Question

A group of researchers has developed a method that they hypothesize will improve the reading ability of at-risk 1st grade students. They randomly selected nine students to participate in their study. The nine students were administered a reading test, then trained for 10 weeks using the new reading program, and then readministered the reading test. The following data are their scores:

Subject   1st Reading 2nd Reading

                  Score            Score

    1               23                19

    2               20                19

    3               24                 21

    4               24                24

    5               24                22

    6               19                17

    7               17                15

    8               18                16

    9               20                18

What would be your statistical conclusion using a directional alternative hypothesis test with an alpha = .01?

Answer 1

NULL HYPOTHESIS H0:

ALTERNATIVE HYPOTHESIS Ha:

From the sample data, it is found that the corresponding sample means are:

Xˉ1=21
Xˉ2=19
Also, the provided sample standard deviations are:

s1 =2.784 and s2 =2.915
and the sample size is n = 9. For the score differences we have

Dˉ=2
sD=1.118
(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:μD= 0

Ha: μD< 0

This corresponds to a left-tailed test, for which a t-test for two paired samples be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.01, and the degrees of freedom are df=8.

Hence, it is found that the critical value for this left-tailed test is tc =−2.896, for α=0.01 and df=8.

The rejection region for this left-tailed test is R={t:t<−2.896}.

(3) Test Statistics

The t-statistic is computed as shown in the following formula:

t= Dbar/Sd/sqrt(n)= 2/1.118/sqrt(9)= 5.367
(4) Decision about the null hypothesis

Since it is observed that t=5.367≥tc =−2.896, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.9997, and since p=0.9997≥0.01, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1 is less than μ2, at the 0.01 significance level.

There is no enough evidence to claim that training has impact on reading scores.