CH14

The big idea is that the Sampling Distribution of M tells us how close to μ the Sample Mean M is likely to be

A Confidence Interval turns that information around to say how close to M the unknown Population Mean μ is likely to be

Statistically Significant

Small P-values are evidence against H₀term-43 - Because they say that the observed result would be unlikely to occur if H₀ was true • That is, small P-values indicate that the data are inconsistent with H₀, so we should REJECT H₀

Check Your Skills 14.27: Average human gestation time is 266 days, when counted from conception - A hospital gives a 90% confidence interval for the mean gestation time from conception among its patients: 264 ± 5 days Is the mean gestation time in that hospital significantly different from 266 days? a. It is not significantly different at the 10% level and, therefore, is also not significantly different at the 5% level b. It is not significantly different at the 10% level but might be significantly different at the 5% level c. It is significantly different at the 10% level

A Conducted Z-Test on calc - P = 0.2881 0.2881 > 0.1; n.s. - 0.2881 > 0.05; n.s.

Confidence Intervals Take the Following Form:

(unbiased estimate) ± (critical value)(standard deviation of the distribution of the estimate)

Statisticians argue over the relative merits of computing a Confidence Interval or a P-value

A Hypothesis Test is obviously particularly well-suited when you have a CLEARLY DEFINED SET OF HYPOTHESES The P-value will say how strong the evidence against the Null Hypothesis is - However, if you reject H₀, you are left not knowing what kind of value the parameter might have - And if you fail to reject H₀, there are still other likely values for the parameter besides that defined by H₀ - That is, the P-VALUE GIVES NO INFORMATION ABOUT EFFECT SIZE All things being equal, a CONFIDENCE INTERVAL PROVIDES MORE USEFUL INFORMATION than a P-value alone

"Significant" in the Statistical Sense

Does not mean "important" - It means simply "not likely to happen just by chance because of random variations from sample to sample, if the Null Hypothesis was true"

Significance Level (α)

An ARBITRARY THRESHOLD that can be USED WHEN REPORTING WHETHER a P-VALUE is STATISTICALLY SIGNIFICANT - The most common fixed values are 0.05 & 0.01

A Level C (i.e., 90%; 95%; 99%) Confidence Interval for a Parameter has 2 Parts:

An INTERVAL CALCULATED FROM the DATA, usually of the form: - estimate ± margin of error - The ESTIMATE is a SAMPLE STATISTIC (M) - The MARGIN OF ERROR (m) represents the ACCURACY OF OUR GUESS FOR the PARAMETER It is usually set at 90% or higher because we want to be reasonably sure of our conclusions - Keep in mind that the only way to be 100% confident about the unknown value of a parameter would be to record the entire population; this is rarely an option, especially in the life sciences

Check Your Skills 14.24: The average human gestation time is 266 days from conception - A researcher suspects that proper nutrition plays an important role and that poor women with inadequate food intake would have shorter gestation times even when given vitamin supplements - A random sample of 20 poor women given vitamin supplements throughout their pregnancy has a mean gestation time from conception of M = 256 days What is the researcher's alternative hypothesis for this test? a. Hₐ: μ ≠ 256 b. Hₐ: μ < 266 c. Hₐ: μ < 256

B

Check Your Skills 14.18: The Framingham Heart Study recorded the systolic blood pressure (SBP) of 3,534 adults and reported a 95% confidence interval for the population mean SBP of 126.7 to 127.9 mm Hg The correct interpretation for this confidence interval is that we are 95% confident that... a. The adults in this sample have SBP values between 126.7 and 127.9 mm Hg b. Adults in the population have SBP values between 126.7 and 127.9 mm Hg c. The population mean SBP is a value between 126.7 and 127.9 mm Hg

C

Check Your Skills 14.20: A study reports the mean change in high-density lipoprotein (HDL, also known as "good" cholesterol) of adults eating raw garlic six days a week for six months - The margin of error for a 95% confidence interval is given as plus or minus 6 milligrams per deciliter of blood (mg/dl) This means that... a. We can be certain that the study result is within 6 mg/dl of the true mean HDL change in this population b. We can be certain that the study result is within 6 mg/dl of the true mean HDL change in this population if the conditions for inferences are satisfied c. The study used a method that gives a result within 6 mg/dl of the true mean HDL change in this population for 95% of all samples of 3,534 adults

C

TEST-Statistic

Calculated from the sample data, it is a Statistic that MEASURES HOW FAR the DATA DIVERGE FROM the NULL Hypothesis H₀ - LARGE values show that the DATA are FAR FROM WHAT WE WOULD EXPECT IF H₀ WAS TRUE

Confidence Level (C)

It gives the PROBABILITY THAT the INTERVAL WILL CAPTURE the TRUE PARAMETER VALUE IN REPEATED SAMPLES - That is, it's the SUCCESS RATE FOR the METHOD

Exercise 14.39: Statistical significance is very often mentioned in research reports When asked what it means, a student replies that "results that are statistically significant tell us that they cannot easily be explained by chance variation alone if the hypothesis tested was true" Do you think that this statement is essentially correct? - Explain your answer

It is essentially correct

Exercise 14.45: Exercise 14.30 gave a 95% confidence interval for the mean body length μ of deer mice in a rich forest habitat Exercise 14.44 used the same data to test the hypothesis that deer mice from this habitat have significantly different body lengths from the general population of deer mice Compare the two results and explain why the confidence interval is more informative than the test P-value

It is possible to determine statistical significance using both P-values and confidence intervals - However, P-values only compare the parameter to a hypothesis, while confidence intervals actually give a range of estimates for the parameter

Apply Your Knowledge 14.13: Ex 14.3 gives the IQ test scores of 31 seventh-grade girls in a Midwest school district - IQ scores follow a Normal distribution with standard deviation σ = 15 - Treat these 31 girls as an SRS of all seventh-grade girls in this district - IQ scores in a broad population are supposed to have mean μ = 100 Is there evidence that the mean in this district differs from 100? - Follow the four-step process, as illustrated in Ex 14.9, in your answer

Let μ be the mean IQ for all seventh-grade girls in this district We test H₀: μ = 100 & Hₐ: μ ≠ 100 (two-sided because we have no prior theory why it should be a directional effect) & find: - z = 2.17 - P = 0.0302 There is reasonably strong evidence that the mean IQ differs from 100 (it is greater) in this population

