hypothesis

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One-tailed Tests Of Significance A test is one-tailed when the H1 uses < or > Step 2 for Example5:

H1 : The mean yearly salary earned by full-time realtors is more than $85,000 H0 : µ ≤ $85,000 H1 : µ > $85,000 One tailed test to the right

Another example: H1 : The mean speed of trucks traveling on I-5 in

Washington is less than 60 miles per hour H0 : µ ≥ 60 H1 : µ < 60 One tailed test to the left

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One Tailed Test To The Right; H1 uses Greater Than “>”

Step 3 for Example 5: H0 : µ ≤ 85,000; H1 : µ > 85,000

-3 -2 -1 0 1 2 3 z

a

Reject H0,

accept H1

If our experiment shows a value within this range, we fail to reject H0

Critical value (z 1.65)

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One Tailed Test To The LeftH1 uses Less Than “<“

H0 : µ ≥ some number

H1 : µ < some number

-3 -2 -1 0 1 2 3 z

a

Reject H0,

accept H1

If our experiment shows a value

within this range, we fail to reject H0

Critical value (z or t)

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Two-tailed Tests Of Significance A test is two-tailed when H1 uses Examples:H1 : The mean amount spent by customers at

the Wal-Mart in Georgetown is not equal to $25 H0 : µ = $25 H1 : µ $25

H1 : The mean price for a gallon of gasoline is not equal to $3.55 H0 : µ = $3.55 H1 : µ $3.55

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-3 -2 -1 0 1 2 3 z

/a 2 /a 2

If our experiment shows a value within this range,

we fail to reject H0

Reject H0,

accept H1

Reject H0,

accept H1

Two Tailed TestNot Equal

H0 : µ = some number

H1 : µ some number

Critical value (z or t)

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Step 4: Formulate the Decision Rule based on steps 1-3

State decision rule as: If H1 passes some test, we reject H0 and accept H1,

otherwise H0 is not rejected Step 4 for Example 5:

=.05, z = 1.65 H0 : µ ≤ $85,000 H1 : µ > $85,000 Decision Rule: If test statistic is greater than 1.65, then

we reject H0 and accept H1, otherwise we fail to reject H0

Good statisticians go through all the steps up to writing down the decision rule before they look at the results

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Step 5: Make a decision about the Null Hypothesis based on sample

information. (Take a random sample, compute the test statistic, compare it to critical value, and make decision to reject or not reject

null hypotheses). Interpret the results of the test.

Take sample, calculate mean and standard deviation, then calculate the test statistic and come to a conclusion

The two possible conclusions are: “Do not reject H0

or Reject H0, accept H1

State decision as: “ The experiment supports the claim that H1” “ The experiment does not support the claim that H1” “Evidence indicates…” “Data shows…”

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Compute The Test Statistic:Testing For The Population Mean

zX

/ n ns

Xt

/

σ known σ unknown

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Failing to reject H0 does not prove H0, it simply means that we have failed to disprove H0

Rejecting H0 does not prove H1, it simply means that H0 is not reasonable

Step 5: Make a decision about the Null Hypothesis based on sample

information. (Take a random sample, compute the test statistic, compare it to critical value, and make decision to reject or not reject

null hypotheses). Interpret the results of the test.

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Sample of Real Estate Salaries1 855002 890003 920004 890005 845006 751007 862008 1050009 92100

10 68200 Xbar = 8859511 81230 Mu = 8500012 63050 s = 1254913 74200 n = 3614 8950015 98000 z = 1.71916 8450017 6945018 10050019 9452020 8920021 8750022 9160023 9000024 8850025 8900026 9350027 11060028 10200029 9400030 8400031 8650032 7950033 6845034 13050035 8960036 93430

Calculations

Step 5 for Example 5:

$88,595 $85,0001.72

$12,549 36

Xzsigma n

Because 1.72 is greater than 1.645 we reject H0 and accept H1. The evidence suggests that the mean yearly salary is greater than $85,000.00. We can be reasonably sure that the mean salary is greater than $85,000.00. Because 0.0428 is less than 0.05 we reject H0 and accept H1

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P-value In Hypothesis Testing Assuming that the null hypothesis is true, a p-Value

is the probability of finding a value of the test statistic at least as extreme as the computed value for the test

p-Value < significance level: H0 is rejected If H0 is very small, there is little likelihood that H0

is true

p-Value > significance level: H0 is not rejected If H0 is very large, there is little likelihood that H0

is false

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Computation Of The P-value One-Tailed Test:

p-Value =P( z ≥ absolute value of the computed test statistic value) and compare to alpha

Two-Tailed Test: For a two-tail test, use alpha divided by 2 (alpha/2) and compare that

to the p-value from one side of the two tails p-Value = P( z ≥ absolute value of the computed test statistic value)

and compare p to alpha/2

OR For a two-tail test, use (2* p-value) and compare to alpha

p-Value = 2*P( z ≥ absolute value of the computed test statistic value) and compare p to alpha

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Interpreting The Weight Of Evidence Against H0

If the p-value is less than:

.10, we have some evidence that H0 is not true

.05, we have strong evidence that H0 is not true

.01, we have very strong evidence that H0 is not

true

.001, we have extremely strong evidence that H0

is not true

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Conduct A Test Of Hypothesis About A Population Mean

Hypothesis Testing Example 1 The processors of Fries’ Catsup indicate on the label

that the bottle contains 16 ounces of catsup The standard deviation of the process is 0.5 ounces A sample of 36 bottles from last hour’s production

revealed a mean weight of 16.12 ounces per bottle At the .05 significance level is the process out of

control? Can we conclude that the mean amount per bottle is

different from 16 ounces?

