Question

When subjects were treated with a​ drug, their systolic blood pressure readings​ (in mm​ Hg) were measured before and after the drug was taken. The results are given in the table below. Assume that the paired sample data is a simple random sample and that the differences have a distribution that is approximately normal. Using a 0.050 significance​ level, is there sufficient evidence to support the claim that the drug is effective in lowering systolic blood​ pressure?

Before: 205 159 208 195 205 210 207 157 189 167 164 169

After: 176 163 183 176 143 179 189 152 155 177 148 186

In this​ example, Md is the mean value of the differences d for the population of all pairs of​ data, where each individual difference d is defined as the systolic blood pressure reading before the drug was taken minus the reading after the drug was taken. What are the null and alternative hypotheses for the hypothesis​ test?

Thank you in advance.

Answer 1

Here, we have given that,

The Paired sample data Xi and Yi and the difference Di is approximately normally distributed.

Xi: Before the drug measurement of Systolic blood pressure

Yi: After the drug measurement of Systolic blood pressure

Xi (Before)	Yi (After)	Difference Di=Xi-Yi
205	176	29
159	163	-4
208	183	25
195	176	19
205	143	62
210	179	31
207	189	18
157	152	5
189	155	34
167	177	-10
164	148	16
169	186	-17

n=number of observations= 12

= sample mean of differences =

=

=

Sd= sample standard deviation of differences

=

=

= 21.73

we are using paired t-test to test the hypothesis.

Claim: To check whether the drug is effective in lowering systolic blood pressure.

The null and alternative hypothesis is as follows,

v/s

where = the mean value of the differences d for the population of all pairs of​ data.

Now,

Now, we can find the test statistic

Now, first, we find the

=

=6.272

Now,

=

= 2.764

we get the test statistics is 2.764

Now, we want to find the p-value

Degrees of freedom=n-1=12-1=11

= level of significance=0.05

This is left one-tailed test as our interest is to see the drug in lowering the systolic blood pressure,

P-value =1- P( T > statistics)

= 1- 0.0092( using EXCEL =TDIST( |t-statistics |= 2.020,D.F=11,tail=1))

=0.9908

we get the p-value is 0.9908

Decision:

P-value (0.9908) > 0.05 ()

Conclusion:

That is we fail to reject Ho (Null Hypothesis)

There is not sufficient evidence to support the claim the drug is effective in lowering systolic blood pressure.