Z-Test Statistic

Measures how far the observed sample mean M deviates from the hypothesized population value μ₀, in relative terms This relative measurement is in the standard scale obtained by dividing by the standard deviation of M When H₀: is true, it has the standard Normal distribution

Inconclusive

P-values that are not small fail to provide evidence against H₀ - H₀ CANNOT BE REJECTED - Failing to find evidence against H₀ means only that the data are not inconsistent with H₀, which does not at all imply that H₀ is true

Ex 14.9: STATE: When you search the internet for what constitutes "normal" or healthy body temperature, you will probably find that an oral temperature of 98.6 degrees Fahrenheit (°F), or 37.0 degrees Celsius, is considered normal This widely quoted value comes from a paper published in 1868 by German physician Carl Wunderlich, in which he reported more than a million body temperature readings - In this paper, Wunderlich stated that the mean body temperature of healthy adults is 98.6°F More than a century later, a study was designed to evaluate this claim - It found a mean oral temperature of M = 98.25 °F from 130 adults Does this study provide significant evidence that Wunderlich's claim of a mean adult body temperature of 98.6 °F is not correct?

PLAN: The null hypothesis is "no difference" from the accepted mean μ₀ = 98.6 °F - The alternative is two-sided because the study did not have a particular direction in mind before examining the data Thus, the hypotheses about the unknown mean μ of the population of healthy adults are: - H₀: μ = 98.6 - Hₐ: μ ≠ 98.6 SOLVE: Check the conditions for inference - Here are the "simple conditions" stated on page 348 1. SRS: The most important condition is that the 130 healthy adults in the sample are an SRS from the population of all healthy adults - We should check this requirement by asking how the data were obtained - The researchers measured the baseline oral temperature of volunteers participating in various vaccine clinical trials conducted at the University of Maryland - If these individuals enrolled in the clinical trial because of a specific medical concern, for example, the data may not be representative of the whole healthy-adult population and could be biased - The researchers assure us that the volunteers were all healthy - Because the subjects were volunteers, the sample is a random sample but not a true SRS - This is very common in the life sciences, especially when dealing with human subjects 2. Normal distribution: We should examine the distribution of the 130 observations to look for signs that the population distribution is not Normal - A histogram in the published report indicates that the data were roughly Normal 3. Known σ: It is unrealistic to suppose that we know that σ = 0.6 °F - This value is based on Wunderlich's original report of a very large number of body temperatures SOLVE: Obtain the test statistic and P-value - Figure 14.8 shows the output for a one-sample z test using statistical software JMP - It gives z = −6.651 and a P-value rounded to zero, based on which the output specifies that we should "reject the null hypothesis" CONCLUDE: If μ really was 98.6 °F, it would be nearly impossible to obtain an SRS of size 130 from the healthy adult population with sample mean at least as far from 98.6 °F as that obtained here - The observed M = 98.25 °F is therefore very strong evidence that the true mean body temperature of healthy adults is not 98.6 °F The study findings are highly statistically significant (P ≈ 0) - This doesn't imply that the true mean body temperature in the population of healthy adults is very different from 98.6 °F - Rather, it just says that we have very strong evidence to reject H₀ - Because the sample size is rather large (n = 130), the sampling distribution of M is quite narrow - The observed sample mean M = 98.25 °F is therefore much further away from a hypothetical population mean of 98.6 °F than would be expected under such a narrow sampling distribution The computations performed by any statistical software in this example involve the following steps: - Test statistic: z = (M - μ₀)/(σ/√n) = (98.25 - 98.6)/(0.6/√130) = -6.65 - P-value: P = 2P(Z ≤ |z|) = 2P(Z ≤ -6.65) ≈ 0, computed using the standard Normal curve

Statistical Inference

Provides methods for DRAWING CONCLUSIONS ABOUT a POPULATION FROM SAMPLE DATA - It USES the language of PROBABILITY TO SAY HOW TRUSTWORTHY OUR CONCLUSIONS ARE

Statistical Hypotheses should...

REFLECT a LEGITIMATE INQUIRY GROUNDED IN EXISTING KNOWLEDGE ON the TOPIC, not some arbitrary choice - In particular, the hypotheses should express the hopes or suspicions we have BEFORE we see the data - It is cheating to first look at the data and then frame hypotheses to fit what the data show Thus, the fact that a random sample of aspirin tablets in Ex 14.6 gave a sample average M larger than the population mean μ should not influence our choice of Hₐ When you DO NOT HAVE a SPECIFIC DIRECTION firmly in mind in advance, you must use a TWO-sided alternative

Ex 14.5: The study in Ex 14.4 is seeking evidence for a lower inorganic phosphorus level in the elderly, on average - The Null Hypothesis says that "there is no difference from the adult mean of 1.2 mmol/l," on average, in a large population of elderly individuals between the ages of 75 and 79 years - The Alternative Hypothesis says that "their mean is lower than 1.2 mmol/l"

So the Hypotheses are: - H₀: μ = 1.2 - Hₐ: μ < 1.2 The Alternative Hypothesis is one-sided because we are interested only in whether the elderly have a lower inorganic phosphorus level, on average - This choice of Alternative Hypothesis is based on the biological hypothesis that some physiological processes such as mineral absorption may become less efficient with age, leading to lower blood levels

Exercise 14.41: Sketch the standard Normal curve for the z test statistic and mark off areas under the curve to show why a value of z that is significant at the 1% level in a one-sided test is always significant at the 5% level If z is significant at the 5% level, what can you say about its significance at the 1% level?

