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Hypothesis Testing: One Sample Mean or Proportion. P - Values, Sections Section 10.3 Testing a Proportion, Section 10.6 Power of a Test, Section 10.7. P-value Approach. State null and alternative hypotheses H 0 :  = 8000 H 1 :   8000 (two sided test)

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hypothesis testing one sample mean or proportion

Hypothesis Testing: One Sample Mean or Proportion

P - Values, Sections Section 10.3

Testing a Proportion, Section 10.6

Power of a Test, Section 10.7

p value approach
P-value Approach
  • State null and alternative hypotheses
    • H0:  = 8000
    • H1:  8000 (two sided test)
  • Set level of significance, ⍺ = 0.05
    • Do not find the critical values
  • Calculate appropriate test statistic
    • Z = (7814.1-8000)/94.28 = -1.97

PP 3

p value
P-value
  • The p-value tells us the probability of observing a test statistic as extreme or more extreme than the one we observed if the null hypothesis is true
  • In this example, the p-value tells us the probability of observing the test statistic of -1.97 or one more extreme if  = 8000
    • Think more extreme as |1.97|

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what is the p value of 1 97

Sampling Distribution of Sample Means, Normally Distributed

.4756

8185.5

7814.1

-1.97

0

1.97

What is the p-value of |1.97|?
  • Need P(0 < z < 1.97) = 0.4756
    • Standard normal table
  • In one tail = .5000 - .4756 = .0244
  • In two tails = .0244*2 = .0488
  • p-value = .0488
  • This tells us that if the null hypothesis is true we would observe a test statistic as extreme as 1.97 in 4.88 samples out of 100 samples (4.88% of the time)

.0244

.0244

Z

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making the statistical decision
Making the Statistical Decision
  • Decide whether the p-value is small and implies the sample mean is a rare event if the null hypothesis is true or whether the p-value is large and implies the sample mean could be expected to occur if the null hypothesis is true
  • Use our conventional levels of alpha (levels of significance) as benchmarks
  • Compare the p-value we calculate from the data with the level of significance

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compare the p value with
Compare the p-value with ⍺
  • If  = .05, then
    • If p-value < .05, reject the null hypothesis
    • If p-value ≥ .05, do not reject the null hypothesis
  • If  = .01, then
    • If p-value < .01, reject the null hypothesis
    • If p-value ≥ .01, do not reject the null hypothesis

PP 3

p value and critical values consistent
P-value and Critical Values - Consistent
  • If the test statistic (|1.97|) is greater than the critical value (|1.96|), the p-value (.0488) must be smaller than .05
  • We will reject the null hypothesis whether we use a critical value or a p-value approach

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observed significance level
Observed Significance Level
  • A p-value is:
    • the exact level of significance of the test statistic
    • the smallest value a can be and still allow us to reject the null hypothesis
    • the amount of area left in the tail beyond the test statistic for a one-tailed hypothesis test or
    • twice the amount of area left in the tail beyond the test statistic for a two-tailed test.
    • the probability of getting a test statistic from another sample that is at least as far from the hypothesized mean as this sample statistic is

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one tail tests
One Tail Tests
  • The researcher is interested in whether the population parameter is lower (or higher) than the stated value in the null
  • Candy manufacturer claims there are 100 candies on average in each $1.00 bag
  • After sampling a number of bags, you are concerned that this claim may be an overstatement
  • Do we care if there are more than 100 candies in a bag on average?

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example of lower tail test
Example of Lower Tail Test
  • Form the alternative hypothesis first, since it embodies the challenge
    • H0: μ = 100 or H0: μ ≥ 100 (No evidence to reject the manufacturer’s claim)
      • The equality is always in the null
    • H1: μ < 100 (The manufacturer’s claim is rejected; Mean number of candies is less than 100)

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lower tail test
Lower Tail Test
  • The rejection region is in the lower tail of the distribution
  • All of ⍺ is placed in the one tail
  • Reject the null
    • If we observe a small sample mean that is unlikely to occur if μ = 100

Sampling Distribution under the null hypothesis

reject

Do not reject

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example of upper tail test
Example of Upper Tail Test
  • A company is thinking about setting up an on-site day-care program for its employees. The CEO has stated that she will do so only if more than 80% of the employees favor such a decision.
    • When will an action take place?
      • H0: π ≤ 0.80
      • H1: π > 0.80

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two sided tests
Two Sided Tests
  • Always use = and ≠ in the statistical hypotheses
  • Directionless in that the alternative hypothesis allows for either the greater than (>) or less than (<) possibility

