ACMS 20340Statistics for Life Sciences

Chapter 14:Introduction to Inference

Sampling Distributions

For a population distributed as N(µ, σ) the statistic x̄ calculatedfrom a sample of size n has the distribution N(µ, σ/

√n).

We would like to use x̄ to estimate µ.

Unfortunately, while x̄ is likely to be close to µ, they are unlikely tobe exactly equal.

We will make things easier and only guess an interval whichcontains µ instead of its exact value.

Inference Assumptions

We will make the following (possibly unrealistic) assumptions:

I The population is normally distributed N(µ, σ).

I We do not know µ, but we do know σ.

I We have a random sample of size n.

Later we will see how to handle the common case where we do notknow σ.

To what extent can we determine µ?

Since the population is distributed as N(µ, σ), we know x̄ has thedistribution N(µ, σ/

√n).

For example, heights of 8 year old boys are normally distributedwith σ = 10. The population also has a mean µ, but we do notknow 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 x̄ iswithin 1.4 of µ.

P(µ− 1.4 < x̄ < 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 x̄ is within 1.4 of µ is 0.95.

In other words, for 95% of all samples, 1.4 is the maximumdistance between x̄ and µ.

So if we estimate that µ lies in the interval [x̄ − 1.4, x̄ + 1.4], wewill be right 95% of the time we take a sample.

Confidence Intervals

We say the interval [x̄ − 1.4, x̄ + 1.4] is a 95% confidence intervalfor µ, because 95% of the time, the interval we construct containsµ.

The 95% is the confidence level.

In general we write the interval as

x̄ ± 1.4

Of course, we could ask for different confidence levels. Othercommon choices are 90%, and 97%, 98%, 99%.

A 100% confidence interval would be the range [−∞,∞], which isnot useful at all. So we must allow the possibility of being wrong.

Confidence IntervalsThe interval x̄ ± 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 thetime, but 5% of the time it will be wrong.

Confidence Intervals

For any given samplewe construct an interval.We only know about thelong run probability ofour sample giving a goodinterval.

We do not know, without further information, whether the intervalfrom our particular sample is one of the 95% which contains µ, orone of the 5% which don’t.

Summing Up The Main Idea

The sampling distribution of x̄ tells us how close to µ the samplemean x̄ is likely to be.

A confidence interval turns that information around to say howclose to x̄ 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 x̄by constructing a level C confidence interval.The interval will look like

x̄ ± 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% 1.64595% 1.96099% 2.576

For any confidence level C , the critical value z∗ is the number forwhich

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:

I The population is normal with distribution N(µ, σ)

I We know the value of σ, but do not know µ.

I We have a SRS.

How much can we relax these assumptions?

I We always need a SRS, otherwise x̄ is not a random variable.

I This method requires us to know σ. (There are technicalproblems with estimating σ by s)

I We only needed the population to be normal to ensure thesampling distribution was normal. In practice we can fudgethis, especially if the sample sizes are large enough. Then thecentral limit theorem says the sampling distribution isapproximately 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 almostnever make only 8 out of 20!”

The basis for our response: If Charlie’s claim were true and werepeated the sample of 20 shots many times, then he would almostnever 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 goodevidence that the claim is NOT true.

As with confidence intervals, we ask what would happen if werepeated the sample or experiment many times.

For now, we will assume that we have a perfect SRS from anexactly Normal population with standard deviation σ known to us.

Phosphorus in the blood

Levels of inorganic phosphorus in the blood of adults are Normallydistributed with mean µ = 1.2 and standard deviation σ = 0.1mmol/L.

Does inorganic phosphorus blood level decrease with age?

A retrospective chart review of 12 men and women between theages of 75 and 79 yields:

1.26 1.00 1.19 1.39 1.10 1.291.00 0.87 1.03 1.00 1.23 1.18

The sample mean is x̄ = 1.128 mmol/L.

The Question

Do these data provide good evidence that, on average,inorganic phosphorus levels among adults of ages 75 to79 are lower than in the whole adult population?

To answer this question, here’s how we proceed:

I We want evidence that the mean blood level of inorganicphosphorus in adults of ages 75 to 79 is less than 1.2 mmol/L.

I Thus the claim we test is that the mean for people ages 75 to79 is 1.2 mmol/L.

Answering the Question (I)

If the claim that the population mean µ for adults aged 75 to 79 is1.2 mmol/L were true,

then sampling distribution of x̄ from 12 individuals ages 75 to 79would be Normal

with mean µx̄ = 1.2 and standard deviation

σx̄ =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 thepopulation 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 chancewhen the population mean is µ = 1.2.

Answering the Question (III)

In our case, the sample mean x̄ = 1.128 mmol/L is very far fromthe population mean µ = 1.2.

An observed value this small would rarely occur just by chance ifthe 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.

I The test is designed to determine the strength of the evidenceagainst the null hypothesis.

I Usually the null hypothesis is a statement of “no effect” or“no difference.”

The claim about the population that we are trying to find evidencefor is called the alternative hypothesis.

