power and sample size in testing one mean

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Power and Sample Size in Testing One Mean

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Type I & Type II Error

• Type I Error: reject the null hypothesis when it is true. The probability of a Type I Error is denoted by .

• Type II Error: accept the null hypothesis when it is false and the alternative hypothesis is true. The probability of a Type II Error is denoted by .

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Null hypothesis

Decision True False

Reject H0 Type I Error Correct Decision

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Do not reject H0 Correct Decision

Type II Error

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The serum cholesterol levels for all 20- to 24-year-old males of a certain

population was claimed to be normally distributed with

a mean of 180 mg/100ml and

standard deviation is 46 mg/100ml.

A Population

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Is the mean cholesterol level of the population of 20- to 74-year-olds

higher than 180 mg/100ml ?

H0: = 180 mg/100ml (or 180 mg/100ml)

Ha: > 180 mg/100ml

Research Hypothesis

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Power of the Test

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Power of the TestThe power of the statistical test is the ability of the study as designed to distinguish between the hypothesized value and some specific alternative value. That is, the power is the probability of rejecting the null hypothesis if the null hypothesis is false.

Power = P(reject H0 | H0 is false)Power = 1

The power is usually calculated given a simple alternative hypothesis: Ha: = 211 mg/100ml

Power = P(reject H0 | Ha is true)

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0 1.645

Rejection region

z

Right tail area is 0.05

Assuming that we use a sample of size 25 and test the hypothesis with a level of significant = 0.05, what would be the and the power of the test, i.e. 1 ?

If we take the critical value approach, the critical value would be 1.645. At the level of significant = 0.05, we would reject H0 if z 1.645.

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If a sample of size 25 is selected, what would be the critical value in terms of the mean cholesterol level?

0 1.645

Rejection region

z

Right tail area is 0.05

x180 ?

Rejection region

1.19525

46645.1180 195.1

)25

46,180( XXNX

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If the alternative hypothesis Ha: = a = 211 mg/100ml, what would be the probability of NOT rejecting H0, i.e. ?

H0: = 180 v.s. Ha: = 211

That is to say “how powerful can this test detect a 31 mg/100ml increase in average cholesterol level?”

x

= ?

180 195.1 211

Power of the test when sample size is 25:

Power = 1 – = 1 – 0.042 = 0.958.

0.042

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Assuming that we use a sample of size 100 and test the hypothesis with a level of significant = 0.05, what would be the and the power of the test, i.e. 1 ?

0 1.645

Rejection region

zRight tail area is 0.05

6.187100

46645.1180

x180 ?

Rejection region

187.6

)100

46,180( XXNX

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If we assume the actual population mean is 211 mg/100ml for Ha, what is the probability of accepting H0, i.e. ?

x

= ?

180 187.6 211

= .05

Power of the test when sample size is 100: Power = 1 – ?

Recall that the power was .958 when sample size was 25.

0.00

1 – 0 1

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x 180 211

n = 25

187.6

n = 100

195.1

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Sample Size Estimation

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H0: = 180 mg/100ml (or 180 mg/100ml) v.s.HA: > 180 mg/100ml,

to perform a test at the power level of 0.95, with the level of significance of 0.01, how large a sample do we need?

(Power of 0.95 means we want to risk a 5% chance of failing to reject the null hypothesis if the true mean is as large as 211 mg/100ml. Or, to say, we want to have a 95% chance of rejecting the null hypothesis if the true mean is as large as 211 mg/100ml.)

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x

180 211

= 0.05 = .01

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nx

4632.2180*

If = .01, then the critical value in z-score would be 2.32. The critical value in average cholesterol level would be

If the true mean is 211 mg/100ml, with a power of 0.95, i.e. = .05, we would accept the null hypothesis when the sample average is less than

nx

46645.1211*

= 0.05

Let the unknown sample size to be n.

x

= .01

118 211*x

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nn

46645.1211

4632.2180

6.34180211

)46)(645.132.2( 2

n

Set the two equations equal to each other:

Solve for n:

So, the sample size needed is 35.

x

118 211*x

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2

0

)(

a

zzn

Formula: The estimated sample size for a one-sided test with level of significance and a power of 1 to detect a difference of a – 0 is,

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2

0

2/ )(

a

zzn

For a two-sided test, the formula is

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Example:

For testing hypothesis that

H0: = 0 = 70 v.s. Ha: 70 (Two-sided Test)

with a level of significance of = .05. Find the sample size so that one can have a power of 1 = .90 to reject the null hypothesis if the actual mean is a = 80. The standard deviation, , is approximately equal 15.

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2

0

2/ )(

a

zzn

= .051 = .90 = .10

The sample size needed is 24.

z/2 = z.025 = 1.96

z = z.1 = 1.28

2462.237080

15)28.196.1(2

n

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Example: For testing hypothesis that H0: = 0 = 70 v.s. Ha: < 70 (One-sided Test)with a level of significance of = .05. Find the sample size so that one can have a power of 1 .95 to reject the null hypothesis if the actual mean is a 60 (or the actual difference in means is 10). The standard deviation, , is approximately equal 15.

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Example: For testing hypothesis that

H0: = 0 = 70 v.s. Ha: 70 (Two-sided Test)with a level of significance of = .05. Find the sample size so that one can have a power of 1 .95 to reject the null hypothesis if the actual mean is a 75. The standard deviation, , is approximately equal 15.