Chapter PVII, Problem 23RE

(a)

To determine

To explain did the groups differ in average math score at the start of this study.

Answer to Problem 23RE

There is no evidence that the two groups differed in ability at the start of the study.

Explanation of Solution

It is given in the question the table for the instruction methods and the tests as:

		Acc. Math
	Pre-test	Post-test	Individual gain
Mean	560.01	637.55	77.53
SD	84.29	82.9	78.01
n	231	231	231
		Control
	Pre-test	Post-test	Individual gain
Mean	549.65	588.76	39.11
SD	74.68	83.24	66.25
n	245	245	245

Let us assume all the assumptions and conditions are met. Thus, for testing the hypothesis we will use the calculator $\text{TI}-89$ as, first we will go in the STAT TESTS menu, choose $4:2-$SampTTest. Then you must specify if you are using data stored in two lists or if you will enter the means, standard deviations, and sizes of both samples. You must also indicate whether to pool the variances (when in doubt, say no) and specify whether the test is to be two-tail, lower-tail, or upper-tail. So, now we will define the hypothesis as:

Null hypothesis: The groups do not differ in average math score at the start of this study.

Alternative hypothesis: The groups differ in average math score at the start of this study.

Thus, by using the calculator $\text{TI}-89$ , the test statistics and the P-value is as:

  $\begin{array}{l}df=459.3\\ P=0.1574\\ t=1.42\end{array}$

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

  $P>0.05\Rightarrow \text{Fail to Reject }{H}_0$

Thus, we conclude that there is no evidence that the two groups differed in ability at the start of the study.

(b)

To determine

To explain did the groups taught using the accelerated Math program show a significant improvement in test scores.

Answer to Problem 23RE

The groups taught using the accelerated Math program show a significant improvement in test scores.

Explanation of Solution

It is given in the question the table for the instruction methods and the tests as:

		Acc. Math
	Pre-test	Post-test	Individual gain
Mean	560.01	637.55	77.53
SD	84.29	82.9	78.01
n	231	231	231
		Control
	Pre-test	Post-test	Individual gain
Mean	549.65	588.76	39.11
SD	74.68	83.24	66.25
n	245	245	245

Let us assume all the assumptions and conditions are met. Thus, for testing the hypothesis we will use the calculator $\text{TI}-89$ as, first we will go in the STAT TESTS menu, choose $4:2-$ SampTTest. Then you must specify if you are using data stored in two lists or if you will enter the means, standard deviations, and sizes of both samples. You must also indicate whether to pool the variances (when in doubt, say no) and specify whether the test is to be two-tail, lower-tail, or upper-tail. So, now we will define the hypothesis as:

Null hypothesis: The groups taught using the accelerated Math program do not show a significant improvement in test scores.

Alternative hypothesis: The groups taught using the accelerated Math program show a significant improvement in test scores.

Thus, by using the calculator $\text{TI}-89$ , the test statistics and the P-value is as:

  $\begin{array}{l}P<0.0001\\ t=15.11\end{array}$

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

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

Thus, we conclude that the groups taught using the accelerated Math program show a significant improvement in test scores.

(c)

To determine

To explain did the control group show a significant improvement in test scores.

Answer to Problem 23RE

The control group show a significant improvement in test scores.

Explanation of Solution

It is given in the question the table for the instruction methods and the tests as:

		Acc. Math
	Pre-test	Post-test	Individual gain
Mean	560.01	637.55	77.53
SD	84.29	82.9	78.01
n	231	231	231
		Control
	Pre-test	Post-test	Individual gain
Mean	549.65	588.76	39.11
SD	74.68	83.24	66.25
n	245	245	245

Let us assume all the assumptions and conditions are met. Thus, for testing the hypothesis we will use the calculator $\text{TI}-89$ as, first we will go in the STAT TESTS menu, choose $4:2-$SampTTest. Then you must specify if you are using data stored in two lists or if you will enter the means, standard deviations, and sizes of both samples. You must also indicate whether to pool the variances (when in doubt, say no) and specify whether the test is to be two-tail, lower-tail, or upper-tail. So, now we will define the hypothesis as:

Null hypothesis: The control group do not show a significant improvement in test scores.

Alternative hypothesis: The control group show a significant improvement in test scores.

Thus, by using the calculator $\text{TI}-89$ , the test statistics and the P-value is as:

  $\begin{array}{l}P<0.0001\\ t=9.24\end{array}$

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

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

Thus, we conclude that the control group show a significant improvement in test scores.

(d)

To determine

To explain were gains significantly higher for the accelerated math group than for the control group.

Answer to Problem 23RE

The gains were significantly higher for the accelerated math group than for the control group.

Explanation of Solution

It is given in the question the table for the instruction methods and the tests as:

		Acc. Math
	Pre-test	Post-test	Individual gain
Mean	560.01	637.55	77.53
SD	84.29	82.9	78.01
n	231	231	231
		Control
	Pre-test	Post-test	Individual gain
Mean	549.65	588.76	39.11
SD	74.68	83.24	66.25
n	245	245	245

Let us assume all the assumptions and conditions are met. Thus, for testing the hypothesis we will use the calculator $\text{TI}-89$ as, first we will go in the STAT TESTS menu, choose $4:2-$SampTTest. Then you must specify if you are using data stored in two lists or if you will enter the means, standard deviations, and sizes of both samples. You must also indicate whether to pool the variances (when in doubt, say no) and specify whether the test is to be two-tail, lower-tail, or upper-tail. So, now we will define the hypothesis as:

Null hypothesis: The gains were significantly same for the accelerated math group than for the control group.

Alternative hypothesis: The gains were significantly higher for the accelerated math group than for the control group.

Thus, by using the calculator $\text{TI}-89$ , the test statistics and the P-value is as:

  $\begin{array}{l}P<0.0001\\ t=5.78\end{array}$

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

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

Thus, we conclude that the gains were significantly higher for the accelerated math group than for the control group.