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Section 8.5Testing a claim about a mean

(σ unknown)

Objective

For a population with mean µ (with σ unknown), use a sample to test a claim about the mean.

Testing a mean (when σ known) uses the t-distribution

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Notation

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(1) The population standard deviation σ is unknown

(2) One or both of the following:

Requirements

The population is normally distributed

or

The sample size n > 30

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Test StatisticDenoted t (as in t-score) since the test uses the t-distribution.

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People have died in boat accidents because an obsolete estimate of the mean weight (of 166.3 lb.) was used.

A random sample of n = 40 men yielded the mean

x = 172.55 lb. and standard deviation s = 26.33 lb.Do not assume the population standard deviation is known.

Test the claim that men have a mean weight greater than 166.3 lb. using 90% confidence.

What we know: µ0 = 166.3 n = 40 x = 172.55 s = 26.33

Claim: µ > 166.3 using α = 0.1

Note: Conditions for performing test are satisfied since n >30

Example 1

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What we know: µ0 = 166.3 n = 40 x = 172.55 s = 26.33

Claim: µ > 166.3 using α = 0.1

H0 : µ = 166.3

H1 : µ > 166.3 right-tailed test

Initial Conclusion: Since t in critical region, Reject H0

Final Conclusion: Accept the claim that the mean weight is greater than 166.3 lb.

t in critical region(df = 39)

Using Critical RegionsExample 1

tα = 1.304

t = 1.501

Test statistic:

Critical value:

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Stat → T statistics → One sample → with summary

Calculating P-value for a Mean(σ unknown)

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Then hit Next

Enter the Sample mean (x)

Sample std. dev. (s)

Sample size (n)

Calculating P-value for a Mean(σ unknown)

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Then hit Calculate

Select Hypothesis Test

Enter the Null:mean (µ0)

Select Alternative (“<“, “>”, or “≠”)

Calculating P-value for a Mean(σ unknown)

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Test statistic (t)

P-value

Calculating P-value for a Mean(σ unknown)

The resulting table shows both the test statistic (t) and the P-value

Initial Conclusion

Since P-value < α (α = 0.1), reject H0

Final Conclusion

Accept the claim the mean weight greater than 166.3 Ib

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Using StatCrunch

Using the P-valueExample 1

Stat → T statistics→ One sample → With summary

Null: proportion=

Alternative

Sample mean:

Sample std. dev.:

Sample size:

● Hypothesis Test172.55

37.8

40

166.3

>

P-value = 0.0707

What we know: µ0 = 166.3 n = 40 x = 172.55 s = 26.33

Claim: µ > 166.3 using α = 0.1

Initial Conclusion: Since P-value < α, Reject H0

Final Conclusion: Accept the claim that the mean weight is greater than 166.3 lb.

H0 : µ = 166.3

H1 : µ > 166.3

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P-Values

A useful interpretation of the P-value: it is observed level of significance

Thus, the value 1 – P-value is interpreted as observed level of confidence

Recall: “Confidence Level” = 1 – “Significance Level”

Note: Only useful if we reject H0

If H0 accepted, the observed significance and confidence are not useful.

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P-Values

From Example 1:

P-value = 0.0707 1 – P-value = 0.9293

Thus, we can say conclude the following:

The claim holds under 0.0707 significance.

or equivalently…

We are 92.93% confident the claim holds

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Loaded Die

When a fair die (with equally likely outcomes 1-6) is rolled many times, the mean valued rolled should be 3.5

Your suspicious a die being used at a casino is loaded (that is, it’s mean is a value other than 3.5)

You record the values for 100 rolls and end up with a mean of 3.87 and standard deviation 1.31

Using a confidence level of 99%, does the claim that the dice are loaded?

What we know: µ0 = 3.5 n = 100 x = 3.87 s = 1.31

Claim: µ ≠ 3.5 using α = 0.01

Note: Conditions for performing test are satisfied since n >30

Example 2

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H0 : µ = 3.5

H1 : µ ≠ 3.5

What we know: µ0 = 3.5 n = 100 x = 3.87 s = 1.31

Claim: µ ≠ 3.5 using α = 0.01

two-tailed test

Example 2

t in critical region(df = 99)

Test statistic:

Critical value:z = 3.058

zα = -2.626 zα = 2.626

Using Critical Regions

Initial Conclusion: Since P-value < α, Reject H0

Final Conclusion: Accept the claim the die is loaded.

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Using StatCrunch

Using the P-valueExample 2

Null: proportion=

Alternative

Sample mean:

Sample std. dev.:

Sample size:

● Hypothesis Test3.87

1.31

100

3.5

≠

P-value = 0.0057

Initial Conclusion: Since P-value < α, Reject H0

Final Conclusion: Accept the claim the die is loaded.

H0 : µ = 3.5

H1 : µ ≠ 3.5

What we know: µ0 = 3.5 n = 100 x = 3.87 s = 1.31

Claim: µ ≠ 3.5 using α = 0.01

We are 99.43% confidence the die are loaded

Stat → T statistics→ One sample → With summary

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Section 8.6Testing a claim about a

standard deviation

Objective

For a population with standard deviation σ, use a sample too test a claim about the standard deviation.

