Chapter 10.2, Problem 67E

(a)

To determine

To explain why did the researchers randomly assign the subjects to the two treatment groups.

Explanation of Solution

The researchers randomly assign the subjects to the two treatment groups such that the groups were as similar as possible prior to the treatments.

This is necessary to do because else we would not be sure that the differences after the treatments are due to the treatment themselves or due to the difference among the groups that were already existing prior to the treatments.

(b)

To determine

To estimate and interpret the P -value.

Answer to Problem 67E

There is $1\%$ probability of getting similar sample results or more extreme results when there is no difference in the mean creativity rating of the intrinsic rewards and the mean creativity rating of the extrinsic rewards.

Explanation of Solution

It is given that:

  $\begin{array}{l}{\overline{x}}_1=19.883\\ {\overline{x}}_2=15.739\end{array}$

Given claim is that: Higher mean creativity rating for intrinsic group.

The claim is either null hypothesis or an alternative hypothesis.

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

Where,

  ${\mu }_1=$ true mean creativity rating for Intrinsic group.

  ${\mu }_2=$ true mean creativity rating for Extrinsic group.

The difference in the sample means is then,

  $\begin{array}{l}{\overline{x}}_1-{\overline{x}}_2=19.883-15.739\\ =4.144\end{array}$

As we know that the P -value is the probability of obtaining the sample results or results more extreme when the null hypothesis is true.

Thus, we note that $1\text{ of the 100}$ dots lies to the right of $4.144$ , then, we have,

  $\begin{array}{l}P\text{-value}=\frac{1}{100}\\ =0.01\\ =1\%\end{array}$

Thus, we conclude that there is $1\%$ probability of getting similar sample results or more extreme results when there is no difference in the mean creativity rating of the intrinsic rewards and the mean creativity rating of the extrinsic rewards.

(c)

To determine

To explain what conclusion would you make.

Answer to Problem 67E

There is convincing evidence to support the claim that intrinsic rewards promote creativity more than extrinsic rewards.

Explanation of Solution

From part (b), we have,

  $\begin{array}{l}{\overline{x}}_1=19.883\\ {\overline{x}}_2=15.739\end{array}$

Given claim is that: Higher mean creativity rating for intrinsic group.

The claim is either null hypothesis or an alternative hypothesis.

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

Where,

  ${\mu }_1=$ true mean creativity rating for Intrinsic group.

  ${\mu }_2=$ true mean creativity rating for Extrinsic group.

The difference in the sample means is then,

  $\begin{array}{l}{\overline{x}}_1-{\overline{x}}_2=19.883-15.739\\ =4.144\end{array}$

As we know that the P -value is the probability of obtaining the sample results or results more extreme when the null hypothesis is true.

Thus, we note that $1\text{ of the 100}$ dots lies to the right of $4.144$ , then, we have,

  $\begin{array}{l}P\text{-value}=\frac{1}{100}\\ =0.01\\ =1\%\end{array}$

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 to support the claim that intrinsic rewards promote creativity more than extrinsic rewards.

(d)

To determine

To explain what type of error could you have made- Type I error or Type II error, based on the conclusion in part (c).

Answer to Problem 67E

Type I error.

Explanation of Solution

We conclude in part (c) that,

There is convincing evidence to support the claim that intrinsic rewards promote creativity more than extrinsic rewards.

A type I error occurs if we reject a null hypothesis when the null hypothesis is true. And the Type II error occurs if we fails to reject the null hypothesis when the null hypothesis is false.

Thus, in this case we reject the null hypothesis then it is a Type I error.