9_t_Test_Single_Mean

9_t_Test_Single_Mean - t Test for a Single Population Mean...

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Unformatted text preview: t Test for a Single Population Mean t Test for a Single Population Mean • z test vs. t test – Known population standard deviation – Unknown population standard deviation • t test • Degrees of freedom • t distributions • Assumptions • Measures of effect size Example: Bills Fans’ IQs • We know that the national average for IQ scores is 100 with a standard deviation of 16. • We sample 5 people at a tailgate party at Ralph Wilson Stadium and test their IQs. • Are Bills fans’ IQs different? Example: Bills Fans’ IQs H : μ = 100 H 1 : μ ≠ 100 α = .05 z crit = ± 1.96 Example: Bills Fans’ IQs H : μ = 100 H 1 : μ ≠ 100 α = .05 z crit = ± 1.96 Example: Bills Fans’ IQs H : μ = 100 H 1 : μ ≠ 100 α = .05 z crit = ± 1.96 Example: Bills Fans’ IQs H : μ = 100 H 1 : μ ≠ 100 α = .05 z crit = ± 1.96 (Two-tailed Test) Example: Bills Fans’ IQs 115 110 120 135 115 Example: Bills Fans’ IQs 115 110 120 135 115 M = 119 z test μ hyp = 100 σ = 16 z = X − μ σ X = 119 − 100 16 5 = 2.66 z obt = 2.66 > z crit = 1.96 Re ject H : μ = 100 μ hyp = 100 σ = 16 z = X − μ σ X = 119 − 100 16 5 = 2.66 z obt = 2.66 > z crit = 1.96 Re ject H : μ = 100 z test Based on the null hypothesis ( H ) z test μ hyp = 100 σ = 16 z = X − μ σ X = 119 − 100 16 5 = 2.66 z obt = 2.66 > z crit = 1.96 Re ject H : μ = 100 μ hyp = 100 σ = 16 z = X − μ σ X = 119 − 100 16 5 = 2.66 z obt = 2.66 > z crit = 1.96 Re ject H : μ = 100 z test z = M – µ σ M μ hyp = 100 σ = 16 z = X − μ σ X = 119 − 100 16 5 = 2.66 z obt = 2.66 > z crit = 1.96 Re ject H : μ = 100 z test z = M – µ = 119 – 100 = 2.66 σ M 16 / √ 5 μ hyp = 100 σ = 16 z = X − μ σ X = 119 − 100 16 5 = 2.66 z obt = 2.66 > z crit = 1.96 Re ject H : μ = 100 z test z = M – µ = 119 – 100 = 2.66 σ M 16 / √ 5 μ hyp = 100 σ = 16 z = X − μ σ X = 119 − 100 16 5 = 2.66 z obt = 2.66 > z crit = 1.96 Re ject H : μ = 100 z test z = M – µ = 119 – 100 = 2.66 σ M 16 / √ 5 μ hyp = 100 σ = 16 z = X − μ σ X = 119 − 100 16 5 = 2.66 z obt = 2.66 > z crit = 1.96 Re ject H : μ = 100 z test z = M – µ = 119 – 100 = 2.66 σ M 16 / √ 5 The sample mean of 119 is 2.66 standard error units above the hypothesized population mean of Bill’s fans. μ hyp = 100 σ = 16 z = X − μ σ X = 119 − 100 16 5 = 2.66 z obt = 2.66 > z crit = 1.96 Re ject H : μ = 100 z test z = M – µ = 119 – 100 = 2.66 σ M 16 / √ 5 The sample mean of 119 is 2.66 standard error units above the hypothesized population mean of Bill’s fans. Is this a significant effect? μ hyp = 100 σ = 16 z = X − μ σ X = 119 − 100 16 5 = 2.66 z obt = 2.66 > z crit = 1.96 Re ject H : μ = 100 z test z = M – µ = 119 – 100 = 2.66 σ M 16 / √ 5 z observed = 2.66 > z crit = 1.96 μ hyp = 100 σ = 16 z = X − μ σ X = 119 − 100 16 5 = 2.66 z obt = 2.66 > z crit = 1.96 Re ject H : μ = 100 z test z = M – µ = 119 – 100 = 2.66 σ M 16 / √ 5 z observed...
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This note was uploaded on 08/22/2011 for the course PSY 207 taught by Professor Pfordesher during the Fall '07 term at SUNY Buffalo.

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9_t_Test_Single_Mean - t Test for a Single Population Mean...

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