Hypothesis Tests Regarding a Parameter

Hypothesis Tests Regarding a Parameter
Chapter 10 Hypothesis Tests Regarding a Parameter

The Language of Hypothesis Testing
Section 10.1 The Language of Hypothesis Testing

A hypothesis is a statement regarding a characteristic of one or more populations.
Hypothesis testing is a procedure, based on sample evidence and probability, used to test statements regarding a characteristic of one or more populations. 3

Steps in Hypothesis Testing
Make a statement regarding the nature of the population. Collect evidence (sample data) to test the statement. Analyze the data to assess the plausibility of the statement. 4

The null hypothesis, denoted H0, is a statement to be tested
The null hypothesis, denoted H0, is a statement to be tested. The null hypothesis is a statement of no change, no effect, or no difference and is assumed true until evidence indicates otherwise. The alternative hypothesis, denoted H1, is a statement that we are trying to find evidence to support. 5

In this chapter, there are three ways to set up the null and alternative hypotheses:
Equal versus not equal hypothesis (two-tailed test) H0: parameter = some value H1: parameter ≠ some value Equal versus less than (left-tailed test) H1: parameter < some value Equal versus greater than (right-tailed test) H1: parameter > some value 6

“In Other Words” The null hypothesis is a statement of status quo or no difference and always contains a statement of equality. The null hypothesis is assumed to be true until we have evidence to the contrary. We seek evidence that supports the statement in the alternative hypothesis. 7

Parallel Example 2: Forming Hypotheses
For each of the following claims, determine the null and alternative hypotheses. State whether the test is two-tailed, left-tailed or right-tailed. In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. Using an old manufacturing process, the standard deviation of the amount of wine put in a bottle was 0.23 ounces. With new equipment, the quality control manager believes the standard deviation has decreased. 8

Solution In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. The hypothesis deals with a population proportion, p. If the percentage participating in charity work is no different than in 2008, it will be 0.62 so the null hypothesis is H0: p=0.62. Since the researcher believes that the percentage is different today, the alternative hypothesis is a two-tailed hypothesis: H1: p≠0.62. 9

Solution b) According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. The hypothesis deals with a population mean, μ. If the mean call length on a cellular phone is no different than in 2006, it will be 3.25 minutes so the null hypothesis is H0: μ = 3.25. Since the researcher believes that the mean call length has increased, the alternative hypothesis is: H1: μ > 3.25, a right-tailed test. 10

Solution c) Using an old manufacturing process, the standard deviation of the amount of wine put in a bottle was 0.23 ounces. With new equipment, the quality control manager believes the standard deviation has decreased. The hypothesis deals with a population standard deviation, σ. If the standard deviation with the new equipment has not changed, it will be 0.23 ounces so the null hypothesis is H0: σ = 0.23. Since the quality control manager believes that the standard deviation has decreased, the alternative hypothesis is: H1: σ < 0.23, a left-tailed test. 11

Four Outcomes from Hypothesis Testing
Reject the null hypothesis when the alternative hypothesis is true. This decision would be correct. Do not reject the null hypothesis when the null hypothesis is true. This decision would be correct. Reject the null hypothesis when the null hypothesis is true. This decision would be incorrect. This type of error is called a Type I error. Do not reject the null hypothesis when the alternative hypothesis is true. This decision would be incorrect. This type of error is called a Type II error. 12

Parallel Example 3: Type I and Type II Errors
For each of the following claims, explain what it would mean to make a Type I error. What would it mean to make a Type II error? In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. 13

Solution In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. A Type I error is made if the researcher concludes that p ≠ 0.62 when the true proportion of Americans 18 years or older who participated in some form of charity work is currently 62%. A Type II error is made if the sample evidence leads the researcher to believe that the current percentage of Americans 18 years or older who participated in some form of charity work is still 62% when, in fact, this percentage differs from 62%. 14

Solution b) According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. A Type I error occurs if the sample evidence leads the researcher to conclude that μ > 3.25 when, in fact, the actual mean call length on a cellular phone is still 3.25 minutes. A Type II error occurs if the researcher fails to reject the hypothesis that the mean length of a phone call on a cellular phone is 3.25 minutes when, in fact, it is longer than 3.25 minutes. 15

= P(rejecting H0 when H0 is true) β = P(Type II Error)
α = P(Type I Error) = P(rejecting H0 when H0 is true) β = P(Type II Error) = P(not rejecting H0 when H1 is true) 16