Something that occurs "less than once in 100 repetitions" also occurs "less than 5 times in 100 repetitions," but not vice versa

ONE-Sided

The Alternative Hypothesis states that a Parameter is LARGER than or that it is SMALLER than the Null Hypothesis value - Alternative Hypothesis is directional

TWO-Sided

The Alternative Hypothesis states that the Parameter is DIFFERENT from the Null value (it could be either SMALLER or LARGER)

Interpreting a Confidence Interval

The Confidence Level is the success rate of the method that produces the interval We don't know whether the 95% Confidence Interval from a particular sample is: - One of the 95% of intervals that capture μ - One of the unlucky 5% of intervals that miss To say that we are 95% confident that the unknown value of μ is between 131.1 & 133.9 cm is shorthand for "WE GOT THESE NUMBERS USING A METHOD THAT GIVES CORRECT RESULTS NINETY-FIVE PERCENT (95%) OF THE TIME"

Exercise 14.29: Here is an explanation from the Associated Press concerning one of its opinion polls - Explain briefly but clearly in what way this explanation is incorrect For a poll of 1,600 adults, the variation due to sampling error is no more than three percentage points either way - The error margin is said to be valid at the 95 percent confidence level - This means that, if the same questions were repeated in 20 polls, the results of at least 19 surveys would be within three percentage points of the results of this survey

The Margin of Error says that 95% of surveys should contain the true population proportion, not necessarily the proportion obtained by this survey

ONE-Tailed P-Value

The P-Value obtained for a one-sided Hₐ - Includes only one of these "tails"

ALTERNATIVE Hypothesis (Hₐ or H₁)

The more general claim about the population that we are trying to find evidence FOR Ex 14.4: - The mean inorganic phosphorus blood level μ in the elderly population is less than 1.2 mmol/l - The Parameter is lower than the mean of the adult population, a consequence of aging It can be: - ONE-sided - TWO-sided

NULL Hypothesis (H₀)

The specific claim about a population of interest tested by a Hypothesis Test - Usually it is a STATEMENT OF "NO EFFECT" or "NO DIFFERENCE" A Hypothesis Test is designed to assess the strength of the evidence AGAINST this hypothesis Ex 14.4: - The mean inorganic phosphorus blood level μ in the elderly population is 1.2 mmol/l - There is no difference from the adult mean of 1.2 mmol/l

Simple Conditions for Inference About a Mean:

To make the reasoning as clear as possible, we consider the simplest of situations for Statistical Inference about a Population Mean 1. We have an SRS from the Population of interest - There is no nonresponse or other practical difficulty 2. The Quantitative Variable we measure has a perfectly Normal distribution N(μ, σ) in the Population 3. We don't know the Population Mean μ - But we do know the Population Standard Deviation σ

Reasoning of Statistical Estimation

We learned in Chapter 13 that the statistic M is an unbiased estimator of the parameter μ, but how far from could a given M be? 1. To estimate the unknown population mean μ, use the mean M = 132.5 of the random sample - We don't expect M to be exactly equal to μ, so we want to say how accurate this estimate is 2. We know the sampling distribution of M - In repeated samples, M has the Normal distribution with mean μ & standard deviation σ/√n - So the sampling distribution of the average height M of 217 eight-year-old American boys has standard deviation: • σ/√n • 10/√217 • 0.7 cm (round off) 3. The 95 part of the approximate 68-95-99.7 rule for Normal distributions says that M is within 1.4 cm (that's 2 standard deviations) of the mean μ in about 95% of all samples - That is, for 95% of all samples, 1.4 cm is the maximum distance separating M & μ - Therefore, if we estimate that the value of μ is somewhere in the interval from M - 1.4 to M + 1.4, we'll be right 95% of the times we take a sample - For this particular NHANES sample, this interval is: • From M - 1.4 = 132.5 - 1.4 = 131.1 • To M + 1.4 = 132.5 + 1.4 = 133.9 Because we got the interval 131.1 to 133.9 cm from a method that captures the population mean μ 95% of the time: - We say that we are 95% confident that the mean height of all eight-year-old boys in the United States is some value in that interval, perhaps as low as 131.1 cm or as high as 133.9 cm

Statistically Significant at Level α

When the P-value is as small as or smaller than α

Confidence Interval for the Mean of a Normal Population

Draw an SRS of size n from a Normal population having unknown mean μ and known standard deviation σ A level C confidence interval for μ is: - M ± z*(σ/√n) Any technology with basic statistical inference capabilities will perform all necessary computations and produce the lower & upper boundaries of the confidence interval

Ex 14.1: A recent NHANES reports that the mean height of a Sample of 217 eight-year-old boys was M = 132.5 centimeters (52.2 inches) - On the basis of this Sample, we want to estimate the mean height μ in the Population of more than a million American eight-year-old boys

To match the "simple conditions," we will treat: - The NHANES Sample as a perfect SRS of all American eight-year-old boys - The height in this population as having an exactly Normal distribution with Standard Deviation σ = 10 cm

Confidence Interval

Used for ESTIMATING the VALUE OF a POPULATION PARAMETER

Hypothesis Testing

Used to ASSESS the EVIDENCE AGAINST a CLAIM ABOUT a POPULATION PARAMETER

Hypothesis Tests Have a Different Goal:

To ASSESS the STRENGTH OF the EVIDENCE PROVIDED BY DATA AGAINST some CLAIM CONCERNING a POPULATION - It is designed to assess the strength of the evidence AGAINST the NULL Hypothesis HYPOTHESES ALWAYS REFER TO a POPULATION, not to a particular outcome

Use a Confidence Interval when your goal is...

To ESTIMATE a POPULATION PARAMETER

The usual reason for taking a Sample is not to learn about the individuals in the Sample, but rather...

To INFER from the Sample data SOME CONCLUSION ABOUT the WIDER POPULATION that the Sample represents

Exercise 14.35: The Institute of Medicine sets the tolerable upper level of sodium intake for adults and teenagers at 2300 milligrams (mg) per day A random sample of 3727 adolescents representative of the U.S. adolescent population found a mean sodium intake of 3486 mg per day Does this finding suggest that U.S. adolescents consume sodium in excess of the tolerable upper level? - State H₀ & Hₐ for a test to answer this question - State carefully what the parameter μ in your hypotheses is

μ is the mean daily sodium intake (in mg) in the American teenage population - H₀: μ = 2300 - Hₐ: μ > 2300

TWO-Tailed P-Value

The P-value obtained for a two-sided Hₐ - Includes both "tails" of the Normal sampling distribution

2 Most Common Types of Inference

1. CONFIDENCE INTERVALS 2. HYPOTHESIS TESTS Both are based on the SAMPLING DISTRIBUTIONS of Statistics - That is, both REPORT PROBABILITY that STATE WHAT WOULD HAPPEN IF WE USED the INFERENCE METHOD MANY TIMES

2 Things Influence the Size of a Confidence Interval:

1. Confidence Level (C) - 99%; 95%; 90% 2. Sample Size (n)

Apply Your Knowledge 14.17: A 95% Confidence Interval for a Population Mean is 31.5 ± 3.5 a. With a two-sided alternative, can you reject the Null Hypothesis that μ = 34 at the 5% Significance Level? - Why? b. With a two-sided alternative, can you reject the Null Hypothesis that μ = 36 at the 5% Significance Level? - Why?

