Chapter 10.2, Problem 57E

(a)

To determine

To explain do these data provide convincing evidence that fish oil helps reduce blood pressure more on average than regular oil or not.

Answer to Problem 57E

There is convincing evidence that fish oil helps reduce blood pressure more on average than regular oil.

Explanation of Solution

In this study the researcher wants to know that fish oil helps reduce blood pressure more on average than regular oil or not.

It is given that:

  $\alpha =0.05$

The given claim that: Mean is higher for the fish oil.

Now, we have to find out the appropriate hypotheses for performing a significance test.

Thus, the claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population proportions are equal. If the null hypothesis is the claim then the alternative hypothesis states the opposite of the null hypothesis.

Therefore, the appropriate hypotheses for this is:

  $\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_a:{\mu }_1>{\mu }_2\end{array}$

Where we have,

  ${\mu }_1=$ the true mean reduction in blood pressure for subjects using fish oil.

  ${\mu }_2=$ the true mean reduction in blood pressure for subjects using regular oil.

Now, the mean is as follows:

  $\begin{array}{l}{\overline{x}}_1=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}\\ =\frac{8+12+10+14+2+0+0}{7}\\ =6.5714\\ {\overline{x}}_2=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}\\ =\frac{-6+0+1+2-3-4+2}{7}\\ =-1.1429\end{array}$

And the standard deviation is as follows:

  $\begin{array}{l}{s}_1=\sqrt{\frac{{\displaystyle \sum {(x-\overline{x})}^2}}{n-1}}\\ =\sqrt{\frac{{(8-6.5714)}^2+{(12-6.5714)}^2++{(10-6.5714)}^2+\mathrm{....}+{(0--6.5714)}^2}{7-1}}\\ =5.8554\\ s_2=\sqrt{\frac{{\displaystyle \sum {(x-\overline{x})}^2}}{n-1}}\\ =\sqrt{\frac{{(-6+1.1429)}^2+{(0+1.1429)}^2+{(1+1.1429)}^2+\mathrm{....}+{(2+1.1429)}^2}{7-1}}\\ =3.1848\end{array}$

Now, find the test statistics:

  $\begin{array}{l}t=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{6.5714-(-1.1429)-0}{\sqrt{\frac{{5.8554}^2}{7}+\frac{{3.1848}^2}{7}}}\\ =3.062\end{array}$

Now, the degree of freedom will be:

  $\begin{array}{l}df=\min (n_1-1,n_2-1)\\ =\min (7-1,7-1)\\ =6\end{array}$

So the P -value will be:

  $0.01<P<0.02$

By the calculator we calculate the P -value that will be: $2\times \text{tcdf(3}\text{.062,1E99,6)}$ which results in the P -value of $0.01108$ .

And we know that if the P -value is less than or equal to the significance level then the null hypothesis is rejected, then,

  $P<0.05\Rightarrow \text{Reject }{H}_0$

Thus, we conclude that there is convincing evidence that fish oil helps reduce blood pressure more on average than regular oil.

(b)

To determine

To interpret the P -value from part (a) in the context of this study.

Answer to Problem 57E

There is $1.108\%$ chance of obtaining similar results or more extreme when the mean reduction in blood pressure is the same for those using fish oil and those using regular oil.

Explanation of Solution

From part (a), we have that,

So the P -value will be:

  $0.01<P<0.02$

By the calculator we calculate the P -value that will be: $2\times \text{tcdf(3}\text{.062,1E99,6)}$ which results in the P -value of $0.01108$ .

And we know that if the P -value is less than or equal to the significance level then the null hypothesis is rejected, then,

  $P<0.05\Rightarrow \text{Reject }{H}_0$

Thus, we conclude that there is convincing evidence that fish oil helps reduce blood pressure more on average than regular oil.

Thus we can say that there is $1.108\%$ chance of obtaining similar results or more extreme when the mean reduction in blood pressure is the same for those using fish oil and those using regular oil.