9-5

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9-5

Day 1 Testing Paired Differences (independent samples)

What do you do if samples are independent?

Comparing period 5 Calc average on the midterm and period 6 Calc average midterm?

Get a random sample of seniors and compare GPAs of girls to GPA of boys… If the samples are independent then you compare mean differences using a new formula

σ Known or Unknown

1  1  1  1  2 2 2 2     0 μ 1 0 μ 1 0 μ 1 0 μ 1     μ μ 2 μ 2 μ 2 2 Left Tailed Right Tailed Two Tailed Customary…

And the usuals…

x 1 and x 2 should have normal distributions with mean μ 1 and μ 2 . If both n 1 and n 2 are larger than 30, then the CLT assures you that the distributions of the xbars are normal.

How do I do it?

So glad you asked!! Using the same pattern 1. State null and alternate hypotheses 2. Compute your test statistic 3. Find the P-value that corresponds to the sample test statistic 4. Conclude 5. State your conclusion

σ known

z   x 1  x 2 z  σ 1 2 n 1  x 1   σ 2 2 n 2 x 2   σ 1 2 n 1  σ 2 2 n 2 2 

σ unknown

t   x 1  x 2 t  s 1 2 n 1  x 1   s 2 2 n 2 x 2  s 1 2 n 1  s 2 2 n 2  2 

Example

A random sample of n 1 =12 winter days in For Englewood (a suburb of Denver) a random sample of n 2 =14 winter days gave a pollution index is normally distributed, and previous studies show σ 1 = 21 and σ 2 = 15.

Does this information suggest that the mean population pollution index of Englewood is different from Denver in the winter?

Use 1% level of significance.

Example

A random sample of n 1 =16 communities in western Kansas gave an average rate of hay fever (per 1000, under 25 yrs of age) of 109.50 with a sample standard deviation of 15.41. A random sample of n 2 =14 regions in western Kansas gave an average rate of hay fever (per 1000, over 50 yrs of age) of 99.36 with a sample standard deviation of 11.57.

Assuming that the hay fever rate of each group is approximately normal, does the data suggest that the 50+ has a lower rate of hay fever?

Use a 5% level of confidence.

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