Sample size calculation for a proportion 1.96 2 * p * q n w2 w  ( p  0.5) HYPOTHESIS TEST FOR A PROPORTION THE PROCEDURE IS BASICALLY THE SAME AS FOR THE MEAN. THE ONLY DIFFERENCE IS IN THE FORMULA FOR Z, SINCE WE ARE USING THE NORMAL APPROXIMATION TO THE BINOMIAL DISTRIBUTION z p̂  p and    pq n SINCE WE ARE TESTING AN HYPOTHESIS CONCERNING THE POPULATION PROPORTION, p, WE USE THAT VALUE TO CALCULATE SIGMA. [REMEMBER THE REQUIREMENT FOR USING THE NORMAL APPROXIMATION: np and nq must be greater than 5.] H 0 : p  0.36 1. po p1 p q n sigma a Z alpha H1 : p  0.36   .05 Z  1.96 2. 3. 4. z p̂  p pˆ  z    pq  n .36 *.64  0.03236 220 96  0.4364 220 .4364  .36  2.361 0.03236 5. COMPARE THE ACTUAL AND CRITICAL Z VALUES: SINCE THE ACTUAL Z VALUE OF 2.361 IS GREATER THAN THE CRITICAL VALUE OF 1.96, IT IS IN THE REJECTION REGION. 6. THEREFORE, WE DO REJECT THE NULL HYPOTHESIS. AND WE CONCLUDE THAT THE PROPORTION IS NOT EQUAL TO 0.36. II. P-VALUE APPROACH 0.990887 0.009113 ACCORDING TO THE NORMAL CURVE THE AREA ABOVE A Z-VALUE OF 2.361 = .009 COMPARE THE P-VALUE WITH THE ALPHA VALUE, AND THE P-VALUE OF .009 IS LESS THAN THE ALPHA VALUE OF .025 OR, DOUBLING THE P-VALUE OF .009 = .018. COMPARE THIS TO ALPHA OF .05. THEREFORE, WE DO REJECT THE NULL HYPOTHESIS. AND WE CONCLUDE THAT THE PROPORTION IS NOT EQUAL TO 0.36. R THE MEAN. THE ONLY E ARE USING THE APPROXIMATION:  0.03236 0.700 x n 0.7 p-hat 0.3 225 Z 0.0306 p-value 0.05 -1.645 0.04 S GREATER THAN HE REJECTION REGION. HYPOTHESIS. RTION IS NOT EQUAL ABOVE A Z-VALUE 152 225 0.676 -0.800 0.7882 0.800 1.576 2-tail WEEK 10 - LECTURE NOTES STATISTICAL APPLICATIONS - HYPOTHESIS TESTS I. HYPOTHESIS TESTS APPLIED TO CONTINUOUS DATA - AVERAGES A. TESTS FOR ONE AVERAGE (MEAN) NULL HYPOTHESIS: POPULATION AVERAGE = ALTERNATIVE HYPOTHESIS: POPULATION AVERAGE DOES NOT = t Distribution is used. x  0 sx t sx  s n Using the example of page 340: 488.50  480.00  2.55 3.333 p  0.03 t ZA sx  10.54  3.333 10 1) Assumptions This test assumes a normal distribution for the population or process. B. TESTS FOR COMPARING TWO AVERAGES NULL HYPOTHESIS: AVERAGE 1 = AVERAGE 2 ALTERNATIVE HYPOTHESIS: AVERAGE 1 DOES NOT = AVERAGE 2 t Distribution is used. t x1  x2 sx1  x2 s x1  x2  s 2p n1  s 2p n2 9 9 1 1  n1 n2  sp Using the example on page 341: 488.5  492.9  0.7145 6.158 p  0.48 t 10.5 16.4 110.25 268.96 379.21 189.605 sx1  x2  s 2p  110.25 268.96 2.439546 992.25 2420.64 3412.89 18 189.605 10 18.9605 2 37.921 6.158003 -4.4 6.158003 -0.71452 189.605 189.605   6.158 10 10 (n1  1) s12  (n2  1) s22 9 *10.52  9 *16.42   189.605 n1  n2  2 18 C. TESTS FOR COMPARING SEVERAL AVERAGES NULL HYPOTHESIS: ALL AVERAGES ARE EQUAL ALTERNATIVE HYPOTHESIS: AT LEAST ONE AVERAGE IS DIFFERENT USING THE EXAMPLE ON PAGES 310, 311 AND 342: Analysis of variance Source df SS Brand 2 Error 27 