31.5 + 3.5 = 35 31.5 - 3.5 = 28 a) No - Because 34 is within the C.I. b) Yes - Because 36 falls outside the C.I.

Check Your Skills 14.23: The average human gestation time is 266 days from conception - A researcher suspects that proper nutrition plays an important role and that poor women with inadequate food intake would have shorter gestation times even when given vitamin supplements - A random sample of 20 poor women given vitamin supplements throughout their pregnancy has a mean gestation time from conception of M = 256 days What is the null hypothesis for the researcher's test? a. H₀: μ = 266 b. H₀: μ = 256 c. H₀: μ < 256

A

Check Your Skills 14.19: What is the margin of error for the cited confidence interval of 126.7 to 127.9 mm Hg? a. 0.6 mm Hg b. 1.2 mm Hg c. 2.4 mm Hg

A - (126.7 + 127.9)/2 - 254.6/2 - M = 127.3 m = z*(σ/√n) - 127.3 - 126.7 - 0.6

We can Judge Whether any Observed M is Surprising by Locating it on the Sampling Distribution

A Sample Mean M CLOSE to the hypothesized Population Mean, WITHIN the CENTRAL BULK the hypothesized sampling distribution, would indicate that: - Such a Sample Mean could easily occur just by CHANCE when the population mean is μ = 1.2 - The NULL Hypothesis (H₀) is TRUE In contrast, a Sample Mean M located on the EDGES of the hypothesized Sampling Distribution would indicate that: - This OUTCOME is SOMEWHAT UNLIKELY TO OCCUR JUST BY CHANCE when the Population Mean is μ = 1.2 - The ALTERNATIVE Hypothesis (Hₐ) is TRUE; the NULL is REJECTED The idea here is that OBSERVING an OUTCOME THAT WOULD RARELY HAPPEN IF a HYPOTHETICAL CLAIM WAS TRUE is good evidence that this claim is NOT true

Confidence Intervals & Two-Sided Tests

A Two-Sided Test at Significance Level α can be carried out from a Confidence Interval with Confidence Level: - C = 1 - α A level α Two-Sided Hypothesis Test rejects a hypothesis H₀: μ = μ₀ exactly when the value μ₀ falls outside a level 1 - α Confidence Interval for μ

Check Your Skills 14.26: You use software to run a hypothesis test - The program tells you that the P-value is 0.031 - This result is a. Not significant at the 5% level b. Significant at the 5% level but not at the 1% level c. Significant at the 1% level

B 0.031 < 0.05; significant 0.031 > 0.01; n.s.

Check Your Skills 14.21: A laboratory scale is known to have a standard deviation of σ = 0.005 grams in repeated weighings - Scale readings in repeated weighings are Normally distributed, with the mean of all possible measurements equal to the true weight of the specimen A specimen is weighed 4 times on this scale - The average weight is 4.53 grams - What is a 99% confidence interval for the true weight of this specimen? a. (4.521, 4.539) b. (4.524, 4.536) c. (4.527, 4.533)

B M = 4.53 g n = 4 m = z*(σ/√n) - (2.326)(0.005/√4) - (2.326)(0.005/2) - (2.326)(0.0025) - 0.005815 4.53 + 0.005815 = 4.536 4.53 - 0.005815 = 4.524

Check Your Skills 14.22: A laboratory scale is known to have a standard deviation of σ = 0.005 grams in repeated weighings - Scale readings in repeated weighings are Normally distributed, with the mean of all possible measurements equal to the true weight of the specimen Suppose that we needed a 95% confidence level for this specimen using the same data - What would the margin of error be? a. 0.0025 grams b. 0.0049 grams c. 0.0069 grams

B m = z*(σ/√n) - (1.96)(0.005/√4) - (1.96)(0.005/2) - (1.96)(0.0025) - m = 0.0049

Check Your Skills 14.25: The average human gestation time is 266 days from conception - A researcher suspects that proper nutrition plays an important role and that poor women with inadequate food intake would have shorter gestation times even when given vitamin supplements - A random sample of 20 poor women given vitamin supplements throughout their pregnancy has a mean gestation time from conception of M = 256 days Human gestation times are approximately Normal with standard deviation σ = 16 days - The P-value for the researcher's test is... a. More than 0.1 b. Between 0.01 and 0.1 c. Less than 0.01

B σ/√n - 16/√20 - 16/4.472 - 3.5778 z = (M - μ)/σ - (256 - 266)/3.5778 - (-10)/3.5778 - (-2.795) ≈ (-2.80); p = 0.0026 on Table B Can also get the same values by performing a Z-Test on TI-84 - stat - TESTS - Z-Test (1) - Stats; enter them in - Draw

Ex 14.8: Suppose that we know that the aspirin content of the tablets described in Example 14.6 follows a Normal distribution with standard deviation σ = 5 mg If the manufacturing process is well calibrated, the mean aspirin content of all tablets is μ = 325 mg - This is our null hypothesis - The alternative hypothesis says simply that "the mean of all tablets is not 325 mg" H₀: μ = 325 Hₐ: μ ≠ 325 Data from a random sample of 10 aspirin tablets give μ = 326.9 mg

Because the alternative is two-sided, the P-value is the probability of getting an M at least as far from the hypothesized μ = 325 in either direction as the observed M = 326.9 Enter the information for this example into the P-Value of a Test of Significance applet and click "Update" - Figure 14.7 shows the applet output as well as the information we entered - The P-value is the sum of the two shaded areas under the Normal curve: P = 0.2295 Values as far from 325 as M = 326.9 (in either direction) would happen 23% of the time if the true population mean was μ = 325 An outcome that would occur so often, if H₀ was true, is not good evidence against H₀ We conclude that the data fail to provide statistically significant evidence (P = 0.2295) that the mean aspirin content of all manufactured tablets is not 325 mg - The conclusion here is NOT THAT H₀ IS TRUE - The study looked for evidence against H₀: μ = 325 and FAILED TO FIND STRONG ENOUGH EVIDENCE AGAINST this hypothesis; that is all we can say - The mean μ for the population of all aspirin tablets could, for instance, be a value close but not exactly equal to 325 mg

