20. In the late 1980s, the pharmaceutical company Upjohn received approval from

Question

20. In the late 1980s, the pharmaceutical company Upjohn received approval from the Food and Drug Administration to market Rogaine'' a 2% minoxidil solution, for the treatment of androgenetic alopecia (male pattern hair loss). Upjohn's advertising campaign for Rogaine included the results of a double-blind randomized clinical trial, conducted with 1431 patients in 27 centers across the United States. The results of this study at the end of four months are summarized in the 2 × 5 contingency table below, where the two row categories represent the treatment arm and control arm respectively, and each column represents a response category the degree of hair growth reported. [Source: Ronald L. Iman, A Data-Based Approach to Statistics, Duxbury Press] Degree of Hair Growth Minimal ModerateDense Growth Growth Growth 58 29 87 ew Total Vellus 172 Rogaine Placebo Total Growth 301 423 724 178 114 292 322 (a) Conduct a Chi-squared Test of the null hypothesis Ho: TRogaine Tplacebo versus the alternative hypothesis HA·TRogane 1 bo across the five hair growth categories (That is, and NoGrowth | Rogaine No Growth | Placebo |Rogaine-New Vellus | Placebo and New Vellus e TDense Growth | Placebo) Infer whether or not we can reject the null and hypothesis at the -.01 significance level. Interpret in context: At the -.01 significance level, what exactly has been demonstrated about the efficacy of Rogaine versus placebo? (b) Form a 2 × 2 contingency table by combining the last four columns into a single column labeled Growth. Conduct a Chi-squared Test for the null hypothesis Ho: TRogaine- Tplacebo versus the alternative H,ogame binary response categories. (That is, Ho: Growth! Rogaine- .) Inter whether or not we can reject the null hypothesis at the = .01 significance level. Interpret in context At the = .01 significance level, what exactly has been demonstrated about the efficacy of 7Placebo between the resulting No Growth versus Growth Growth | Placebo ogaine versus placebo? (c) Calculate the p-value using a two-sample Z-test of the null hypothesis in part (b), and show that the square of the corresponding z-score is equal to the Chi-squared test statistic found in (b). Verify that the same conclusion about Ho is reached, at the = .01 significance level.

Explanation / Answer

PART A..

Given table data is as below

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calculation formula for E table matrix

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expected frequecies calculated by applying E - table matrix formulae

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calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above

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set up null vs alternative as

null, Ho: no relation b/w rogaine and placebo OR rogaine and placebo are independent

alternative, H1: exists a relation b/w rogaine and placebo OR rogaine and placebo are dependent

level of significance, = 0.01

from standard normal table, chi square value at right tailed, ^2 /2 =13.2767

since our test is right tailed,reject Ho when ^2 o > 13.2767

we use test statistic ^2 o = (Oi-Ei)^2/Ei

from the table , ^2 o = 48.4159

critical value

the value of |^2 | at los 0.01 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 5 - 1 ) = 1 * 4 = 4 is 13.2767

we got | ^2| =48.4159 & | ^2 | =13.2767

make decision

hence value of | ^2 o | > | ^2 | and here we reject Ho

^2 p_value =0

ANSWERS

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null, Ho: no relation b/w rogaine and placebo OR rogaine and placebo are independent

alternative, H1: exists a relation b/w rogaine and placebo OR rogaine and placebo are dependent

test statistic: 48.4159

critical value: 13.2767

p-value:0

decision: reject Ho

PART B.

Given table data is as below

------------------------------------------------------------------

calculation formula for E table matrix

------------------------------------------------------------------

expected frequecies calculated by applying E - table matrix formulae

------------------------------------------------------------------

calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above

------------------------------------------------------------------

set up null vs alternative as

null, Ho: no relation b/w rogaine and placebo OR rogaine and placebo are independent

alternative, H1: exists a relation b/w rogaine and placebo OR rogaine and placebo are dependent

level of significance, = 0.01

from standard normal table, chi square value at right tailed, ^2 /2 =6.6349

since our test is right tailed,reject Ho when ^2 o > 6.6349

we use test statistic ^2 o = (Oi-Ei)^2/Ei

from the table , ^2 o = 40.5817

critical value

the value of |^2 | at los 0.01 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 2 - 1 ) = 1 * 1 = 1 is 6.6349

we got | ^2| =40.5817 & | ^2 | =6.6349

make decision

hence value of | ^2 o | > | ^2 | and here we reject Ho

^2 p_value =0

ANSWERS

---------------

null, Ho: no relation b/w rogaine and placebo OR rogaine and placebo are independent

alternative, H1: exists a relation b/w rogaine and placebo OR rogaine and placebo are dependent

test statistic: 40.5817

critical value: 6.6349

p-value:0

decision: reject Ho

MATRIX col1 col2 col3 col4 col5 TOTALS row 1 301 172 178 58 5 714 row 2 423 150 114 29 1 717 TOTALS 724 322 292 87 6 N = 1431

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