The probability of making a Type I error, α, is chosen by the researcher before the sample data is collected. The level of significance, α, is the probability of making a Type I error. “In Other Words” As the probability of a Type I error increases, the probability of a Type II error decreases, and vice-versa. 17

CAUTION! We never “accept” the null hypothesis, because, without having access to the entire population, we don’t know the exact value of the parameter stated in the null. Rather, we say that we do not reject the null hypothesis. This is just like the court system. We never declare a defendant “innocent”, but rather say the defendant is “not guilty”. 18

Parallel Example 4: Stating the Conclusion
According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. Suppose the sample evidence indicates that the null hypothesis should be rejected. State the wording of the conclusion. Suppose the sample evidence indicates that the null hypothesis should not be rejected. State the wording of the conclusion. 19

Solution Suppose the sample evidence indicates that the null hypothesis should be rejected. State the wording of the conclusion. The statement in the alternative hypothesis is that the mean call length is greater than 3.25 minutes. Since the null hypothesis (μ = 3.25) is rejected, there is sufficient evidence to conclude that the mean length of a phone call on a cell phone is greater than 3.25 minutes. 20

Solution b) Suppose the sample evidence indicates that the null hypothesis should not be rejected. State the wording of the conclusion. Since the null hypothesis (μ = 3.25) is not rejected, there is insufficient evidence to conclude that the mean length of a phone call on a cell phone is greater than 3.25 minutes. In other words, the sample evidence is consistent with the mean call length equaling 3.25 minutes. 21

Hypothesis Tests for a Population Proportion
Section 10.2 Hypothesis Tests for a Population Proportion

A researcher obtains a random sample of 1000 people and finds that 534 are in favor of the banning cell phone use while driving, so = 534/1000. Does this suggest that more than 50% of people favor the policy? Or is it possible that the true proportion of registered voters who favor the policy is some proportion less than 0.5 and we just happened to survey a majority in favor of the policy? In other words, would it be unusual to obtain a sample proportion of or higher from a population whose proportion is 0.5? What is convincing, or statistically significant, evidence? 23

When observed results are unlikely under the assumption that the null hypothesis is true, we say the result is statistically significant. When results are found to be statistically significant, we reject the null hypothesis. 24

To determine if a sample proportion of 0
To determine if a sample proportion of is statistically significant, we build a probability model. Since np(1 – p) = 1000(0.5)(1 – 0.5) = 250 ≥ 10 and the sample size (n = 1000) is sufficiently smaller than the population size, we can use the normal model to describe the variability in . The mean of the distribution of is and the standard deviation is 25

Sampling distribution of the sample proportion
26

The Logic of the Classical Approach
We may consider the sample evidence to be statistically significant (or sufficient) if the sample proportion is too many standard deviations, say 2, above the assumed population proportion of 0.5. 27

standard deviations above the hypothesized proportion of 0.5.
Recall that our simple random sample yielded a sample proportion of 0.534, so standard deviations above the hypothesized proportion of 0.5. Therefore, using our criterion, we would reject the null hypothesis. 28

Why does it make sense to reject the null hypothesis if the sample proportion is more than 2 standard deviations away from the hypothesized proportion? The area under the standard normal curve to the right of z = 2 is 29

and only 2.28% of the sample proportions will be more than 0.532.
If the null hypothesis were true (population proportion is 0.5), then 1 – = = 97.72% of all sample proportions will be less than .5 + 2(0.016) = 0.532 and only 2.28% of the sample proportions will be more than 30

Hypothesis Testing Using the P-Value Approach
Definition A P-value is the probability of observing a sample statistic as extreme or more extreme than one observed under the assumption that the statement in the null hypothesis is true. Hypothesis Testing Using the P-Value Approach If the probability of getting a sample statistic as extreme or more extreme than the one obtained is small under the assumption the statement in the null hypothesis is true, reject the null hypothesis. 31

The Logic of the P-Value Approach
A second criterion we may use for testing hypotheses is to determine how likely it is to obtain a sample proportion of or higher from a population whose proportion is 0.5. If a sample proportion of or higher is unlikely (or unusual), we have evidence against the statement in the null hypothesis. Otherwise, we do not have sufficient evidence against the statement in the null hypothesis. 32

