Ch 9 Hypothesis Testing with One Sample - Part II
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Transcript of Ch 9 Hypothesis Testing with One Sample - Part II
HYPOTHESIS TESTING WITH ONE SAMPLEChapter 9 – Part II
HYPOTHESIS TESTING STEPS So, how do we do hypothesis testing? Here are the steps in more detail:
1. State the problem2. Formulate the null and alternative hypotheses3. Choose the significance level (i.e. choose the probability of rejecting H0 when H0 is true)4. Determine the appropriate test statistic5. Compute the test statistic6. Find critical value(s) at the appropriate level of significance (traditional method) or
compute the p-value (p-value method)7. Compare test statistics to critical value(s) or p-value to level of significance 8. Based on this comparison reject or fail to reject H0
9. State your conclusions How do we do all this?
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
DISTRIBUTIONS NEEDED FOR HYPOTHESIS TESTING Recall the following three distributions
If you are testing a single population mean and the standard deviation is KNOWN then
If you are testing a single population mean and the standard deviation is UNKNOWN then
If you are testing a single population proportion then
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
CALCULATING THE TEST STATISTIC Test Statistic: A value computed from the sample data that is used in
making the decision of rejecting H0 (or not) When the hypothesis test is about the MEAN and the population standard
deviation is KNOWN, then the test statistic is
When the hypothesis test is about the MEAN and the population standard deviation is UNKNOWN, then the test statistic is
When the hypothesis test is about a PROPORTION, then the test statistic is
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
CALCULATING THE TEST STATISTIC The cost of a daily newspaper varies from city to city. However, the
variation among prices remains steady with a population standard deviation of 20 cents. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. 12 newspapers were bought and the mean cost was found to be 95 cents. Solution: This problem asks us to test if the mean cost of a daily newspaper is
$1.00. Further, we know the population standard deviation (Ç). So, we use z:¿.95−1.2
√12
¿−0.050.0577 ¿−0.866
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
CALCULATING THE TEST STATISTIC Registered nurses earned an average salary of $69,110. For that same
year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than $69,110. The sample average was $71,121, with a sample standard deviation of $7,489. Solution: This problem asks us to test if the mean salary for nurses is higher than
$69,110. Here, we do NOT know the population standard deviation (Ç). So, we use t:
s ¿71121−69110
7489√ 41
¿2011
1169.585 ¿1.719
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
CALCULATING THE TEST STATISTIC The US Department of Energy reported that 51.7% of homes were heated
by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Solution: This problem asks us to test if the proportion of households is equal to
51.7%. Since this is asking about a proportion, the test statistic is:
¿0.52−0.517
√ 0.517(1−0.517)221
¿0.0030.0336 .0892
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
CALCULATING THE TEST STATISTIC A particular brand of tires claims that its deluxe tire averages at least
50,000 miles before it needs to be replaced. From past studies of this tire, the population standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles.
Question: Which distribution do we want to use?
S = 9800
¿46500−50000
8000√28
¿−35001151.858 2.315
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
CALCULATING THE CRITICAL VALUES The Critical Region is the set of all values of the test statistic that cause
us to reject the null hypothesis A critical value is any value that separates the critical region (where we
reject H0) from the values of the test statistic that do not lead to the rejection of H0
Common choices for the significance level are 0.10, 0.05, and 0.01 So what does a critical region look like?
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
CRITICAL REGION
When Ha contains a ‘not equal to’; i.e.
So the red area is the critical region
If our test statistic is inside the red area, then we reject H0
If our test statistic is inside the blue area, then we fail to reject H0
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
CRITICAL REGION
When Ha contains a ‘less then’;
i.e. Note that tells us which tail: <
So the red area is the critical region
If our test statistic is inside the red area, then we reject H0
If our test statistic is inside the blue area, then we fail to reject H0
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
CRITICAL REGION
When Ha contains a greater then; i.e. Note that tells us which tail: >
So the red area is the critical region
If our test statistic is inside the red area, then we reject H0
If our test statistic is inside the blue area, then we fail to reject H0
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
CRITICAL REGION So the critical region gives us a rule for when we should reject (or fail to reject the
null hypothesis) If our test statistic is inside the critical region (the red) than we REJECT the null
hypothesis If our test statistic is not inside the critical region then we FAIL TO REJECT the null
hypothesis This is equivalent to saying if the absolute value of our test statistic is bigger than
the absolute value of the critical value then we REJECT the null hypothesis, i.e.