One-sided vs. two-sided alternative hypotheses

The alternative hypothesis is one-sided if it states that aparameter is larger than or that it is smaller than the nullhypothesis value.

The alternative hypothesis is two-sided if it states that theparameter is merely different from the null value.

Hypothesis Notation

Null hypothesis: H0

Alternative hypothesis: Ha

Remember that these are always hypotheses about somepopulation parameter, not some particular outcome.

Back to the phosphorus example

Nullhypothesis:

“No differencefrom adult meanof 1.2 mmol/L.”

H0 : µ = 1.2

Alternativehypothesis:

“Their mean islower than 1.2mmol/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 isfine as long as the production has mean µ = 325 mg.

Let’s test the accuracy of the statement on the label:

H0 : µ = 325mgHa : µ 6= 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 haveprior to our seeing the data.

We shouldn’t first look at the data and then frame hypotheses tofit what the data show.

The P-value of a test

Starting with a null hypothesis, we consider the strength of theevidence against this hypothesis.

The number that measures the strength of the evidence against anull hypothesis is called a P-value.

How statistical tests work

A test statistic calculated from the sample data measures how farthe data diverge from the null hypothesis H0.

Large values of the statistic show that the data are far from whatwe would expect if H0 were true.

The probability, assuming that H0 is true, that the test statisticwould take a value as or more extreme than the observed value iscalled the P-value of the test.

The smaller the P-value, the stronger the evidence provided by thedata is against H0.

Interpreting P-values

Small P-values ⇒ Evidenceagainst H0

Why? Small P-values say the observed result would be unlikely tooccur if H0 were true.

Large P-values ⇒ Fail to pro-vide evidenceagainst H0

One-sided P-value

In the inorganic phosphorus levels example, we tested thehypotheses

H0 : µ = 1.2Ha : µ < 1.2.

Values of x̄ less than 1.2 favor Ha over H0.

The 12 individuals of ages 75 to 79 had mean inorganicphosphorus level x̄ = 1.128.

Thus the P-value is the probability of getting an x̄ as small as1.128 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 aTest 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 previousexample follows a Normal distribution with σ = 5 mg.

H0 : µ = 325mg

Ha : µ 6= 325mg

Data from a random sample of 10 aspirin tablets yields x̄ = 326.9.

The alternative hypothesis is two-sided, so the P-value is theprobability of getting a sample whose mean x̄ at least as far fromµ = 325 mg in either direction as the observed x̄ = 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 strongevidence against the null hypothesis.

In the aspirin example, a P-value of 0.2302 did not give convincingevidence.

How small should a P-value be for us to reject the null hypothesis?

Unfortunately, there is no general rule as this ultimately dependson the specific circumstances.

Statistical Significance

However, there are fixed values commonly used as evidence againsta 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 asample would give evidence this strong just by chance when thenull 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 arestatistically significant at level α.

“signficant” 6= “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 thesample mean x̄ with the claimed population mean stated in thenull hypothesis H0.

The P-value shows how likely (or unlikely) an x̄ 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 hasunknown 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 =x̄ − µ0

σ/√

n.

z-scores and P-values

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!"#$%&'#()*+(,"-)./'#

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 byGerman physician Carl Wunderlich, based on over a millionbody-temperature readings.

Suppose we claim that this value is not correct.

Example: State Hypotheses

The null hypothesis is “no difference” from the accepted meanµ0 = 98.6◦F.

H0 : µ = 98.6

The alternative hypothesis is two-sided because we have noparticular direction in mind prior to examining the data.

Ha : µ 6= 98.6

Example: z Test Statistic

Suppose we know that individual body temperatures follow aNormal distribution with standard deviation σ = 0.6◦F.

We take a sample of 130 adults and the mean oral temperature isx̄ = 98.25◦F.

The one-sample z test statistic is

z =x̄ − µ0

σ/√

n=

98.25− 98.6

0.6/√

130

= −6.65

Example: Finding the P-value

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40')516789')+6)8::)/0')70#9/)8,)4#;&')<=

68))))))))))))))))))))))))+6)'66',/+#&&>)5'98?)! !""#"#"$$The z-score is off the chart on Table B, so P(Z ≤ −6.65) isessentially zero.

Example: Conclusion

We are doing a two-sided test, so the probability that we comparewith 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 ofhealthy adults is not 98.6◦F.

Hypothesis Tests from Confidence Intervals

Confidence intervals and tests of significance have similarities.

I Both start with a sample mean x̄ .

I Both rely on Normal probabilities.

In fact, a two-sided test at significance level α can be carried outfrom 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 forµ.

Let’s look back at our body temperature example.

Example: Body Temperature

Recall we had a sample of 130 adults with a mean bodytemperature of x̄ = 98.25◦F.

Also recall that µ0 = 98.6◦F, and the population standarddeviation σ = 0.6◦F.

Now we will construct a 99% confidence interval.

Example: Body Temperature

The confidence interval is

x̄ ± z∗σ√n.

Plugging in x̄ , 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 theconfidence interval [98.11, 98.39].

Thus we reject the null hypothesis at a significance level ofα = 0.01.