We can compute the probability of obtaining a sample proportion of 0
We can compute the probability of obtaining a sample proportion of or higher from a population whose proportion is 0.5 using the normal model. 33

Recall So, we compute The value is called the P-value, which means about 2 samples in 100 will give a sample proportion as high or higher than the one we obtained if the population proportion really is 0.5. Because these results are unusual, we take this as evidence against the statement in the null hypothesis. 34

Using the P-value Approach
Hypothesis Testing Using the P-value Approach If the probability of getting a sample proportion as extreme or more extreme than the one obtained is small under the assumption the statement in the null hypothesis is true, reject the null hypothesis. 35

Recall: The best point estimate of p, the proportion of the population with a certain characteristic, is given by where x is the number of individuals in the sample with the specified characteristic and n is the sample size. 36

Recall: The sampling distribution of is approximately normal, with mean and standard deviation provided that the following requirements are satisfied: The sample is a simple random sample. np(1 – p) ≥ 10. The sampled values are independent of each other. 37

Testing Hypotheses Regarding a Population Proportion, p
To test hypotheses regarding the population proportion, we can use the steps that follow, provided that: the sample is obtained by simple random sampling. np0(1 – p0) ≥ 10. the sampled values are independent of each other. 38

Step 1: Determine the null and alternative. hypotheses
Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways: 39

Step 2: Compute the test statistic
Note: We use p0 in computing the standard error rather than This is because, when we test a hypothesis, the null hypothesis is always assumed true. 40

Use Table V to determine the critical value.
Classical Approach Use Table V to determine the critical value. Two-Tailed Left-Tailed Right-Tailed (critical value) (critical value) (critical value) 41

Step 3: Compare the critical value with the test statistic:
Classical Approach Step 3: Compare the critical value with the test statistic: Step 4: State the conclusion. 42

Use calculator to determine p-value. or
P-Value Approach Step 3: Use calculator to determine p-value. or Use Standard Table to determine correct p-value. If the P-value < α, reject the null hypothesis. Step 4: State the conclusion. 43

Parallel Example 1: Testing a Hypothesis about a
Parallel Example 1: Testing a Hypothesis about a Population Proportion: Large Sample Size In 1997, 46% of Americans said they did not trust the media “when it comes to reporting the news fully, accurately and fairly”. In a 2007 poll of 1010 adults nationwide, 525 stated they did not trust the media. At the α = 0.05 level of significance, is there evidence to support the claim that the percentage of Americans that do not trust the media to report fully and accurately has increased since 1997? Source: Gallup Poll 44

Solution We want to know if p > First, we must verify the requirements to perform the hypothesis test: This is a simple random sample. np0(1 – p0) = 1010(0.46)(1 – 0.46) = > 10 Since the sample size is less than 5% of the population size, the assumption of independence is met. 45

Step 1: H0: p = 0.46 versus H1: p > 0.46
Solution Step 1: H0: p = versus H1: p > 0.46 The level of significance is α = 0.05. Step 2: The sample proportion is The test statistic is then 46

Solution: Classical Approach
Since this is a right-tailed test, we determine the critical value at the α = 0.05 level of significance to be z0.05 = Step 3: Since the test statistic, z0 = 3.83, is greater than the critical value 1.645, we reject the null hypothesis. 47

Solution: P-Value Approach
Step 3: Since this is a right-tailed test, the P value is the area under the standard normal distribution to the right of the test statistic z0= That is, P-value = P(Z > 3.83) ≈ 0. Since the P-value is less than the level of significance, we reject the null hypothesis. 48

Solution Step 4: There is sufficient evidence at the α = 0.05 level of significance to conclude that the percentage of Americans that do not trust the media to report fully and accurately has increased since 1997. 49

Example The two major college entrance exams that a majority of colleges accept for admission are the SAT and ACT. ACT looked at historical records and established 22 as the minimum ACT math score for a student to be considered prepared for college mathematics. An official wonders whether less than half of the students in her state are prepared for College Algebra. She obtains a simple random sample of 500 records of students who have taken the ACT and finds that 219 are prepared for college mathematics. Does this represent significant evidence that less than half of Illinois students who have taken the ACT are prepared for college mathematics? Use α = 0.05 level of significance. 50

Two-Tailed Hypothesis Testing Using Confidence Intervals
When testing H0: p = p0 versus H1: p ≠ p0 if a (1 – α) • 100% confidence interval contains p0, do not reject the null hypothesis. However, if the confidence interval does not contain p0, conclude that p ≠ p0 at the level of significance, α. 51

