ACMS 20340 Statistics for Life Sciences Chapter 14: Introduction - - PowerPoint PPT Presentation

ACMS 20340 Statistics for Life Sciences

Chapter 14: Introduction to Inference

Sampling Distributions

For a population distributed as N(µ, σ) the statistic ¯ x calculated from a sample of size n has the distribution N(µ, σ/√n). We would like to use ¯ x to estimate µ. Unfortunately, while ¯ x is likely to be close to µ, they are unlikely to be exactly equal. We will make things easier and only guess an interval which contains µ instead of its exact value.

Inference Assumptions

We will make the following (possibly unrealistic) assumptions:

◮ The population is normally distributed N(µ, σ). ◮ We do not know µ, but we do know σ. ◮ We have a random sample of size n.

Later we will see how to handle the common case where we do not know σ.

To what extent can we determine µ?

Since the population is distributed as N(µ, σ), we know ¯ x has the distribution N(µ, σ/√n). For example, heights of 8 year old boys are normally distributed with σ = 10. The population also has a mean µ, but we do not know it. The population distribution is N(µ, 10). Samples of size 217 are distributed as N(µ, 0.7). Why? σ/√n = 10/ √ 217 ≈ 10/14.73 ≈ 0.6788 ≈ 0.7.

To what extent can we determine µ?

Using the normal tables, we can calculate the probability that ¯ x is within 1.4 of µ. P(µ − 1.4 < ¯ x < µ + 1.4) = P µ − 1.4 − µ 0.7 < Z < µ + 1.4 − µ 0.7

• = P(−2 < Z < 2)

= 0.954

To what extent can we determine µ?

Thus, the probability that ¯ x is within 1.4 of µ is 0.95. In other words, for 95% of all samples, 1.4 is the maximum distance between ¯ x and µ. So if we estimate that µ lies in the interval [¯ x − 1.4, ¯ x + 1.4], we will be right 95% of the time we take a sample.

Confidence Intervals

We say the interval [¯ x − 1.4, ¯ x + 1.4] is a 95% confidence interval for µ, because 95% of the time, the interval we construct contains µ. The 95% is the confidence level. In general we write the interval as ¯ x ± 1.4 Of course, we could ask for different confidence levels. Other common choices are 90%, and 97%, 98%, 99%. A 100% confidence interval would be the range [−∞, ∞], which is not useful at all. So we must allow the possibility of being wrong.

Confidence Intervals

The interval ¯ x ± 1.4 is not 100% reliable. The exact interval we will get depends on the sample we chose. All the intervals will have length 2.8, but their centers will vary. Saying we are 95% confident means the interval we constructed will contain µ 95% of the time, but 5% of the time it will be wrong.

Confidence Intervals

For any given sample we construct an interval. We only know about the long run probability of

• ur sample giving a good

interval. We do not know, without further information, whether the interval from our particular sample is one of the 95% which contains µ, or

• ne of the 5% which don’t.

Summing Up The Main Idea

The sampling distribution of ¯ x tells us how close to µ the sample mean ¯ x is likely to be. A confidence interval turns that information around to say how close to ¯ x the unknown population mean µ is likely to be.

General Method to Construct a Confidence Interval

We estimate parameter µ of a normal population N(µ, σ) using ¯ x by constructing a level C confidence interval. The interval will look like ¯ x ± z∗ σ √n.

margin of error

z∗ is called the critical value and depends only on C.

Confidence Levels

Common z∗ values are Confidence Level z∗ 90% 1.645 95% 1.960 99% 2.576 For any confidence level C, the critical value z∗ is the number for which P(Z < −z∗) = 1 − C 2 We can find this using a table look-up.

Critical Value in Tables

Or, common values of z∗ are listed in table C in the textbook.

Assumptions

Remember the assumptions we made at the beginning:

◮ The population is normal with distribution N(µ, σ) ◮ We know the value of σ, but do not know µ. ◮ We have a SRS.

How much can we relax these assumptions?

◮ We always need a SRS, otherwise ¯

x is not a random variable.

◮ This method requires us to know σ. (There are technical

problems with estimating σ by s)

◮ We only needed the population to be normal to ensure the

sampling distribution was normal. In practice we can fudge this, especially if the sample sizes are large enough. Then the central limit theorem says the sampling distribution is approximately normal.

A Story About Basketball

Charlie claims that he makes free throws at an 80% clip. To test his claim, we ask Charlie to take 20 shots. Unfortunately, Charlie only makes 8 out of 20. We respond, “Someone who makes 80% of his shots would almost never make only 8 out of 20!” The basis for our response: If Charlie’s claim were true and we repeated the sample of 20 shots many times, then he would almost never make just 8 out of 20 shots.

