Test for a Mean

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Test for a Mean
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Example

• A city needs 32,000 in annual revenue from
  parking fees. Parking is free on weekends and
  holidays there are 250 days in which parking is
  not free. This implies that the daily revenue
  must be at least 128.
• Officials initially set rates on the low end,
  hoping to attract more shoppers. After a trial
  period of 25 days, they will test the claim that
  too little revenue is being collected. The
  significance level is set at ? 0.05.

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Test of a Mean

• H0 ? 128 H1 ? lt 128
• Assuming sampling is done randomly
• For a sample from a confirmed Normal population
  (or a sufficiently large sample), the test
  statistic t follows a (an) approximate t
  distribution with (n 1) DF.

P-value computations depend on the orientation
(right-, left- two-tailed) of H1.
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Data

• 115.34 142.43 137.10 123.65 87.00
• 123.77 118.83 117.48 141.30 128.54
• 125.44 110.43 133.78 96.86 121.94
• 130.73 133.67 119.11 137.56 124.70
• 99.63 121.20 117.66 121.98 132.37

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Example

• H0 ? 128

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Test Statistic / P-value
H0 ? 128 H1 ? lt 128

• Since the test is left tailed, we need the area
  below -2.037 for a t distribution with 24 DF.

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Sampling Distribution of T24 When H0is True
From Data
From Table
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Sampling Distribution of T24 When H0is True
More than 0.025
From Data
From Table
9
Sampling Distribution of T24 When H0is True
From Data
From Table
10
Sampling Distribution of T24 When H0is True
Less than 0.03
From Data
From Table
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Test Statistic / P-value

• Since the test is left tailed, we need the area
  below -2.037 for a t distribution with 24 DF.
• That area is between 0.025 and 0.03.
• T 2.06 1.97 2.037
• Area 0.025 0.03

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Test Statistic / P-value

• Since the test is left tailed, we need the area
  below -2.037 for a t distribution with 24 DF.
• That area is ________ .
• T 2.06 1.97 2.037
• Area 0.025 0.03 0.026
• by interpolation

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Test Statistic / P-value

• Since the test is left tailed, we need the area
  below -2.037 for a t distribution with 24 DF.
• That area is 0.026.
• T 2.06 1.97 2.037
• Area 0.025 0.03 0.026
• by interpolation

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Test Statistic / P-value

• T 2.06 1.97 2.037
• Area 0.025 0.03 0.026
• You neednt interpolate. Write either
• 0.025 lt P-value lt 0.03 (between 0.025 0.03)
• OR
• P-value 0.03 (better P-value ?
  0.03) (always take the larger probability)

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Interpretation of P-value

• Suppose revenue has population mean 128 per day.
• (Only) 2.6 of all possible samples of size 25
  give a T statistic as low as the observed -2.037
  (which comes from the sample mean of 122.50,
  with standard deviation 13.50).

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Conclusion

• Since P-value 0.026 lt 0.05 we reject the null
  hypothesis.
• Simple, non technical interpretation of the
  conclusion
• There is sufficient evidence in the data to
  conclude that the mean daily revenue is less than
  the needed 128 per day.

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Quiz

• H0 ? 30 H1 ? gt 30
• n 19 DF 18
• T 1.96 P-value 0.033
• True or false?
• 3.3 of the data is above 30.
• FALSE

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Quiz

• H0 ? 30 H1 ? gt 30
• n 19 DF 18
• T 1.96 P-value 0.033
• True or false?
• 3.3 of the data is below 30.
• FALSE

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Quiz

• H0 ? 30 H1 ? gt 30
• n 19 DF 18
• T 1.96 P-value 0.033
• True or false?
• 3.3 of the cars go 30.
• FALSE

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Quiz

• H0 ? 30 H1 ? gt 30
• n 19 DF 18
• T 1.96 P-value 0.033
• True or false?
• The probability the null hypothesis is true is
  0.033.
• FALSE
• but if you think this way youll make correct
  decisions

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What a P-value is

• H0 ? 30 H1 ? gt 30
• n 19 DF 18
• T 1.96 P-value 0.033
• Assume (pretend) the null hypothesis is true.
  Consider (pretend) the study is redone. The
  probability of a T at least as high as 1.96 is
  0.033.
• T 1.96 because the sample mean is 1.96 standard
  deviations above 30. Not so likely.