Example When asked about gun ownership, 46% of all Americans said protecting the right to own guns is more important then controlling gun ownership. The Pew Research Center surveyed 1267 randomly selected Americans with at least a bachelor’s degree and found that 559 believed that protecting the right to own guns is more important. Does this result suggest the proportion of Americans with at least a bachelor’s degree feel different than the general American population when it comes to gun control? Use α = level of significance. 52

Example A study found that 34% of teenagers text while driving. Does a recent survey conducted, which found that 353 of 1200 randomly selected teens had texted while driving, suggest that the proportion of teens who text while driving has changed? Use 95% confidence interval to answer the question. 53

Hypothesis Tests for a Population Mean
Section 10.3 Hypothesis Tests for a Population Mean

follows Student’s t-distribution with n –1 degrees of freedom.
To test hypotheses regarding the population mean assuming the population standard deviation is unknown, we use the t-distribution rather than the Z-distribution. When we replace σ with s, follows Student’s t-distribution with n –1 degrees of freedom. 55

Testing Hypotheses Regarding a Population Mean
To test hypotheses regarding the population mean, we use the following steps, provided that: The sample is obtained using simple random sampling. The sample has no outliers, and the population from which the sample is drawn is normally distributed or the sample size is large (n ≥ 30). The sampled values are independent of each other. 56

Step 1: Determine the null and alternative hypotheses. Again, the hypotheses can be structured in one of three ways: 57

which follows the Student’s t-distribution with n – 1 degrees of freedom. Use Table VII to determine the critical value. 58

Classical Approach Step 3: Compare the critical value with the test statistic: Step 4: State conclusion 59

The procedure is robust, which means that minor departures from normality will not adversely affect the results of the test. However, for small samples, if the data have outliers, the procedure should not be used. 60

Parallel Example 1: Testing a Hypothesis about a Population Mean, Large Sample Assume the resting metabolic rate (RMR) of healthy males in complete silence is 5710 kJ/day. Researchers measured the RMR of 45 healthy males who were listening to calm classical music and found their mean RMR to be with a standard deviation of At the α = 0.05 level of significance, is there evidence to conclude that the mean RMR of males listening to calm classical music is different than 5710 kJ/day? 61

Solution We assume that the RMR of healthy males is 5710 kJ/day. This is a two-tailed test since we are interested in determining whether the RMR differs from 5710 kJ/day. Since the sample size is large, we follow the steps for testing hypotheses about a population mean for large samples. 62

Solution Step 1: H0: μ = 5710 versus H1: μ ≠ 5710
The level of significance is α = 0.05. Step 2: The sample mean is = and the sample standard deviation is s = The test statistic is 63

Since this is a two-tailed test, we determine the critical values at the α = 0.05 level of significance with n –1 = 45 – 1 = 44 degrees of freedom to be approximately –t0.025 = –2.021 and t0.025 = Step 3: Since the test statistic, t0 = –0.013, is between the critical values, we fail to reject the null hypothesis. 64

Solution Step 4: There is insufficient evidence at the α = level of significance to conclude that the mean RMR of males listening to calm classical music differs from 5710 kJ/day. 65

Parallel Example 1: Testing a Hypothesis about a Population Mean, Large Sample The mean height of American males is 69.5 inches. The heights of the 43 male U.S. presidents have a mean inches and a standard deviation of 2.77 inches. Treating the 43 presidents as a simple random sample, determine if there is evidence to suggest that U.S. presidents are taller than the average American male. Use the α = 0.05 level of significance. 66

Parallel Example 3: Testing a Hypothesis about a Population Mean, Small Sample According to the United States Mint, quarters weigh 5.67 grams. A researcher is interested in determining whether the “state” quarters have a weight that is different from 5.67 grams. He randomly selects 18 “state” quarters, weighs them and obtains the following data (assume population is normal). At the α = 0.05 level of significance, is there evidence to conclude that state quarters have a weight different than grams? 67

Solution Step 1: H0: μ = 5.67 versus H1: μ ≠ 5.67
The level of significance is α = 0.05. Step 2: From the data, the sample mean is calculated to be and the sample standard deviation is s = The test statistic is 68