Tests of a standard deviation use the 2-distribution

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Notation

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Notation

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(1) The sample is a simple random sample

(2) The population is normally distributed

Very strict condition!!!

Requirements

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Test StatisticDenoted 2 (as in 2-score) since the test uses the 2 -distribution.

n Sample size

s Sample standard deviation

σ0 Claimed standard deviation

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Critical Values

Right-tailed test “>“ Needs one critical value (right tail)

Use StatCrunch: Chi-Squared Calculator

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Critical Values

Left-tailed test “<” Needs one critical value (left tail)

Use StatCrunch: Chi-Squared Calculator

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Critical Values

Two-tailed test “≠“ Needs two critical values (right and left tail)

Use StatCrunch: Chi-Squared Calculator

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Statisics Test Scores

Tests scores in the author’s previous statistic classes have followed a normal distribution with a standard deviation equal to 14.1. His current class has 27 tests scores with a standard deviation of 9.3.

Use a 0.01 significance level to test the claim that this class has less variation than the past classes.

Example 1

What we know: σ0 = 14.1 n = 27 s = 9.3

Claim: σ < 14.1 using α = 0.01

Note: Test conditions are satisfied since population is normally distributed

Problem 14, pg 449

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What we know: σ0 = 14.1 n = 27 s = 9.3

Claim: σ < 14.1 using α = 0.01

H0 : σ = 14.1

H1 : σ < 14.1 Left-tailed

Using Critical RegionsExample 1

2 in critical region

(df = 26)

Initial Conclusion: Since 2 in critical region, Reject H0

Final Conclusion: Accept the claim that the new class has less variance than the past classes

2

2L

Test statistic:

Critical value:

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Calculating P-value for a Variance

Stat → Variance → One sample → with summary

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Then hit Next

Enter the Sample variance (s2)

Sample size (n)

Calculating P-value for a Variance

s2 = 9.32 = 86.49

NOTE: Must use Variance

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Then hit Calculate

Select Hypothesis Test

Enter the Null:variance (σ02)

Select Alternative (“<“, “>”, or “≠”)

Calculating P-value for a Variance

σ02 = 14.12 = 198.81

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Test statistic (2)

P-value

The resulting table shows both the test statistic (2) and the P-value

Calculating P-value for a Variance

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What we know: σ0 = 14.1 n = 27 s = 9.3

Claim: σ < 14.1 using α = 0.01

Using Critical RegionsExample 1

Using StatCrunch

Initial Conclusion: Since P-value < α (α = 0.01), Reject H0

Final Conclusion: Accept the claim that the new class has less variance than the past classes

We are 99.44% confident the claim holds

Stat → Variance → One sample → With summary

Null: proportion=

Alternative

Sample variance:

Sample size:

86.49

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198.81

<

● Hypothesis Test

P-value = 0.0056s2 = 86.49

σ02 = 198.81

H0 : σ2 = 198.81

H1 : σ2 < 198.81

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BMI for Miss America

Listed below are body mass indexes (BMI) for recent Miss America winners. In the 1920s and 1930s, distribution of the BMIs formed a normal distribution with a standard deviation of 1.34.

Use a 0.01 significance level to test the claim that recent Miss America winners appear to have variation that is different from that of the 1920s and 1930s.

Example 2

What we know: σ0 = 1.34 n = 10 s = 1.186

Claim: σ ≠ 1.34 using α = 0.01

Note: Test conditions are satisfied since population is normally distributed

Problem 17, pg 449

19.5 20.3 19.6 20.2 17.8 17.9 19.1 18.8 17.6 16.8

Using StatCrunch: s = 1.1862172

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Using Critical RegionsExample 2

2 not in critical region(df = 26)

Test statistic:

Critical values: 2

0.005

2R

2L

Initial Conclusion: Since 2 not in critical region, Accept H0

Final Conclusion: Reject the claim recent winners have a different variations than in the 20s and 30s

Since H0 accepted, the observed significance isn’t useful.

What we know: σ0 = 1.34 n = 10 s = 1.186

Claim: σ ≠ 1.34 using α = 0.01

H0 : σ = 1.34

H1 : σ ≠ 1.34 two-tailed

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Using P-valueExample 2

Using StatCrunch

Null: proportion=

Alternative

Sample variance:

Sample size:

1.407

10

1.796

<

● Hypothesis Test

P-value = 0.509

Initial Conclusion: Since P-value ≥ α (α = 0.01), Accept H0

Final Conclusion: Reject the claim recent winners have a different variations than in the 20s and 30s

Since H0 accepted, the observed significance isn’t useful.

s2 = 1.407

σ02 = 1.796

What we know: σ0 = 1.34 n = 10 s = 1.186

Claim: σ ≠ 1.34 using α = 0.01

H0 : σ2 = 1.796

H1 : σ2 < 1.796

Stat → Variance → One sample → With summary