Question

In: Statistics and Probability

A teacher gives a third grade class of n = 16 a vocabulary test at the...

A teacher gives a third grade class of n = 16 a vocabulary test at the beginning of the school year. To evaluate the changes that occur during the year, students are tested again at the end of the year. Their test scores revealed an average improvement of MD = 5.7 points with s2 = 144. Does this constitute a significant improvement? Use a one tail, related-samples t test with an alpha level of .05.

find standard error find degrees of freedom

Enter the tabled critical value of t (tcrit)

Calculate the t statistic

Complete the APA formatted statistic:

Calculate Cohen's d

Solutions

Expert Solution

Solution :

The null and alternative hypotheses are as follows:

Where, represents the population mean score for the test held at the end of the year and represents the population mean score for the test held at the beginning of the school year.

The standard error is given as follows :

Where, s​​​​​​2 is sample variance of differences and n is sample size.

We have, s​​​​​​2 = 144 and n = 16

The standard error is 3.

Degrees of freedom = (n - 1) = (16 - 1) = 15

Significance level = 0.05

Since, our test is right-tailed test, therefore we shall obtain right-tailed critical t-value at 0.05 significance level and 15 degrees of freedom, which is given as follows:

The right-tailed critical value of t is 1.7530.

To test the hypothesis we shall use related sample t-test. The value of test statistic is given as follows:

Where, is sample mean of the differences, s​​​​​​2 is sample variance of differences and n is sample size.

We have, s​​​​​​2 = 144 and n = 16

The value of the test statistic is 1.9.

For right-tailed t-test, we make decision rule as follows :

If value of the test statistic is greater than the right-tailed critical value of t, then we reject the null hypothesis (H​​​​​​0).

If value of the test statistic is less than the right-tailed critical value of t, then we fail to reject the null hypothesis (H​​​​​​0).

Value of the test statistic = 1.9

Right-tailed critical value of t = 1.7530

(1.9 > 1.7530)

Since, value of the test statistic is greater than the right-tailed critical value of t, therefore we shall reject the null hypothesis (H​​​​​​0) at 0.05 level of significance.

Conclusion : At significance level of 0.05, there is sufficient evidence to conclude that there is significant improvement in test score.

Cohen's d effect size is given as follows:

Cohen's d is 0.475.

Please rate the answer. Thank you.


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