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one tailed tests
One tailed tests
  • Are always directional
    • The alternative hypothesis uses either the greater than (>) or less than (<) sign
  • Only used when the researcher
    • Knows for certain that the outcome of an experiment is going to occur only in one direction
    • Is only interested in one direction of the experiment
  • In one-tailed problems, the researcher is trying to demonstrate that something is
    • Older, younger, higher, lower, more, less, greater, less than, has increased, has decreased and so on
  • The directionality to be demonstrated is placed in the alternative hypothesis

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problem lifetime of light bulbs
Problem – Lifetime of Light Bulbs
  • The advertising campaign of a light bulb manufacturer claims that the mean lifetime of their bulbs is 1600 hours
    • The known population standard deviation is  = 120 hours
  • In your job for Consumer Report you have been asked to check whether the company is making false claims
  • You obtain a sample of 100 bulbs
    • The mean lifetime of the sample is 1,570 hours
  • Let  = .05
  • Sample means are normally distributed via the CLT

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problem lifetime of light bulbs1
Problem – Lifetime of Light Bulbs
  • H0: ≥ 1600
    • The company’s claim appears valid
  • H1:  < 1600
    • The company’s claim appears to be an overstatement
  • Find critical value Z.05 = -1.64
    • Look for probability of .4500 in standard normal table
    • Look for probability of .05 in t table for infinite df
  • DR: if (Z test statistic ≥ -1.64) Do not reject H0
  • Z = (1570 - 1600)/(120/100) = -2.50

PP 3

lifetime of light bulbs critical value approach

Sampling Distributionof the sample mean under the null hypothesis

⍺ = 0.05

Do not reject

1570

1600

-2.50

-1.64

Z

Lifetime of Light Bulbs: Critical Value Approach
  • DR: if (Z test statistic ≥ -1.64) do reject H0
  • Calculate test statistic
    • Z = (1570 - 1600)/(120/100) = -2.50
  • Reject the H0
  • The company appears to be overstating the lifetime of its bulbs

-1.64

PP 3

lifetime of light bulbs p value approach
Lifetime of Light Bulbs: p-value Approach
  • H0: ≥ 1600 (The company’s claim appears valid)
  • H1:  < 1600 (The company’s claim appears to be an overstatement)
  • Sample means are normally distributed via the CLT
  • Let ⍺ = 0.05
  • Find test statistic:
    • Z = (1570 - 1600)/(120/100) = -2.50

PP 3

lifetime of light bulbs p value approach1

-2.50

0

Lifetime of Light Bulbs: p-Value Approach
  • How likely are we to observe a test statistic of -2.50 or one more extreme, if the null hypothesis is true?
  • Standard Normal Table
    • P(0 < Z < 2.50) =.4938
  • Calculate area in tail
    • .5000 - .4938 = .0062
  • Compare with level of significance
    • .0062 < .05, reject null

Sampling Distributionof the sample mean under the null hypothesis

p-value = .0062

.4938

1570

1600

Z

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the t test statistic
The t-Test Statistic
  • Consider the population standard deviation, , is unknown
  • Sampling distribution of Sample Means is normal because
    • Population follows the normal distribution
    • Or the sample size is greater than 30 (CLT)
  • How do we estimate the standard error of the sample means?
    • Substitute S, the sample standard deviation, for 

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t test problem soft drink
T-test Problem: Soft Drink
  • A soft drink manufacturer wants to test whether the mean rating for a new flavor that it has just developed equals 60. From many previous studies it is known that its old flavor has a mean rating of 60.
  • If the mean rating for the new flavor is higher than 60, the manufacturer will substitute the new flavor for the old and stop producing the latter.
  • The manufacturer samples 16 people and obtains ratings on a scale of 1 to 90. The sample mean is 62.38 and the sample standard deviation is 7.6.
  • Should the manufacturer continue to produce the old flavor or should it shift production to the new flavor?
  • The ratings are normally distributed in the population. Let  = .05

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t test problem
T-test Problem
  • Where is the action?
    • If the mean rating for the new flavor is higher than 60, the manufacturer will substitute the new flavor for the old and stop producing the latter
    • H0: μ ≤ 60 (The new is the same or worse than the old. Do not switch)
    • H1: μ > 60 (The new is better than the old. Switch)
      • One sided, upper tail test

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t test problem1

Sampling Distributionof the sample mean under the null hypothesis

62.38

1.25

1.753

T-test Problem
  • Estimate the standard error
  • Calculate the test statistict15 = (62.38 - 60)/1.9 = 1.25
  • Find the CV – t table
    • t15,.05 = 1.753
  • Do not reject
  • Do not switch to new

reject

⍺ = .05

Do not reject

60

Z

PP 3

testing hypotheses about the population proportion
Testing Hypotheses about the Population Proportion
  • Proportions are used for qualitative data
    • Observations fall into one category or another
      • Will you vote for the candidate: yes or no
      • Do you like the flavor: yes or no
    • We are interested in some characteristic of the observation
  • Population proportion is of interest
    • π = number of observations with characteristic/ population size = x/N

