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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

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

- 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

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

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

- 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

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

PP 3

p

Normal Distribution as an Approximation to the BinomialSampling 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

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

Reject

E(p) = π0

p

-Z⍺/2

Z⍺/2

Z

Standard Error =

Hypothesis about the Population ProportionAssuming 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

p

Proportion Problem - OTC MedSampling 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

Look at t- table

PP 3

.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

Standard Normal

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

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