Z Test for a Population Mean

Draw an SRS of size n from a Normal population that has unknown mean μ & known standard deviation σ A test statistic & a P-value are obtained to TEST the NULL HYPOTHESIS THAT μ HAS A SPECIFIED VALUE - H₀: μ = μ₀ The ONE-sample z test statistic is: - z = (M - μ₀)/(σ/√n) As the variable Z follows the standard Normal distribution, the P-value for a test of H₀ against: - Hₐ: μ > μ₀ is (Z ≥ z) - Hₐ: μ < μ₀ is (Z ≤ z) - Hₐ: μ ≠ μ₀ is 2P(Z ≥ |z|) Any technology with basic statistical inference capabilities will perform all necessary computations and produce both z & the P-value

P-Value

The PROBABILITY that MEASURES the STRENGTH OF the EVIDENCE AGAINST a NULL Hypothesis - The probability (computed assuming that H₀ is true) that the Test Statistic would take a value at least as extreme (in the direction of Hₐ) as that which is actually observed "There is a __________% chance that the Null Hypothesis is true" - The SMALLER it is, the STRONGER the EVIDENCE AGAINST H₀ Its ORDER OF MAGNITUDE MATTERS MORE than its exact value - That of 0.03 & 0.0004 are both significant at the α = 0.05 level, but one of 0.0004 offers stronger evidence against H₀

Ex 14.4: Phosphorus is a mineral that is essential to a whole range of metabolic processes as well as to bone stiffening - Levels of inorganic phosphorus in the blood are known to vary among adults Normally with mean 1.2 millimoles per liter (mmol/l) and standard deviation 0.1 mmol/l A study was conceived to examine inorganic phosphorus blood levels in older individuals to see if it decreases with age, since we know that older people are more prone to bone fractures

Here are data from a retrospective chart review of 12 men and women between the ages of 75 and 79 years: 1.26 1.39 1.00 1.00 1.00 1.10 0.87 1.23 1.19 1.29 1.03 1.18 Some of these values are above the adult population mean and some are below it The average inorganic phosphorus level in the study is given by the sample mean M = 1.128 mmol/l - Are these data good evidence that, on average, inorganic phosphorus levels among people age 75 to 79 are lower than in the whole adult population? A Statistical Hypothesis Test starts with a careful statement of the claims we want to examine regarding some property of one or more populations - Here, we considered the competing claims that the mean inorganic phosphorus blood level μ in the elderly population is either 1.2 mmol/l (no difference from the adult mean of 1.2 mmol/l) or less than 1.2 mmol/l (lower than the mean of the adult population, a consequence of aging) = These competing claims are formally expressed as a NULL Hypothesis and an ALTERNATIVE Hypothesis

In Ex 14.4 we sought evidence that the mean blood level of inorganic phosphorus in people age 75 to 79 is LESS THAN 1.2 mmol/l (Hₐ), so the specific claim we test is that the mean for people age 75 to 79 IS 1.2 mmol/l (H₀)

If the claim that μ = 1.2 is true, the sampling distribution of M from 12 individuals age 75 to 79 is Normal with mean μ = 1.2 and standard deviation: - σ/√n - 0.1/√12 - 0.0289 Figure 14.5 shows this sampling distribution - We can judge whether any observed M is surprising by locating it on this distribution The sample of 12 individuals aged 75 to 79 produced M = 1.128 mmol/l - That is far out on the Normal curve in Figure 14.5 • So far out that AN OBSERVED VALUE THIS SMALL OR EVEN SMALLER WOULD RARELY OCCUR JUST BY CHANCE if the true population mean μ was 1.2 mmol/l - The observed sample mean is good evidence that the true value of μ is, in fact, less than 1.2 mmol/l • That is, the mean inorganic phosphorus blood level of elderly individuals between the ages of 75 and 79 years is lower than the known mean of 1.2 mmol/l in the general population

Estimate

It is the CENTER OF the INTERVAL - It is our UNBIASED GUESS FOR the VALUE OF the UNKNOWN PARAMETER, BASED ON the SAMPLE DATA

Margin of Error (m)

It shows HOW ACCURATE WE BELIEVE OUR GUESS IS, BASED ON the VARIABILITY OF the ESTIMATE It decreases when: - Sample Size (n) increases - Confidence Level (C) decreases

Critical Values (z*, t*, etc.)

NUMBERS that MARK OFF SPECIFIC AREAS OF a DISTRIBUTION - They are easily obtained with software functions for inverse Normal calculations

Ex 14.3: STATE: We are interested in the mean IQ score of seventh-grade girls in a Midwest school district - Here are the scores for 31 randomly selected seventh-grade girls in the district: 114 100 104 89 102 91 114 114 103 105 108 130 120 132 111 128 118 119 86 72 111 103 74 112 107 103 98 96 112 112 93

PLAN: We will estimate the mean IQ score μ for all seventh-grade girls in this Midwest school district by giving a 95% confidence interval SOLVE: Check the conditions for inference. Here are the "simple conditions": 1. SRS: The girls are an SRS from the population of all seventh-grade girls in the school district 2. Normal distribution: IQ score is a quantitative variable taking numerical values that are typically rounded to the nearest integer when provided - We expect from experience that IQ scores in a homogeneous population with similar developmental age will follow approximately a Normal distribution - We can't look at the population, but we can examine the sample - Figure 14.4 shows a dotplot of the data - Its shape is slightly left-skewed, but there are no extreme outliers - Thus, we have no reason to doubt that the population distribution is close to Normal - Additionally, the central limit theorem tells us that a sample size of 31 is large enough to overcome moderate departures from Normality for performing inference 3. Known σ: - Most IQ tests are designed so that σ = 15, so it is not unrealistic to assume that we know that σ = 15 for our calculations SOLVE: Obtain the confidence interval - Using statistical software and inputting the raw data, we obtain a 95% confidence interval of (100.56, 111.12) for the mean IQ score μ in this population CONCLUDE: We are 95% confident that the mean IQ score for all seventh-grade girls in this Midwest school district is between 100.6 and 111.1 The computations performed by any statistical software in this example involve the following steps: - Sample mean: M = 105.84 - Critical value for a 95% confidence level: z* = 1.960 - Margin of error: m = z*(σ/√n) = (1.96)(15)/31 = 5.28 - 95% confidence interval for μ: M ± m = 105.84 ± 5.28, or (100.56, 111.12)

