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Outline

Small Sample Tests

1. Hypothesis Test for – Small Samples2. t-test Example 13. t-test Example 24. Hypothesis Test for the Population

Proportion p – Large Samples5. Population Proportion Example 16. Population Proportion Example 2

2Hypothesis Testing – Small Samples

Small Sample Tests

Central Limit Theorem (review):

For large enough samples, sampling distribution of the mean will be normal even if the raw data are not normally-distributed.

But what do we do when our sample size is not “large enough?”

3For n < 30, we face two problems:

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a) Central Limit Theorem does not apply

Z does not give probability of finding X in some range relative to μO, when n < 30.

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When sampling distribution of the mean is normal, the Z table gives us the probability that we will find a sample mean in some range (for samples of size n, with μ = μ0).

X

X

μ0

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But when n is < 30, we cannot be sure that the sampling distribution of the mean is normal. So, how do we obtain the probability that a sample mean will be in a given range?

X

μ0

X

6For n < 30, we face two problems:

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b) s is not a good estimator of .

That is, variability in the sample is not a good source of information about variability in the population.

7Hypothesis Testing – Small Samples

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To measure the probability of finding the mean of a small sample in a given range relative to μO, we use a different probability distribution – the t distribution.

t = – s/√n

X

8Hypothesis test for small samples

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CAUTION: The t distribution changes shape with

sample size, becoming more like the SND as n gets larger.

For a hypothesis test, to find the critical value of t in the t table, you need to know 2 things: α and degrees of freedom.

For the one-sample t test, d.f. = n – 1.

9Hypothesis Testing – Small Samples

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Very important point about testing with n < 30:

If an exam question with n < 30 gives you the population standard deviation, , then use Z.

Large n (≥ 30) use Z Small n, known use Z Small n, unknown use t

10Hypothesis Testing – Small Samples

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H0: = 0

HA: ≤ 0 HA: ≠ 0

or HA: ≥ 0

(One-tailed test) (Two-tailed test)

Test Statistic: t = - 0

s/√n

X

11Hypothesis Testing – Small Samples

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Rejection Region:

One-tailed test: Two-tailed test:

tobt > tα │tobt│ > tα/2

or tobt < -tα

Where tα and tα/2 are based on d.f. = (n – 1) Remember to report your decision

explicitly!!

12Confidence Interval – Small Samples

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C.I. = ± tα/2 (s/√n)

Notes: tα/2 is based on d.f. = (n – 1). Use this C.I. when n < 30 and not known.

X

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Example 1 – t-test

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In a recent pollution report, a team of scientists expressed alarm at the dihydrogen monoxide levels in fish in Ontario lakes. Historically, the average dihydrogen monoxide level has been 2.65 parts per thousand (ppt). This year, samples of fish from 20 lakes in Ontario turned up an average dihydrogen monoxide count of 2.98 ppt with a variance of 36.

Note: for more information on DHMO, visit http://www.dhmo.org/

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Example 1 – t-test

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a. Does it appear that dihydrogen monoxide levels are increasing in fish in Ontario lakes? (α = .01)

b. The dihydrogen monoxide levels of fish in 5 lakes in the Timmins area were 2.84, 3.96, 4.40, 1.60, and 2.63. Construct a 95% confidence level for the average dihydrogen monoxide counts in fish in the Timmins area.

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Example 1a – t-test

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H0: = 2.65

HA: > 2.65

Test Statistic: t = - 0

s/√n

Rejection region: tobt > t.01,19 = 2.539

X

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Example 1a – t-test

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= 2.98 s2 = 36 n = 20

tobt = 2.98 – 2.65 √36/20

tobt = 0.246

Decision: Do not reject H0. There is not enough evidence to conclude that dihydrogen monoxide levels are increasing.

X

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Example 1b – t-test

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ΣX = 15.43 = 15.43/5 = 3.086

ΣX2 = 52.584 (ΣX)2 = 15.432 = 238.085

S2 = 52.584 – 238.085 = 1.242 5

4S = √1.242 = 1.114 S = 1.114/√5 = .498

X

X

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Example 1b – t-test

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C.I. = ± tα/2 (s/√n)

For 95% C.I., α = .05. Associated t.025,4 = 2.776

C.I. = 3.086 ± 2.776 (.498) = (1.703 ≤ ≤ 4.469)

X

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Example 2 – t-test

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There have been claims that provincial funding cuts for hospitals have led to an increase in the average waiting time for elective surgery. A search of past records reveals that that average waiting time for elective surgery was 38.5 days prior to the 2003 Ontario election. A random sample of patients scheduled for elective surgery is identified, and the waiting time until surgery for each patient is measured. The data are shown below as # of days intervening between when surgery is ordered and when it occurred.

Is there evidence in these data that average waiting time has increased under the current provincial government? (α = .01)

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Example 2 – t-test

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Patient # Waiting time (days)1 432 283 554 385 306 457 518 39

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Example 2 – t-test

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H0: = 38.5

HA: > 38.5

Test Statistic: t = – 0

s/√n

Rejection region: tobt > t.01,7 = 2.998

X

Why one-tailed?

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Example 2 – t-test

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Σx = 329 n = 8 = 41.125

Σx2 = 14149s2 = 14149 – (329)2

8 7

s2 = 88.411s = √88.411 = 9.402

X

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Example 2 – t-test

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tobt = 41.125 – 38.5 9.402/√8

tobt = 0.787

Decision: Do not reject H0. There is not sufficient evidence to conclude that waiting times have increased.

Note: Be sure to give the full decision.

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Example 3 – t-test

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It is known that the mean # of errors made on a particular pursuit rotor task is 60.9. A physiologist wishes to know if people who have had a spinal cord injury but who are apparently recovered perform less well on this task. In order to test this, a random sample of 8 people who have had spinal cord injuries is chosen and they are administered the pursuit rotor task. The # of errors each made is:

63, 66, 65, 62, 60, 68, 66, 64

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Example 3 – t-test

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a. Is there evidence to support the belief that ‘recovered’ patients are impaired in performing this task? (α = .01)

b. Form the 90% C.I. for the mean number of errors committed by recovered patients.

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Example 3a – t-test

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Note these words:

“It is known…” Population information

“The mean # of errors This is 0.

is … 60.9”

“In order to test this…” Hypothesis Test!!

“a random sample of 8…” n < 30, not known

This calls for a t-test…

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Example 3a – t-test

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H0: = 60.9

HA: > 60.9

Test Statistic: t = - 0

s/√n

Rejection region: tobt > t.01,7 = 2.998

X

Why “greater than?”

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Example 3a – t-test

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= 64.25 s = 2.55 n = 8

tobt = 64.25 – 60.9 2.55/√8

tobt = 3.72

Decision: Reject H0. Recovered patients perform worse than normals on this task.

X

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Example 3b – t-test

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C.I. = ± tα/2 (s/√n)

For 90% C.I., α = .10. Corresponding t.05,7 = 1.895.

C.I. = 64.25 ± 1.895 (2.55/√8)

(62.542 ≤ ≤ 65.958)

X