x 0 L z 2 x z 2 s n x 0 U z 2 x z 2 s n Steps for Calculating for a Large

# X 0 l z 2 x z 2 s n x 0 u z 2 x z 2 s n steps for

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x 0, L 0 z 2 x 0 z 2 s n x 0, U 0 z 2 x 0 z 2 s n
Steps for Calculating for a Large-Sample Test about µ 2. Specify the value of µ a in the alternative hypothesis for which the value of is to be calculated. Then convert the border value(s) of to z -value(s) using the alternative distribution with mean µ a . The general formula for the z -value is z x 0 a x x 0
Steps for Calculating for a Large-Sample Test about µ Sketch the alternative distribution (centered at µ a ) and shade the area in the acceptance (nonrejection) region. Use the z -statistic(s) and Table II in Appendix D to find the shaded area, which is .
Power of Test Probability of rejecting false H 0 Correct decision Equal to 1 – Used in determining test adequacy Affected by True value of population parameter Significance level Standard deviation & sample size n
Two-Tailed z Test Example Does an average box of cereal contain at least 368 grams of cereal? A random sample of 25 boxes had x = 372.5 . The company has specified to be 15 grams. Test at the .05 level of significance. 368 gm.
Finding Power Step 1 x 0 = 368 Reject H 0 Do Not Reject H 0 Hypothesis: H 0 : 0 368 H a : 0 < 368 = .05 Draw 15 25 n
Finding Power Steps 2 & 3 x a = 360 ‘True’ Situation: a = 360 ( H a ) Draw Specify 1– x 0 = 368 Reject H 0 Do Not Reject H 0 Hypothesis: H 0 : 0 368 H a : 0 < 368 = .05 Draw 15 25 n
Finding Power Step 4 363.065 x a = 360 ‘True’ Situation: a = 360 ( H a ) Draw Specify 1– x 0 = 368 Reject H 0 Do Not Reject H 0 Hypothesis: H 0 : 0 368 H a : 0 < 368 = .05 Draw 15 25 n x L 0 z n 368 1.64 15 25 363.065
Finding Power Step 5 363.065 x a = 360 ‘True’ Situation: a = 360 ( H a ) Draw Specify x 0 = 368 Reject H 0 Do Not Reject H 0 Hypothesis: H 0 : 0 368 H a : 0 < 368 = .05 Draw = .154 1– =.846 z Table 15 25 n x L 0 z n 368 1.64 15 25 363.065
Properties of and Power 1. For fixed n and , the value of decreases, and the power increases as the distance between the specified null value µ 0 and the specified alternative value µ a increases.
Properties of and Power 2. For fixed n and values of µ 0 and µ a , the value of increases, and the power decreases as the value of is decreased.
Properties of and Power 3. For fixed n and values of µ 0 and µ a , the value of decreases, and the power increases as the sample size n is increased.
Key Ideas Key Words for Identifying the Target Parameter – Mean, Average p – Proportion, Fraction, Percentage, Rate, Probability 2 – Variance, Variability, Spread
Key Ideas Elements of a Hypothesis Test 1. Null hypothesis ( H 0 ) 2. Alternative hypothesis ( H a ) 3. Test statistic ( z, t, or 2 ) 4. Significance level ( ) 5. p-value 6. Conclusion
Key Ideas Errors in Hypothesis Testing Type I Error = Reject H 0 when H 0 is true (occurs with probability ) Type II Error = Accept H 0 when H 0 is false (occurs with probability ) Power of Test = P (Reject H 0 when H 0 is false) = 1 –
Key Ideas Forms of Alternative Hypothesis Lower-tailed : H a : < 50 Upper-tailed : H a : > 50 Two-tailed : H a : ≠ 50
Key Ideas Using p -values to Decide 1. Choose significance level ( ) 2. Obtain p -value of the test 3. If > p -value, reject H 0
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