invest_3ed.pdf

# Standard deviation equal to sd 2 1 ˆ ˆ p p 2 2 2 1

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standard deviation equal to SD( 2 1 ˆ ˆ p p ± ) = 2 2 2 1 1 1 ) 1 ( ) 1 ( n n S S S S ± ² ± . Under the null hypothesis H 0 : S 1 ± S 2 = 0, the standard deviation simplifies to ¸ ¸ ¹ · ¨ ¨ © § ² ± 2 1 1 1 ) 1 ( n n S S where S is the common population proportion. Technical Conditions: We will consider the normal model appropriate if the sample sizes are large, namely n 1 S 1 > 5, n 1 (1 ± S 1 ) > 5, n 2 S 2 > 5, n 2 (1 ± S 2 ) > 5, and the populations are large compared to the sample sizes. Note: The variability in the differences in sample proportions is larger than the variability of individual sample proportions. In fact, the variances (standard deviation squared) add, and then we take the square root of the sum of variances to find the standard deviation. However, to calculate these values we would need to know S 1 , S 2 , or S . So again we estimate the standard deviation of our statistic using the sample data. Case 1: When the null hypothesis is true, we are assuming the samples come from the same population, so we pool the two samples together to estimate the common population proportion of successes. That is, we estimate S by looking at the ratio of the total number of successes to the total sample size: 2 1 2 1 ˆ n n X X p ² ² = size sample total successes of number total n n p n p n ² ² 2 1 2 2 1 1 ˆ ˆ Then use we use this value to calculate the standard error of 2 1 ˆ ˆ p p ± to be: ¸ ¸ ¹ · ¨ ¨ © § ² ± ± 2 1 2 1 1 1 ) ˆ 1 ( ˆ ) ˆ ˆ ( n n p p p p SE

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Chance/Rossman, 2015 ISCAM III Investigation 3.1 192 (v) Use these theoretical results to suggest the general formula for a test statistic and a method for calculating a p-value to test H 0 : S 1 ± S 2 = 0 (also expressed as H 0: S ¼ S º ) versus the alternative H a : S 1 ± S 2 < 0. (This is referred to as the two-sample z -test or two proportion z -test .) Test statistic = z = ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?−ℎ࠵?࠵?࠵?࠵?ℎ࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵? = (w) Calculate and interpret the value of the test statistic specified in (v) as applied to the hearing loss study. 94 ˆ p ± 06 ˆ p = p ˆ = SE( 94 ˆ p ± 06 ˆ p )= test statistic: z = interpretation: (x) Is the standard error close to the empirical standard deviation from your simulation results? (y) Compute the p-value for this test statistic and compare it to your simulation results. Confidence Interval Case 2: When we make no assumptions about the populations (for example, when we are not testing a particular null hypothesis but only estimating the parameter), we will use a different formula to approximate the standard deviation of 2 1 ˆ ˆ p p ± . (z) Using the above CLT result, suggest a general formula for a confidence interval for the difference in population proportions S 1 S 2 .
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