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Chapter 10

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Chapter 10

One-Sample Tests of Hypothesis

- Define a hypothesis and hypothesis testing
- Describe the five step hypothesis testing procedure
- Distinguish between a one-tailed and a two-tailed test of hypothesis

- Conduct a test of hypothesis about a population mean
- Conduct a test of hypothesis about a population proportion

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!

- 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

- Hypothesis 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

- 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 differenceor
- Is it sampling error

- Five steps to hypothesis testing

- 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.

- 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 thatH0 is true)

- Example: H0 : μ ≤ $85,000

- 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 replacesH0

- If H1 rejected:
- Fail to reject H0
- Assumption, H0, holds

- Example: H1 : μ > $85,000

- 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

- If someone claims: “The mean yearly salary earned by full-time realtors is more than $85,000”

- 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

- 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)

- 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”

- Is the critical value and the test statistic z or t?
- 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

- 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

- 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

- H1 : The mean yearly salary earned by full-time realtors is more than $85,000
- 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

- H1 : The mean speed of trucks traveling on I-5 in Washington is less than 60 miles per hour

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

Critical value (z 1.65)

H0 : µ ≥ some number

H1 : µ < some number

Critical value (z or t)

- 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

H0 : µ = some number

H1 : µ some number

Critical value (z or t)

- 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

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 H0or
- 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…”

σ known

σ unknown

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.

- 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 for Example 5:

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

- 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

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

- p-Value =
- 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

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

- 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

- 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

- 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?

- Step 1:
- State the null and the alternative hypotheses
- H0 : μ= 16
- H1 : μ 16

- State the null and the alternative hypotheses
- Step 2:
- Select a level of significance
- We select .05 significance level (.05/2 = .025)

- Select a level of significance
- 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

- Identify the test statistic and draw

- 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

- Formulate the decision rule

- 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

- 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…”

- 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?

- Step 1:
- State the null and the alternative hypotheses
- H0: µ ≤ $400
- H1: µ > $400

- State the null and the alternative hypotheses
- Step 2:
- Select a level of significance
- We select .05 significance level

- Select a level of significance
- 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)

- Identify the test statistic and draw

- 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

- Formulate the decision rule
- 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

- 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?

- Step 1:
- State the null and the alternative hypotheses
- H0: µ ≤ 250
- H1: µ > 250

- State the null and the alternative hypotheses
- Step 2:
- Select a level of significance
- We select .05 significance level

- Select a level of significance
- 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

- Identify the test statistic and draw

- 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

- Formulate the decision rule
- 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

- 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

- Test statistic for testing a single population proportion:

- Step 1:
- State the null and the alternative hypotheses
- H0 : ≤ .15
- H1: > .15

- State the null and the alternative hypotheses
- Step 2:
- Select a level of significance
- We select .05 significance level (one-tail to right)

- Select a level of significance
- Step 3:
- Identify the test statistic and draw
- For Proportions that pass binomial test, we use z. .05 z = 1.65

- Identify the test statistic and draw
- 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

- Formulate the decision rule

- 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