Chapter PVII, Problem 30RE

(a)

To determine

To explain on Friday, did the week-long study group have a mean score significantly higher than that of the overnight crammers.

Answer to Problem 30RE

There is insufficient evidence that the week-long study group have a mean score significantly higher than that of the overnight crammers.

Explanation of Solution

In the question the statistics summary for the students who studied week-long, i.e. group first and the data for the students who crammed overnight, i.e. second group are given. So, we have to consider the scores for the Friday exams only for the second group. So, we have to calculate its mean, standard deviation by using excel function as:

  $\begin{array}{l}=\text{AVERAGE(Number 1, Number 2,}\mathrm{..}\text{)}\\ =\text{STDEV(Number 1, Number 2,}\mathrm{..}\text{)}\end{array}$

The calculation will be as:

	Group 2
Mean	=AVERAGE(AK13:AK37)
stdv	=STDEV(AK13:AK37)
n	25

The result will be as:

	Group 2
Mean	42.28
stdv	4.430199
n	25

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:

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

Since this is an upper tail test so by using the calculator $\text{TI}-89$ , the test statistics and the P-value is as:

  $\begin{array}{l}P=0.1865\\ t=0.90\end{array}$

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 thatthere is insufficient evidence that the week-long study group have a mean score significantly higher than that of the overnight crammers.

(b)

To determine

To explain was there a significant difference in the percentages of students who passed the quiz on Friday.

Answer to Problem 30RE

There a significant difference in the percentages of students who passed the quiz on Friday.

Explanation of Solution

In the question the statistics summary for the students who studied week-long, i.e. group first and the data for the students who crammed overnight, i.e. second group are given. Thus, first we will calculate the percentage of kids passing the test on Friday in second group as they have scored above $40$ , so we have,

  $\begin{array}{l}=\frac{18}{25}\\ =0.72\end{array}$

Thus, to do the mechanics of a hypothesis test for equality of population proportions we will use the calculator $\text{TI}-89$ . Then select $6:2-$ PropZTestfrom the STAT Tests menu. Enter the observed counts and sample sizes. Indicate what kind of test you want: one-tail upper tail, lower tail, or two-tail. And specify whether results should simply be calculated or displayed with the area corresponding to the P-value of the test shaded. So, we have,

  $\begin{array}{l}{p}_1=0.33\\ n_1=45\\ p_2=0.72\\ n_2=25\end{array}$

The hypothesis is defined as:

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_0:p_1\ne p_2\end{array}$

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

  $\begin{array}{l}z=-3.13\\ P=0.017\end{array}$

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 there is an evidence that there a significant difference in the percentages of students who passed the quiz on Friday.

(c)

To determine

To explain is there any evidence that when students cram for a test, their learning does not last for three days.

Answer to Problem 30RE

When students cram for a test, their learning does not last for three days.

Explanation of Solution

In the question the statistics summary for the students who studied week-long, i.e. group first and the data for the students who crammed overnight, i.e. second group are given. Thus, we will consider the second group scores for Monday and Friday. 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:

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

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

  $\begin{array}{l}P=0.0002\\ t=3.76\end{array}$

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 evidence that when students cram for a test, their learning does not last for three days.

(d)

To determine

To use a $95\%$ confidence interval to estimate the mean number of words that might be forgotten by crammers.

Answer to Problem 30RE

We are $95\%$ confident that the mean number of words that might be forgotten by crammers lie between $2.346\text{ and }7.734$ .

Explanation of Solution

In the question the statistics summary for the students who studied week-long, i.e. group first and the data for the students who crammed overnight, i.e. second group are given. Thus, we will consider the second group scores for Monday and Friday.Thus, to find the confidence interval for proportion we will use the calculator $\text{TI}-89$ . To calculate a confidence interval for a population proportion: Go to the STAT Intsmenu and select $6:2-$ PropZInt. Enter the observed counts and the sample size for both the samples. Specify a confidence level. Calculate the interval. So, by using the calculator $\text{TI}-89$ , the confidence interval will be as:

  $(2.346,7.734)$

So, we are $95\%$ confident that the mean number of words that might be forgotten by crammers lie between $2.346\text{ and }7.734$ .

(e)

To determine

To explain is there any evidence that how much students forget depends on how much they learned to begin with.

Answer to Problem 30RE

How much students forget depends on how much they learned to begin with.

Explanation of Solution

In the question the statistics summary for the students who studied week-long, i.e. group first and the data for the students who crammed overnight, i.e. second group are given. So, we will consider both the groups. And then we will make the contingency table as:

	Group 1	Group 2
Success	14.85	18	32.85
Failure	30.15	7	37.15
	45	25	70

Thus, to test a hypothesis of homogeneity or independence, we will use the calculator $\text{TI}-89$ so, we need to enter the data as a matrix. From the home screen, press APPS and select $6:$ Data/Matrix Editor, then select $3:$ New. Specify type as Matrix and name the matrix in the Variable box. Specify the number of rows and columns. Type the entries, pressing ENTER after each. Press $2\text{nd ESC}$ to leave the editor. To do the test, choose $\text{8:Chi2 2-way}$ from the STAT Tests menu. So, we define hypothesis by:

Null hypothesis: Students forget is independent on how much they learned to begin with.

Alternative hypothesis: Students forget is dependent on how much they learned to begin with.

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

  $\begin{array}{l}{\chi }^2=9.81\\ df=1\\ P=0.0017\end{array}$

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 thathow much students forget depends on how much they learned to begin with.