Finding Confidence Interval in 4 Steps:

STATE: - What is the practical question that requires estimating a parameter? PLAN: - Identify the parameter and choose a level of confidence. SOLVE: - Carry out the work in two phases: 1. Check the conditions for the interval you plan to use 2. Obtain the confidence interval CONCLUDE: - Return to the practical question to describe your results in this setting

4-Step Process for Conducting a Hypothesis Test:

STATE: What is the practical question that requires a statistical test? PLAN: Identify the parameter, state the null & alternative hypotheses, & choose the type of test that fits your situation SOLVE: Carry out the test in 2 phases: 1. CHECK THE CONDITIONS for the test you plan to use 2. Obtain the TEST STATISTIC & the P-VALUE CONCLUDE: Return to the practical question to describe your results in this setting

Ex 14.2: Figure 14.1 summarizes what we learned in Chapter 13 about the sampling distribution of M - Starting with the population, imagine taking many SRSs of 217 eight-year-old American boys. Perhaps one sample has mean height M = 132.5 cm, a second sample has mean M = 134.2 cm, a third has mean M = 131.8 cm, & so on - If we could collect all these sample means and display their distribution, we would get the Normal distribution with mean equal to the unknown value of the parameter μ and standard deviation 0.7 cm

The approximate 68-95-99.7 rule for Normal calculations says that in 95% of all samples, the sample mean M is within 1.4 cm (two standard deviations) of the population mean μ - Whenever this happens, the interval M ± 1.4 captures the value of μ - The formula M ± 1.4 gives an interval based on each sample; 95% of these intervals capture the unknown value of the population mean μ - That is, a statistical method using the interval M ± 1.4 has a 95% success rate in capturing within that interval the mean height μ of all eight-year-old American boys

Apply Your Knowledge 14.7: Hemoglobin is a protein in red blood cells that carries oxygen from the lungs to body tissues - People with less than 12 grams of hemoglobin per deciliter of blood (g/dl) are anemic A public health official in Jordan suspects that the mean μ for all children in Jordan is less than 12 g/dl because of inadequate health conditions - The official measures hemoglobin levels in a sample of 50 children Define the parameter examined and state the appropriate null and alternative hypotheses

The parameter is μ, the mean hemoglobin level in Jordan's children population - H₀: μ = 12 - Hₐ: μ < 12

Ex 14.6: A pharmaceutical company manufactures aspirin tablets that are sold with the label "active ingredient: aspirin 325 mg" - No production, even machine-made, is ever perfect, and individual tablets do vary a little bit in their actual aspirin content - A small amount of variation is acceptable provided that, on average, the whole production has mean μ = 325 mg

The parameter of interest is the mean aspirin content per tablet, μ The Null Hypothesis says that the population of all aspirin tablets produced by this manufacturer has mean μ = 325 mg: - H₀: μ = 325 mg Proper dosage of prescription medicine is important, and deviations from the expected dose distribution in either direction (too much or too little aspirin per tablet) would need to be identified and then, of course, remedied The alternative hypothesis is therefore two-sided: - Hₐ: μ ≠ 325 mg

Ex 14.7: The study of inorganic phosphorus levels in the elderly in Ex 14.5 tests the following hypotheses: - H₀: μ = 1.2 - Hₐ: μ < 1.2 Because the Alternative Hypothesis says that μ < 1.2, values of M less than 1.2 favor Hₐ over H₀ The 12 individuals age 75 to 79 had a mean inorganic phosphorus level of M = 1.128 The P-value is the probability of getting an M as small as 1.128 or smaller when the Null Hypothesis is really true

We act for now as if the "simple conditions" listed on page 348 are true: We have a perfect SRS from an exactly Normal population with standard deviation σ known to us The P-Value of a Test of Significance applet automates the work of finding P-values under the simple inference conditions - We enter the information for Ex 14.5 into the applet (i.e., hypotheses, σ, n, & M), then we click "Update" Figure 14.6 shows the applet output - It displays the P-value as an area under a Normal curve and its corresponding value P = 0.0063 - (Note that this value is a simple Normal computation like the ones we saw in Chapter 11) The small P-value says that we would rarely observe a sample mean inorganic phosphorus level of 1.128 or lower if H₀ was true - This is strong evidence against H₀ and in favor of the alternative Hₐ: μ < 1.2 We conclude that the data provide strong, statistically significant evidence (P = 0.0063) that the mean inorganic phosphorus level in the population of adults age 75 to 79 is less than 1.2 mmol/l

Apply Your Knowledge 14.5: A manufacturer of pharmaceutical products analyzes each batch of a product to verify the concentration of the active ingredient - The chemical analysis is not perfectly precise - In fact, repeated measurements follow a Normal distribution with mean μ equal to the true concentration and with standard deviation σ = 0.0068 grams per liter (g/l) Three analyses of one batch give concentrations of 0.8403, 0.8363, and 0.8447 g/l - To estimate the true concentration, give a 95% confidence interval for μ - Follow the four-step process as illustrated in Example 14.3

We are 95% confident that the concentration of active ingredient in this batch is between 0.8327 and 0.8481 g/l m = z*(σ/√n) = (1.96)(0.0068/√3) = 0.0077 M = 0.8404 M ± m - 0.8404 + 0.0077 = 0.8481 - 0.8404 - 0.0077 = 0.8327

Ex 14.10: In Ex 14.9, a study found that the mean body temperature for a random sample of 130 healthy adults was M = 98.25 °F - Is this value significantly different from the textbook value μ₀ = 98.6 °F at the 5% significance level?