The basic idea of significance tests

An outcome that would rarely happen if a claim were true is good evidence that the claim is NOT true. As with confidence intervals, we ask what would happen if we repeated the sample or experiment many times. For now, we will assume that we have a perfect SRS from an exactly Normal population with standard deviation σ known to us.

Phosphorus in the blood

Levels of inorganic phosphorus in the blood of adults are Normally distributed with mean µ = 1.2 and standard deviation σ = 0.1 mmol/L. Does inorganic phosphorus blood level decrease with age? A retrospective chart review of 12 men and women between the ages of 75 and 79 yields: 1.26 1.00 1.19 1.39 1.10 1.29 1.00 0.87 1.03 1.00 1.23 1.18 The sample mean is ¯ x = 1.128 mmol/L.

The Question

Do these data provide good evidence that, on average, inorganic phosphorus levels among adults of ages 75 to 79 are lower than in the whole adult population? To answer this question, here’s how we proceed:

◮ We want evidence that the mean blood level of inorganic

phosphorus in adults of ages 75 to 79 is less than 1.2 mmol/L.

◮ Thus the claim we test is that the mean for people ages 75 to

79 is 1.2 mmol/L.

Answering the Question (I)

If the claim that the population mean µ for adults aged 75 to 79 is 1.2 mmol/L were true, then sampling distribution of ¯ x from 12 individuals ages 75 to 79 would be Normal with mean µ¯

x = 1.2 and standard deviation

σ¯

x = σ

√n = 0.1 √ 12 = 0.0289.

Answering the Question (II)

There are two general outcomes to consider:

• 1. A sample mean is close to the population mean.

This outcome could easily occur by chance when the population mean is µ = 1.2.

• 2. A sample mean is far from the population mean.

It is somewhat unlikely for this outcome to occur by chance when the population mean is µ = 1.2.

Answering the Question (III)

In our case, the sample mean ¯ x = 1.128 mmol/L is very far from the population mean µ = 1.2. An observed value this small would rarely occur just by chance if the true µ were equal to 1.2 mmol/L.

Null and Alternative Hypotheses

The claim tested by a statistical test is called the null hypothesis.

◮ The test is designed to determine the strength of the evidence

against the null hypothesis.

◮ Usually the null hypothesis is a statement of “no effect” or

“no difference.” The claim about the population that we are trying to find evidence for is called the alternative hypothesis.

One-sided vs. two-sided alternative hypotheses

The alternative hypothesis is one-sided if it states that a parameter is larger than or that it is smaller than the null hypothesis value. The alternative hypothesis is two-sided if it states that the parameter is merely different from the null value.

Hypothesis Notation

Null hypothesis: H0 Alternative hypothesis: Ha Remember that these are always hypotheses about some population parameter, not some particular outcome.

Back to the phosphorus example

Null hypothesis: “No difference from adult mean

• f 1.2 mmol/L.”

H0 : µ = 1.2 Alternative hypothesis: “Their mean is lower than 1.2 mmol/L.” Ha : µ < 1.2 (one-sided)

Aspirin labels

On an aspirin label, we find the following: “Active Ingredient: Aspirin 325 mg” There will be slight variation in the amount of aspirin, but this is fine as long as the production has mean µ = 325 mg. Let’s test the accuracy of the statement on the label: H0 : µ = 325mg Ha : µ = 325mg Note that this is a two sided alternative hypothesis. Why do we use a two-sided Ha rather than a one-sided Ha?

One last point on hypotheses

Hypotheses should express the expectations or suspicions we have prior to our seeing the data. We shouldn’t first look at the data and then frame hypotheses to fit what the data show.

The P-value of a test

Starting with a null hypothesis, we consider the strength of the evidence against this hypothesis. The number that measures the strength of the evidence against a null hypothesis is called a P-value.

How statistical tests work

A test statistic calculated from the sample data measures how far the data diverge from the null hypothesis H0. Large values of the statistic show that the data are far from what we would expect if H0 were true. The probability, assuming that H0 is true, that the test statistic would take a value as or more extreme than the observed value is called the P-value of the test. The smaller the P-value, the stronger the evidence provided by the data is against H0.

Interpreting P-values

Small P-values ⇒ Evidence against H0 Why? Small P-values say the observed result would be unlikely to

• ccur if H0 were true.

Large P-values ⇒ Fail to pro- vide evidence against H0

One-sided P-value

In the inorganic phosphorus levels example, we tested the hypotheses H0 : µ = 1.2 Ha : µ < 1.2. Values of ¯ x less than 1.2 favor Ha over H0. The 12 individuals of ages 75 to 79 had mean inorganic phosphorus level ¯ x = 1.128. Thus the P-value is the probability of getting an ¯ x as small as 1.128 or smaller when the null hypothesis is really true.