PP 3

testing hypotheses about the population proportion1
Testing Hypotheses about the Population Proportion
  • Have the sample information
    • p = observations with characteristic/ sample size = x/n
  • If we had drawn a different sample, what might p in that sample be?
  • Need to understand the sampling distribution of the sample proportion
  • Can calculate exactly what the sampling distribution looks like

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the sampling distribution of the sample proportion
The Sampling Distribution of the Sample Proportion
  • Suppose I have a large population
    • Fifty percent vote yes on an issue and fifty percent vote no
    • Interested in the proportion voting yes (a success)
      • Let π = 0.50
  • Draw a sample of n = 5
    • What would the distribution of sample proportions look like?
    • Answer lies with the binomial distribution
      • “What is the probability of observing exactly 0 (1, 2, 3, 4, 5) “successes” in 5 trials if the probability of success on a single trial is 0.5?

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the binomial distribution
The Binomial Distribution

Use the Binomial Distribution tables in the appendix of Weiers to find the probability of observing exactly 0 successes in 5 trials, or 1 success in 5 trials and so forth.

x = number of successes

Proportion of successes

Probability of x successes

Using the binomial distribution, we can calculate the probability of observing certain sample proportions, given the population proportion is some specific value π = 0.50 and given the sample size, n = 5.

PP 3

the sampling distribution of the sample proportion1

Sampling Distribution of the sample proportion, p, when n = 5 and π = 0.50

P(x/n=p)

= p

The Sampling Distribution of the Sample Proportion
  • Is a binomial distribution
  • Not a continuous distribution
  • E(x/n) = E(p) = π
  • VAR(p) = π(1 - π)/n

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normal distribution as an approximation to the binomial

E(p) = π

p

Normal Distribution as an Approximation to the Binomial

Sampling Distribution of p when n is “large”

  • Sampling distribution of the proportion will be approximately normal with mean π and standard deviation p
    • Conditions to be met
      • nπ > 5 when π 1/2 and n(1 - π) > 1/2 when π > 1/2
      • Or sample size greater than 100

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hypotheses about the population proportion
Hypotheses about the Population Proportion
  • Formulate the hypotheses
    • H0: π = π0
    • H1: π ≠ π0
      • or H1: π > π0 or H1: π < π0
  • Set ⍺ and look-up Z critical value(s) in table
    • Form the rejection region assuming the null hypothesis is correct
  • Or set ⍺ and use the p-value approach

PP 3

hypothesis about the population proportion

Reject

Reject

E(p) = π0

p

-Z⍺/2

Z⍺/2

Z

Standard Error =

Hypothesis about the Population Proportion

Assuming the null hypothesis is correct, two sided test

  • Calculate the test statistic
    • Use π0in the standard error
  • If the test statistic is extreme, reject the H0
    • DR: if (-Z⍺/2 ≤ Z test statistic ≥ Z⍺/2 ) do not reject H0

PP 3

proportion problem
Proportion Problem
  • The manufacturer of an over-the-counter medicine claims that it is 90% effective in relieving an allergy for a period of 8 hours
  • In a sample of 200 persons who have the allergy, the medicine provided relief for 160 people
  • Determine whether the manufacturer’s claim is legitimate (overstating value?)
    • Let  = .01

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proportion problem otc med
Proportion Problem - OTC Med
  • Characteristic of interest
    • Obtaining relief
  • H0: π ≥ 0.90
    • The claim is correct and any observed sample difference is due to chance
  • H1: π < 0.90
    • The claim is false and the observed sample proportion is unlikely to have come from a population with a π of 0.90

PP 3

proportion problem otc med1

Reject

p

Proportion Problem - OTC Med

Sampling Distribution of p

  • Find critical value
    • Z.01 = -2.33
    • DR: if (Z test statistic ≥ -2.33) Do not reject H0
  • What do we observe in the sample?
    • p = x/n = 160/200 = 0.80
  • Create test statistic
    • Z = (.80 - .90)/ 0212 = -4.71
    • Statistical decision
      • Reject the H0
    • Conclusions
      • The claim is not legitimate

p

PP 3

p value approach1

.80

.90

p

- 4.71

0

Z

p - Value Approach
  • How likely are we to observe a sample proportion of 0.80 or one more extreme if the population proportion is 0.90?
  • Standard normal table
    • P(0 < Z < 4.71) ≈ 0.50
  • There is ≈ .000 in the tail
    • p - value = .000
  • p - value < .01, Reject

.000

≈ 0.50

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weier s z tables
Weier’s Z Tables

≈ 0.500

≈ .000

- 3.10

0

- 3.10

Z

Any Z value beyond 3.09 can be viewed as a percentile point that contains roughly 50% of Z values between it and 0

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