Question

Suppose a researcher is interested in the effectiveness of a new respiratory treatment drug aimed at increasing forced vital capacity (FVC) among children with respiratory illnesses. In this study, each patient treated with the study drug is frequency paired with another patient on a number of characteristics treated with a placebo to insure independence among samples. Assume the following table represents the FVC for each participant and the paired control. Participant Treated with new drug FVC (%) Participant Treated with placebo FVC (%) A 91 A’ 84 B 90 B’ 89 C 87 C’ 87 D 88 D’ 84 E 82 E’ 80 F 86 F’ 81

A) What study type is this?

B) Carry out a paired sample t test to determine if the new therapy is significantly effective at increasing average FVC in children with respiratory illnesses. Write out the null and alternative hypotheses using an α=0.01. Interpret your results.

Answer selection:

A) Case-control B) Ho: µ1=µ2 Ha: µ1<µ2 T is 2.244 Pvalue is 0.0001 Reject the Ho that there is no significant increase in FVC among patients treated in this study.

A) Matched-Pairs B) Ho: µ1=µ2 Ha: µ1<µ2 T is 2.939 Pvalue is 0.016 Reject the Ho that there is no significant increase in FVC among patients treated in this study.

A) Matched-Pairs B) Ho: µ1=µ2 Ha: µ1<µ2 T is 3.112 Pvalue is 0.05 Fail to reject the Ho that there is no significant increase in FVC among patients treated in this study.

Answer 1

Solution:

b) The null and alternative hypotheses are as follows:

Ho: µ_{​​​​​​1} = µ​​​​​_{​​​​​​2} i.e. The population mean of the FVC in children when treated with new drug is equal to the population mean of the FVC in children when treated with placebo.

Ha: µ_{​​​​​​1} > µ_{​​​​​​2}  i.e. The population mean of the FVC in children, when treated with new drug is greater than the population mean of the FVC in children, when treated with placebo.

The test statistic for paired t-test is given as follows:

Where, d-bar is sample mean of the paired differences, s is sample standard deviation of the paired differences and n is sample size.

New drug (X)	Placebo (Y)	di = X - Y
91	84	7
90	89	1
87	87	0
88	84	4
82	80	2
86	81	5

We have, n = 6

The value of the test statistic is t = 2.939.

Since, our test is right-tailed test, therefore we shall obtain right-tailed p-value for the test statistic. The right tailed p-value is given as follows:

p-value = P(T > t)

p-value = P(T > 2.939)

p-value = 0.016

The p-value is 0.016.

We make decision rule as follows:

If p-value is less than the significance level, then we reject the H_{​​​​​​0} at given significance level.

If p-value is greater than the significance level, then we fail to reject the H_{​​​​​​0} at given significance level.

We have, p-value = 0.016 and significance level = 0.01

(0.016 > 0.01)

Since, p-value is greater than the significance level of 0.01, therefore we shall be fail to reject H_{​​​​​​0} at 0.01 significance level.

Conclusion : At significance level of 0.01, there is sufficient evidence to conclude that there is no significant increase in FVC among patients treated in this study.