Computing P-values, or Not

We can compute P-values by means of the applet P-Value of a Test of Significance. Our focus for now: Understanding what a P-value means. Next time we’ll talk about how to compute P-values.

Aspirin revisited

Suppose the aspirin content of aspirin tablets from the previous example follows a Normal distribution with σ = 5 mg. H0 : µ = 325mg Ha : µ = 325mg Data from a random sample of 10 aspirin tablets yields ¯ x = 326.9. The alternative hypothesis is two-sided, so the P-value is the probability of getting a sample whose mean ¯ x at least as far from µ = 325 mg in either direction as the observed ¯ x = 326.9.

Conclusion about aspirin?

We failed to find evidence against H0. This just means that the data are consistent with H0. This does not mean that we have clear evidence that H0 is true.

How small should P-values be?

In the phosphorus level example, a P-value of 0.0064 was strong evidence against the null hypothesis. In the aspirin example, a P-value of 0.2302 did not give convincing evidence. How small should a P-value be for us to reject the null hypothesis? Unfortunately, there is no general rule as this ultimately depends

• n the specific circumstances.

Statistical Significance

However, there are fixed values commonly used as evidence against a null hypothesis. The most common values are 0.05 and 0.01. If P ≤ 0.05, then there is no more than a 1 in 20 chance that a sample would give evidence this strong just by chance when the null hypothesis is true. If P ≤ 0.01, its no more than a 1 in 100 chance. These fixed standards for P-values are called significance levels.

Significance levels

If the P-value is less than or equal to α, we say the data are statistically significant at level α. “signficant” = “important” “signficant” = “not likely to happen just by chance due to ran- dom variations from sample to sample”

Test for a Population Mean

For significance tests of a population mean, we compare the sample mean ¯ x with the claimed population mean stated in the null hypothesis H0. The P-value shows how likely (or unlikely) an ¯ x is if H0 is true. So how do we calculate the P-value (without help from an applet)?

z-test for a Population Mean

Draw an SRS of size n from a Normal population that has unknown mean µ and known standard deviation σ. To test the null hypothesis that µ has a specified value H0 : µ = µ0, calculate the one-sample z test statistic z = ¯ x − µ0 σ/√n .

z-scores and P-values

!

Example: Body Temperature

The normal healthy body temperature is 98.6 degrees Fahrenheit (37.0 degrees Celsius). This widely quoted value is based on a paper published in 1868 by German physician Carl Wunderlich, based on over a million body-temperature readings. Suppose we claim that this value is not correct.

Example: State Hypotheses

The null hypothesis is “no difference” from the accepted mean µ0 = 98.6◦F. H0 : µ = 98.6 The alternative hypothesis is two-sided because we have no particular direction in mind prior to examining the data. Ha : µ = 98.6

Example: z Test Statistic

Suppose we know that individual body temperatures follow a Normal distribution with standard deviation σ = 0.6◦F. We take a sample of 130 adults and the mean oral temperature is ¯ x = 98.25◦F. The one-sample z test statistic is z = ¯ x − µ0 σ/√n = 98.25 − 98.6 0.6/ √ 130 = −6.65

Example: Finding the P-value

$

The z-score is off the chart on Table B, so P(Z ≤ −6.65) is essentially zero.

Example: Conclusion

We are doing a two-sided test, so the probability that we compare with the significance level is 2P(Z ≤ −6.65) ≈ 0. Using α = 0.01, we will reject H0. There is strong evidence that the true mean body temperature of healthy adults is not 98.6◦F.

Hypothesis Tests from Confidence Intervals

Confidence intervals and tests of significance have similarities.

◮ Both start with a sample mean ¯

x.

◮ Both rely on Normal probabilities.

In fact, a two-sided test at significance level α can be carried out from a confidence interval with confidence level C = 1 − α.

Hypothesis Tests from Confidence Intervals

A level α two-sided significance test rejects a hypothesis H0 : µ = µ0 when the value µ0 falls outside a level 1 − α confidence interval for µ. Let’s look back at our body temperature example.

Example: Body Temperature

Recall we had a sample of 130 adults with a mean body temperature of ¯ x = 98.25◦F. Also recall that µ0 = 98.6◦F, and the population standard deviation σ = 0.6◦F. Now we will construct a 99% confidence interval.

Example: Body Temperature

The confidence interval is ¯ x ± z∗ σ √n. Plugging in ¯ x, z∗, σ, and n yields 98.25 ± (2.576) 0.6 √ 130 , that is, 98.25 ± 0.136. Thus our interval is [98.11, 98.39].

Example: Body Temperature

The hypothesized parameter value µ0 = 98.6◦F falls outside the confidence interval [98.11, 98.39]. Thus we reject the null hypothesis at a significance level of α = 0.01.