# Section 8.5 Testing a claim about a mean ( σ unknown )

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### Transcript of Section 8.5 Testing a claim about a mean ( σ unknown )

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Section 8.5Testing a claim about a mean

(σ unknown)

ObjectiveFor 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 distributedor

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 TestEnter 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-ValuesA 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-ValuesFrom 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 DieWhen 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