We can answer this question: - Directly by performing a two-sided test - Indirectly by obtaining a 95% Confidence Interval Software gives the interval (98.15, 98.35) The hypothesized value μ₀ = 98.6 falls outside this Confidence Interval - We can reject H₀: μ = 98.6 at the 5% Significance Level Let's examine a different situation, purely for the sake of argument: - If the population mean temperature quoted in textbooks had been 98.3 °F, the value tested in H₀ would have fallen inside the Confidence Interval obtained with these data - We would not have been able to reject the claim H₀: μ = 98.3 at the 5% level

Confidence Intervals for the Mean μ

We computed a 95% confidence interval for the mean height μ of eight-year-old American boys in the form of M ± (2σ/√n) - This approach relies on the approximate 68-95-99.7 rule for Normal distributions, which tells us that the central 95% of the Normal Sampling Distribution of M spreads out 2 standard deviations in both directions from the mean μ We can use this reasoning to compute a: - 68% confidence interval for μ in the form of M ± (σ/√n) - 99.7% confidence interval for μ in the form of M ± (3σ/√n) In fact, we can choose any Confidence Level C by finding the proper value to catch the central area C under the Normal Sampling Distribution - Critical Values for each Confidence Level C For C = 95%, the Critical Value is actually z* = 1.960 - This is a bit more precise than the approximate value z* = 2 based on the 68-95-99.7 rule If we start at the sample mean M and move outward by z* standard deviations on either side, we get an interval that contains the population mean μ in a proportion C of all samples This interval is: - From M - z*(σ/√n) to M + z*(σ/√n) - M ± z*(σ/√n) - It is a Level C Confidence Interval for μ

95% Confidence Interval

We have this because the interval catches the unknown value of the Parameter (μ in this case) with 95% probability based on the overall success rate of the method - That is, we are 95% confident that the unknown value of the Parameter is a value within the Confidence Interval

Apply Your Knowledge 14.3: A 2012 Gallup survey of a random sample of 1014 American adults indicates that American families spend, on average, $151 per week on food - The report further states that, with 95% confidence, this estimate has a margin of error of ±$7 a. This confidence interval is expressed in the following form: "estimate ± margin of error" - What is the range of values (lower bound, upper bound) that corresponds to this confidence interval? b. What is the parameter captured by this confidence interval? - What does it mean to say that we have "95% confidence" in this interval?

a) (144, 158) dollars - M + m = 151 + 7 = 158 - M - m = 151 - 7 = 144 b) Parameter: mean weekly spending on food μ for all American adults - The procedure results in a 95% chance of capturing the true value μ

Apply Your Knowledge 14.1: Suppose that you measure the heights of an SRS of 400 eight-year-old American girls, a population with mean μ = 140 cm and standard deviation σ = 8 cm - The mean M of the 400 heights will vary if you take repeated samples a. The sampling distribution of M is approximately Normal - It has mean μ = 140 cm - What is its standard deviation? b. Sketch the Normal curve that describes how M varies in many samples from this population - Mark the mean μ = 140 - According to the 68-95-99.7 rule, approximately 95% of all the values of M fall within ___________ cm of the mean - What is the missing number? - Call it m for "margin of error" - Shade the region from the mean minus m to the mean plus m on the axis of your sketch, as in Figure 14.1 c. Whenever M falls in the region you shaded, the true value of the population mean, μ = 140, lies in the interval between M - m and M + m - Draw that interval below your sketch for one value of M inside the shaded region and one value of M outside the shaded region d. In what percent of all samples will the confidence interval M ± m capture the true mean μ = 140?

a) 0.4 - σ/√n - 8/√400 - 8/20 - 0.4 b) 0.8 cm - 2 SD's away from μ = 140; 0.4 + 0.4 = 0.8 - m = z*(σ/√n) = (1.96)(0.4) = 0.784 ≈ 0.8 d) 95%

Exercise 14.31: Deer mice (Peromyscus maniculatus) are small rodents native to North America - Their adult body lengths (excluding tail) are known to vary approximately Normally, with mean μ = 86 millimeters (mm) and standard deviation σ = 8 mm Deer mice are found in diverse habitats and exhibit different adaptations to their environment - A random sample of 14 deer mice in a rich forest habitat gives an average body length of M = 91.1 mm - Assume that the standard deviation σ of all deer mice in this area is also 8 mm Here is the output from the TI-83 graphing calculator for a 95% confidence interval based on these findings: - ZInterval = (86.909, 95.291) Exercise 14.30 gives a 95% confidence interval for the mean body length of deer mice living in a rich forest habitat a. What is the standard deviation of the sampling distribution of mean body lengths M? b. What critical value was used to compute this 95% confidence interval? c. Show the step-by-step computations required to arrive at the interval provided by the graphing calculator d. Would a 90% confidence interval based on the same data be larger or smaller? - Explain your reasoning

a) 2.138 - σ/√n - 8/√14 - 8/3.742 - 2.138 b) 1.96 c) 91.1 ± (1.96)(2.138) - M ± m - M = 91.1 - m = z*(σ/√n) = 4.191 91.1 + 4.191 = 95.29 91.1 - 4.191 = 86.91 d) Smaller - Because the critical value would be smaller (z* = 1.645)

Exercise 14.33: Drug maker GlaxoSmithKline investigated the potential impact of its oral antidiabetic drug Avandia (rosiglitazone maleate) on the blood lipid levels of adults diagnosed with diabetes who might benefit from taking Avandia - The mean baseline low-density lipoprotein (LDL, also known as "bad" cholesterol) level for a random sample of 964 such patients was M = 125.6 milligrams per deciliter (mg/dl) - The distribution of LDL levels in the adult population is known to be close to Normal with standard deviation σ = 30 mg/dl Here is a 99% confidence interval for the mean μ baseline LDL level among adult diabetics who might benefit from taking Avandia, assuming that σ is the same in this population as in the general population - One-Sample Z - The assumed standard deviation = 30 - N = 964 - M = 125.6 - 99% CI: (123.111, 128.089) a. Express this confidence interval as a sentence written in the context of this problem b. One of the "simple conditions" for inference about a population mean is that the population standard deviation σ is known - Explain why this assumption may not be reasonable in this specific example

a) We are 99% confident that the mean baseline LDL level in the population of adults with diabetes is between 123.1 and 128.1 mg/dl b) LDL levels may be different in the general and diabetic adult populations in terms of both center (μ) and variability (σ)

