Hypothesis Testing: One Sample Mean or Proportion

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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

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
• 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
• 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|

PP 3

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

PP 3

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

PP 3

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
• 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

PP 3

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

PP 3

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?

PP 3

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)

PP 3

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

PP 3

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

PP 3

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

PP 3

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

PP 3

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

PP 3

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

⍺ = 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
• 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

-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

PP 3

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 

PP 3

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

PP 3

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

PP 3

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
• 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 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

PP 3

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?

PP 3

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

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

PP 3

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

PP 3

• 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

Reject

Reject

E(p) = π0

p

-Z⍺/2

Z⍺/2

Z

Standard Error =

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
• 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

PP 3

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

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

.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

PP 3

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

PP 3