If not, then we FAIL TO REJECT the null hypothesis
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
EXAMPLES A particular brand of tires claims that its deluxe tire averages at least
50,000 miles before it needs to be replaced. From past studies of this tire, the population standard deviation is known to be 8,000. A survey of owners of that tire is conducted. From 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using » = 0.05, calculate the critical region. Test statistic was z = -2.315 (earlier example)
tells us “Left Tail Test” The value of
REJECT since our test statistic is inside the critical region
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
EXAMPLES Registered nurses earned an average salary of $69,110. For that same
year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than $69,110. The sample average was $71,121, with a sample standard deviation of $7,489. Using » = 0.05, calculate the critical region. Test statistic was t = 1.719 (earlier example)
tells us “Right Tail Test” Find
df = n-1 = 40
REJECT since our test statistic is inside the critical region
=1.645
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
EXAMPLES The US Department of Energy reported that 51.7% of homes were heated by
natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Using , calculate the critical region.
The tells us “Two Tail Test” Test statistic was z = 0.0892
Fail to reject since our test statistic is outside (or not inside) the critical region
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
RARE EVENTS Suppose you make an assumption about a property of the population (this
assumption is the null hypothesis) Then you gather data from a random sample If the sample has properties that would be very unlikely to occur if the
assumption is true, then you would conclude that your assumption about the population is probably incorrect
So we need some sort of way to quantify how unlikely our assumption is if the null hypothesis is true
We will use the sample data to calculate the ACTUAL probability of getting the test result, called the p-value
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
P-VALUES P-value: The p-value is the probability that, if the null hypothesis is true,
the results from another randomly selected sample will be as extreme or more extreme as the results obtained from the given sample
So the p-value, in some sense, measures how likely our assumption about some property of the population is to be true
A p-value is always a number between 0 and 1 A large p-value indicates that we should FAIL TO REJECT the null hypothesis A small p-value indicates that we should REJECT the null hypothesis
If p is small, reject the null How do we calculate the p-value?
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
CALCULATING THE P-VALUE When Ha contains a ‘not equal
to’; i.e. So the red area is the p-value
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
CRITICAL REGION
When Ha contains a ‘less then’;
i.e. So the red area is the p-value
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
CRITICAL REGION
When Ha contains a ‘greater then’; i.e.
So the red area is the p-value
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
P-VALUES So what is a large (or small) p-value? Any p-value that is smaller than (or equal to) our level is considered small Thus, if our p-value then we REJECT the null hypothesis
If p is small, reject the null! Also, if our p-value > then we FAIL TO REJECT the null hypothesis
Remember, we never ACCEPT H0, we always FAIL TO REJECT H0
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
FINDING P-VALUE ON YOUR CALCULATOR STATTESTS Choose your test:
Z-Test (Population Standard Deviation KNOWN) T-Test (Population Standard Deviation UNKNOWN) 1-PropZTest (Proportion)
Unless you are putting the data in, choose STATS Set your value Enter the other data depending on the type of test ( Choose the value for the alternative hypothesis ( Choose ‘Calculate’
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the population standard deviation is known to be 8,000. A survey of owners of that tire is conducted. From 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using » = 0.05, is the data highly consistent with the claim?
The Population SD is known, so Z-test Choose Stats (not Data)
n: 28
Calculate - Enter Since p-value = 0.0104 < , we REJECT (If p is small, reject the null!) That is, our sample has properties that would be very unlikely to occur if the assumption
is true, thus we conclude that our assumption, , is probably incorrect Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is
less than 50,000 miles.
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
Registered nurses earned an average salary of $69,110. For that same year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than $69,110. The sample average was $71,121, with a sample standard deviation of $7,489. Using » = 0.05, calculate the critical region.
Test statistic was t = 1.719 (earlier example)
df = n-1 = 40 p-value = ? (Find it)
0.0436
Since p-value < , we should reject the Conclusion: At the 5% significance level, there is sufficient evidence to conclude
that the mean salary of California registered nurses exceeds $69,110.
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
The US Department of Energy reported that 51.7% of homes were heated by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Using , calculate the critical region.
“Two Tail Test” Test statistic was z = 0.0892 p-value =
0.9203 Since p-value > we should fail to reject Conclusion: There is not enough evidence to conclude that the proportion
of homes in Kentucky that are heated by natural gas is different from 0.517
(0.46015)2=.9203
0.46015 0.46015
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
RULES TO REMEMBER If our test statistic is inside the critical region then we REJECT the null hypothesis If our test statistic is outside the critical region then we FAIL TO REJECT the null
hypothesisALTERNATIVELY If our p-value then we REJECT the null hypothesis If our p-value then we FAIL TO REJECT the null hypothesis
BOTH WAYS LEAD TO THE SAME SOLUTION!
We never ACCEPT , we always FAIL TO REJECT
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing
HOMEWORK Page 507: 62, 68, 69, 70, 75, 78, 104, 112
Null and Alternative Hypothesis Type I and Type II Errors Hypothesis Testing