Apply Your Knowledge 14.11: When our brains store information, complicated chemical changes take place - In trying to understand these changes, researchers blocked some processes in brain cells taken from rats and compared these cells with a control group of normal cells - They report that "no difference was seen" between the two groups (P = 0.45) a. State clearly what the P-value P = 0.45 says about the response that was observed b. It isn't literally true that "no difference was seen" - That is, the computed mean response was not exactly the same in the two groups - Explain what the researchers mean when they give P = 0.45 and say "no difference was seen"

a) A response at least as large as that which was observed would be seen almost half the time (just by chance) even if there was no effect in the entire population of rats b) The differences observed were small enough that they would often happen just by chance if there was no effect in the entire population of rats

Apply Your Knowledge 14.9: Go back to the anemia study of Exercise 14.7 Suppose that the "simple conditions" hold: The 50 children are an SRS from all Jordanian children, and the hemoglobin level in this population follows a Normal distribution with standard deviation σ = 1.6 g/dl a. We seek evidence AGAINST the claim that μ = 12 - What is the sampling distribution of M in many samples of size 50 if, in fact, μ = 12? - Make a sketch of the Normal curve for this distribution - (Sketch a Normal curve, then mark the axis using what you know about locating the mean and standard deviation on a Normal curve) b. The sample mean was M = 11.3 g/dl - Mark this outcome on the sampling distribution - Enter the hypotheses, n, σ, & M for the Jordan study into the P-Value of a Test of Significance applet - What is the P-value for this study? - What do you conclude? c. A different study of 50 children from another country finds a sample mean M = 11.8 g/dl - Use the applet to obtain the P-value for this country - What do you conclude? d. Explain briefly why these P-values tell us that one outcome is strong evidence against the null hypothesis and that the other outcome is not. e. Is the outcome for either study statistically significant at the α = 0.05 level? - Explain briefly why reporting the actual P-value is more relevant to your conclusion than simply reporting whether the findings are statistically significant at the α = 0.05 level

a) N(12 g/dl, 0.2263 g/dl) - σ/√n - 1.6/√50 - 1.6/7.0711 - 0.2263 b) P = 0.0010, reject H₀ - z = (M - μ/σ) - (11.3 - 12)/0.2263 - (-0.7)/0.2263 - (-3.093) ≈ (-3.09); p = 0.0010 on Table B c) P = 0.1884, fail to reject H₀ - z = (M - μ/σ) - (11.8 - 12)/0.2263 - (-0.2)/0.2263 - (-0.8838) ≈ (-0.88); p = 0.1894 on Table B d) The first P-value is very low, which says that the study outcome would be very unlikely if H₀ was true (so we reject H₀) - The second P-value is not low, which is inconclusive (so we fail to reject H₀) e) The first P-value is statistically significant at the α = 0.05 level, but the second is not - However, the actual P-values are more informative, because we can conclude that the first study provided very strong evidence against (P = 0.001)

Exercise 14.37: One of your friends is testing the effect of drinking coffee on the duration of cold symptoms - The common cold lasts, on average, 6 days Your friend starts with no expectations as to whether drinking coffee will have any effect on cold duration After seeing the results of the experiment, in which the average cold duration was less than 6 days, your friend tests a one-sided alternative about the population mean cold duration when drinking coffee, - H₀: μcoffee = 6 - Hₐ: μcoffee < 6 which gives z = −1.68 with one-sided P-value P = 0.0465 a. Explain why your friend should have used a two-sided alternative hypothesis b. What is the correct two-sided P-value for z = -1.68?

a) She had no expectations before she looked at the data - She cannot change her hypothesis just to fit the data b) P = 0.0930 - 0.0465 x 2

Apply Your Knowledge 14.15: The P-value for a two-sided test of the Null Hypothesis H₀: μ = 10 is 0.06 a. Does the 95% Confidence Interval include the value 10? - Why? b. Does the 90% Confidence Interval include the value 10? - Why?

a) Yes - 10 is within the 95% Confidence Interval, because the P-value is NOT low enough to reject at the α = 0.05 level b) No - 10 is not in the 90% Confidence Interval, because the P-value is low enough to reject at the α = 0.10 level

Exercise 14.47: To assess the accuracy of a laboratory scale, a reference weight known to weigh exactly 10 grams (g) is weighed repeatedly - The scale readings are Normally distributed with standard deviation σ = 0.0002 g The reference weight is weighed 5 times on that scale. The mean result is 10.0023 g. Here is the output from the statistical software Minitab for these data: - One-Sample Z - Test of μ = 10 vs μ ≠ 10 - The assumed standard deviation = 0.0002 - N = 5 - Mean = 10.0023 - SE Mean = 0.0001 - 95% CI: (10.0021, 10.0025) - Z = 25.71 - P = 0.000 a. Do the 5 weighings give good evidence that the scale is not well calibrated (that is, its mean μ for weighing this weight is not 10 g)? - Explain your answer b. Give a 95% confidence interval for the mean weight on this scale for all possible measurements of the reference weight - What do you conclude about the calibration of this scale? c. Compare your conclusions in parts a and b - Explain why the confidence interval is more informative than the test result

a) Yes - P-value very close to zero b) 10.0021 to 10.0025 - The scale is slightly biased on the heavy side c) P-values tell us only that the scale is off, whereas confidence intervals tell us how much it is off (not too much, in fact): Confidence intervals are more informative

Exercise 14.43: Researchers in the study described in Exercise 14.42 also examined the perceived gender orientation of cinnamon - The participants gave the scent of cinnamon an average score of 11.76 a. Is this evidence that the scent of cinnamon is not gender-neutral, on average? - State H₀ & Hₐ in the corresponding test, indicating what the parameter in your hypotheses is b. The researchers report that this test was not statistically significant - Explain why we cannot conclude that cinnamon is a gender-neutral scent c. Assuming that the conditions for inference were met, conclude in the context of this study

a) μ is the mean perceived gender orientation of cinnamon in the American adult population - H₀: μ = 12 - Hₐ: μ ≠ 12 b) A non-significant P-value is inconclusive (H₀ may or may not be true) c) The study was inconclusive and failed to find evidence that cinnamon doesn't have a gender-neutral smell (P not statistically significant)