Total 29 5,067.3 5,112 10,179 Variance F 2,534 189 Averages Brand X Brand Y Brand Z 501 410 488 475 398 467 495 391 483 490 412 484 478 391 498 493 404 491 474 382 481 481 411 502 499 407 511 499 430 524 488.5 403.6 492.9 Variance deg.freedom Sum of sq 111.1667 187.8222 268.9889 9 9 9 Total 1000.5 1690.4 2420.9 5112 p 13.38 0.000 The conclusion is that there are significant differences between some of the averages between the 3 groups. The figure on page 311 clearly shows that. II. HYPOTHESIS TESTS APPLIED TO CONTINUOUS DATA - VARIANCES A. TESTS FOR COMPARING TWO VARIANCES (STANDARD DEVIATIONS) NULL HYPOTHESIS: VARIANCE 1 = VARIANCE 2 ALTERNATIVE HYPOTHESIS: VAR. 1 DOES NOT = VAR. 2 F distribution is used F s22 s12 Using the example on page 343: F  16.4 2  2.44 10.52 p = .20 There is a 20% chance that this difference could be do to random variations. B. TESTS FOR COMPARING SEVERAL VARIANCES NULL HYPOTHESIS: VARIANCES ARE ALL EQUAL ALTERNATIVE HYPOTHESIS: AT LEAST ONE VARIANCE IS DIFFERENT. Levene's test statistic is used. Using the example on page 344: Brand X standard deviation: 10.5 Brand Y standard deviation: 13.7 Brand Z standard deviation: 16.4 Levene's test statistice: .399 p = .68 There is a 68% chance that this difference could be due to random variations. III. HYPOTHESIS TESTS APPLIED TO DISCRETE DATA A. COMPARING TWO PROPORTIONS. NULL HYPOTHESIS: PROPORTION 1 = PROPORTION 2 ALTERNATIVE HYPOTHESIS: PROP. 1 DOES NOT = PROP. 2 z test (Normal distribution) is used. z P p1  p2    PQ PQ  n1 n2 P Q 1 n1 p1  n2 p2 n1  n2 Using the example on page 312: .0311  .0769   2.663 .0394 *.9606 .0394 *.9606 0.0172     0.0172 708 156 p  .004 z P 708 *.0311  156 *.0769  0.0394 864 B. COMPARING MORE THAN TWO PROPORTIONS NULL HYPOTHESIS: PROPORTIONS ARE ALL EQUAL ALTERNATIVE HYPOTHESIS: AT LEAST ONE PROPORTION IS DIFFERENT Chi-square test is used. IV. COMPUTATION OF NEEDED SAMPLE SIZES TO ACHIEVE DESIRED RESULTS Various formulas are available and these are shown on pages 348 - 351 SAMPLE SIZE NEEDED FOR A POPULATION PROPORTION: n = (1.96)^2*pq / W^2 W = the desired width of the confidence interval. HOMEWORK HYPOTHESIS TEST NULL HYPOTHESIS: NO DIFFERENCE BETWEEN THE GROUP PROPORTIONS ALTERNATIVE HYPOTHESIS: THERE IS A DIFFERENCE Ho: P = .50 Ha: P > .50 Z = (0.667 - .50) / 0.076 Sigma = Sq.root (p*q/n) p= 0.5 q= 0.5 n= 43 Variance 0.005814 std. dev. 0.0762493 n = (1.96)^2*pq / W^2 constants 3.8416 p 0.5 q 0.5 W^2 0.0225 n 42.68444 Z = 1.295 p = .014 alpha = .05 2.197368421 0.985989777 0.014010223 708 156 864 0.0311 0.0769 -0.0458 0.0172 -2.66275 0.003875 22.0188 11.9964 34.0152 0.039369444 0.960630556 0.037819491 0.037819 708 156 5.34174E-05 0.000242 0.000296 0.0172 Variance of means deg. Of freedom Sum of squares 2533.643 2 5067.287
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