Step 3: Since this is a two-tailed test, we determine the critical values at the α = 0.05 level of significance with n – 1 = 18 – 1 = 17 degrees of freedom to be –t0.025 = –2.11 and t0.025 = 2.11. Since the test statistic, t0 = 2.75, is greater than the critical value 2.11, we reject the null hypothesis. 69

Solution Step 4: There is sufficient evidence at the α = 0.05 level of significance to conclude that the mean weight of the state quarters differs from grams. 70

Example The “fun size” Snickers bar is supposed to weigh 20 grams. Because the penalty for selling candy bars under their advertised weight it severe, the manufacturer calibrates the machine so the mean weight is 20.1 grams. The quality control engineer is concerned about the calibration. He obtains a random sample of 11 candy bars, weighs them, and obtains the following data (assume population normal): Should the machine be shut down and recalibrated? Because this would be expensive, use α = 0.01 level of significance. 71

Hypothesis Tests for a Population Standard Deviation
Section 10.4 Hypothesis Tests for a Population Standard Deviation

Chi-Square Distribution
If a simple random sample of size n is obtained from a normally distributed population with mean μ and standard deviation σ, then has a chi-square distribution with n – 1 degrees of freedom, where s2 is a sample variance. 73

Testing Hypotheses about a Population Variance or Standard Deviation
To test hypotheses about the population variance or standard deviation, we can use the following steps, provided that the sample is obtained using simple random sampling or from a randomized experiment. the population is normally distributed. 74

Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways: 75

Use Table VIII to determine the critical value using n – 1 degrees of freedom. 76

Classical Approach Step 3: Compare the critical value with the test statistic: Step 4: State conclusion 77

Parallel Example 1: Testing a Hypothesis about a Population Standard Deviation A can of 7-Up states that the contents of the can are 355 ml. A quality control engineer is worried that the filling machine is miscalibrated. In other words, she wants to make sure the machine is not under- or over-filling the cans. She randomly selects 9 cans of 7-Up and measures the contents. She obtains the following data. In section 9.2, we assumed the population standard deviation was Test the claim that the population standard deviation, σ, is greater than 3.2 ml at the α = 0.05 level of significance. 78

Step 1: H0: σ = 3.2 versus H1: σ > 3.2.
Solution Step 1: H0: σ = versus H1: σ > 3.2. This is a right-tailed test. The level of significance is α = 0.05. Step 2: From the data, the sample standard deviation is computed to be s = The test statistic is 79

Step 3: Since this is a right-tailed test, we determine the critical value at the α = 0.05 level of significance with degrees of freedom to be χ20.05= Since the test statistic, , is less than the critical value, , we fail to reject the null hypothesis. 80

Solution: Step 4: There is insufficient evidence at the α = 0.05 level of significance to conclude that the standard deviation of the can content of 7-Up is greater than ml. 81

Example The quality control engineer for M&M-Mars tested whether the mean weight of fun-size Snickers was 20.1 grams. Suppose that the standard deviation of the weight of the candy was 0.75 gram before a machine recalibration. The engineer wants to know if the recalibration resulted in more consistent weights. Conduct the appropriate test at α = 0.05 level of significance. 82

Example A machine fills bottles with 64 fluid ounces of liquid. The quality control manager determines that the fill levels are normally distributed with a mean of 64 ounces and a standard deviation of 0.42 ounce. He has an engineer recalibrate the machine to try to lower the standard deviation. After the recalibration, the manager selects 19 bottles from the line and determines that the standard deviation is 0.38 ounce. Is there less variability in the filling machine? Conduct the appropriate test at α = 0.01 level of significance. 83

Putting It Together: Which Method Do I Use?
Section 10.5 Putting It Together: Which Method Do I Use?

Example In October 1945, the Gallup organization asked 1487 randomly sample Americans, “Do you think we can develop a way to protect ourselves from atomic bombs in case other countries tried to use them against us?” with 788 responding yes. Did the majority of Americans feel the United States could develop a way to protect itself from atomic bombs in 1945? Use α = 0.05 level of significance. 86

Example A pharmaceutical company manufactures a 200-milligram pain reliever. Company specifications require that the standard deviation of the amount of the active ingredient must not exceed 5 mg. The quality-control manager selects a random sample of 30 tablets from a certain batch and finds that the sample standard deviation is 7.3 mg. Assume that the amount of the active ingredient is normally distributed. Determine whether the standard deviation of the amount of active ingredient is greater than 5 mg at the α = 0.05 level of significance. 87

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