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Hypothesis Testing Example 1

Step 1: State the null and the alternative hypotheses

H0 : μ = 16

H1 : μ 16

Step 2: Select a level of significance

We select .05 significance level (.05/2 = .025)

Step 3: Identify the test statistic and draw

Because we know the population standard deviation and n ≥ 30, the test statistic is z

. .025 .5 - .025 = .475 leads us to +/-1.96

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Step 4: Formulate the decision rule

If our test statistic is outside the range -1.96 to 1.96, Reject H0 and accept H1, otherwise we fail to reject H0

Hypothesis Testing Example 1

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Step 5: Sample, Compute Test Statistic And Compare

To Critical Value, Reject Or Not Reject Ho

Because 1.44 is less that 1.96 and greater than -1.96, we do not reject H0

The evidence suggests that the mean is not different from 16 oz.

The difference between 16.12 & 16.00 can be attributed to sampling variation

The processor’s claim that the bottles contain 16 ounces of catsup seems reasonable

44.1365.0

00.1612.16

n

Xz

Hypothesis Testing Example 1

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Computation Of The P-value Two tailed test from example 1:

Significance level = .05 z = 1.44

The p-Value = 2P( z ≥ 1.44) = 2(.5-.4251) = .1498

Because .1498 > .05, do not reject H0

***Note, if you use Student’s t Distribution to estimate p-value, from textbook: “The usual practice is to report that the p-value is less than the larger of the two significance levels…”

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Hypothesis Testing Example 2 Roder’s Discount Store chain issues its own credit

card Lisa, the credit manager, wants to find out if the

mean monthly unpaid balance is more than $400 The level of significance is set at .05 A random check of 172 unpaid balances revealed:

Sample mean = $407 Sample standard deviation = $38 Should Lisa conclude that the population mean is greater

than $400, or is it reasonable to assume that the difference of $7 ($407-$400) is due to chance?

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Step 1: State the null and the alternative hypotheses

H0: µ ≤ $400

H1: µ > $400

Step 2: Select a level of significance

We select .05 significance level

Step 3: Identify the test statistic and draw

Because sigma is not known, the test statistic is t .05 leads us to t = 1.65 (one-tail test to right)

Hypothesis Testing Example 2

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Step 4: Formulate the decision rule

If our test statistic is greater than 1.65 (t >1.65), Reject H0 and accept H1, otherwise we fail to reject H0

Step 5: Sample, Compute Test Statistic And Compare To Critical Value,

Reject Or Not Reject Ho

Because 2.42 > 1.65, H0 is rejected and H1 is accepted The evidence indicates that the mean unpaid > $400 Lisa can conclude that the mean unpaid balance is greater

than $400 It is not reasonable to assume that a computed t-score of

2.42 is due to sampling variation

Hypothesis Testing Example 2

42.217238$

400$407$

ns

Xz

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The current rate for producing 5 amp fuses at Neary Electric Co. is 250 per hour

A new machine has been purchased and installed that, according to the supplier, will increase the production rate

A sample of 10 randomly selected hours from last month revealed:

Mean hourly production for new machine = 256 units Sample standard deviation = 6 per hour At the .05 significance level can Neary conclude that the

new machine is faster?

Hypothesis Testing Example 3

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Step 1: State the null and the alternative hypotheses

H0: µ ≤ 250

H1: µ > 250

Step 2: Select a level of significance

We select .05 significance level

Step 3: Identify the test statistic and draw

t distribution because σ is unknown There are 10 – 1 = 9 degrees of freedom .05 and df = 9 leads use to t = 1.833

Hypothesis Testing Example 3

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Step 4: Formulate the decision rule

If our test statistic is greater than 1.833 (t > 1.833), Reject H0 and accept H1, otherwise we fail to reject H0

Step 5: Sample, Compute Test Statistic And Compare To Critical

Value, Reject Or Not Reject Ho

Because 3.162 > 1.833, we reject Ho and accept H1 The evidence suggests that the mean number of amps

produced per hour is greater than 250

162.3106

250256

ns

Xt

Hypothesis Testing Example 3

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Hypothesis Testing Example 4 In the past, 15% of the mail order solicitations for a certain

charity resulted in a financial contribution A new solicitation letter that has been drafted is sent to a

sample of 200 people and 35 responded with a contribution Assume Experiment passes all the binomial tests At the .05 significance level can it be concluded that the

new letter is more effective?

Conduct A Test Of Hypothesis About A Population Proportion

Test statistic for testing a single population proportion: p = Sample proportion = Population proportion

n

pz

)1(

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Step 1: State the null and the alternative hypotheses

H0 : ≤ .15 H1 : > .15

Step 2: Select a level of significance

We select .05 significance level (one-tail to right)

Step 3: Identify the test statistic and draw

For Proportions that pass binomial test, we use z. .05 z = 1.65

Step 4: Formulate the decision rule

If our test statistic is greater than 1.65 (z >1.65), Reject H0 and accept H1, otherwise we fail to reject H0

Hypothesis Testing Example 4

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Step 5: Sample, Compute Test Statistic And Compare To Critical

Value, Reject Or Not Reject Ho

Because .990148 in not greater than 1.65, we fail to reject Ho

The evidence does not suggest that the new letters are more effective

990148.

200)15.1(15.

15.20035

)1(

n

pz

Hypothesis Testing Example 4