Question

The following table summarizes information from a randomized clinical trial that compared two treatments (medication and placebo) from a respiratory disorder. Outcomes were recorded as either favorable or unfavorable.

(a) (2 points) Write null and alternative hypotheses about the possible relationship between the two variables.

(b) (4 points) Calculate the table of expected counts for these data.

(c) (2 points) Calculate the value of the χ 2 statistic for testing your null hypothesis from (a). You must show your work for full credit.

(d) (1 point) Using statistical software, compute the p-value associated with the χ 2 statistic from part (d).

(e) (3 points) Based on the results to (c) and (d), what can you conclude about your hypothesis from part (a)? Give an explanation in the context of this situation.

5. (12 points) The following table summarizes information from a randomized clinical trial that compared two treatments (medication and placebo) from a respiratory disorder. Outcomes were recorded as either favorable or unfavorable. Outcome Treatment Medication Placebo Total Unfavorable Total Favorable 38 24 21 40 (a) (2 points) Write null and alternative hypotheses about the possible relationship between the two vari- ables. (b) (4 points) Calculate the table of expected counts for these data. (c) (2 points) Calculate the value of the x² statistic for testing your null hypothesis from (a). You must show your work for full credit. (d) (1 point) Using statistical software, compute the p-value associated with the x² statistic from part (d). (e) (3 points) Based on the results to (c) and (d), what can you conclude about your hypothesis from part (a)? Give an explanation in the context of this situation.

Answer 1

Answer Date: 29/01/2020 To test the hypothesis is that there is a relationship between treatment and outcome at 5% significance level. The null hypothesis is that there is no relationship between treatment and outcome. The alternative hypothesis is that is a relationship between treatment and outcome. The expected frequency for Medication treatment and Favourable outcome is, Row total x Column total Expected frequency = N 62 x 59 123 = 29.74

The expected frequency for Medication treatment and Unfavourable outcome is, Row total x Column total Expected frequency = N 62 x 64 123 = 32.26 The expected frequency for Placebo treatment and Favourable outcome is, Row total x Column total Expected frequency = N 61x 59 123 = 29.26

The expected frequency for Placebo treatment and Unfavourable outcome is, Row total > Column total Expected frequency = N 61x 64 123 = 31.74 =- A contingency table for expected value is, Total Outcome Favourable | Unfavourable 29.74 32.26 29.26 31.74 59 64 Treatment Medication Placebo Total 62 123 The chi-square test statistic is,

x=(- Sot (38-29.74)2 29.74 = 8.891 (24 -32.262 32.26 + (21-29.262 (40-31.74)2 29.26 31.74 The chi-square test statistic is (8.891). The p-value is, By using Excel software, find p-value with the help of following steps: 1) Import the data. 2) Select Contingency table from Chi-Square in MegaStat option from Add-Ins menu. 3) Select Chi-Square and Expected frequencies in Output options. 4) Click Ok.

Chi-square Contingency Table Test for Independence Unfavorable Total Favorable 38 29.74 24 32.26 62.00 21 40 61 Medication Observed Expected Placebo Observed Expected Observed Expected 29.26 31.74 Total 64 61.00 123 123.00 59.00 64.00 8.891 chi-square 1 df .0029 p-value From the Excel output, the p-value is 0.0029]. Based on result parts to (c) and (d), the conclusion is that the significant p-value is less than 0.05 which is 0.0029, so the null hypothesis is rejected at 5% level of significance.

There is sufficient evidence to indicate that there is a relationship between treatment and outcome. The result is statistically significant.