A teacher wants to see if a new unit on factoring is helping students learn. She has five randomly selected students take a pre-test and a post test on the material. The scores are out of 20. Has there been improvement? (pre-post) What value of t would you use for the 95% confidence interval?

Using the paired t-test sample test, we can conclude that there has been no improvement in the scores and the value of t would be 2.776

The difference in pretest and post test scores :

Using a calculator :

The test statistics :

Test statistic = -3.6 ÷ (2.88/√5)

Test statistic = -2.79

The Pvalue, t value for a 95% confidence interval :

Using a Pvalue calculator :

T(0.05, 4) = 2.776 (2 tailed)

Since |Pvalue| > α = 0.05

We fail to reject the Null and conclude that there has been no improvement.

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The t-value used for the 95% confidence interval of paired data is 2.776.

Step-by-step explanation:

The confidence interval formula for mean difference for a paired data is as follows:

Here,

= sample mean of the difference,

= sample standard deviation of the difference,

n = sample size (both samples are of same size).

= critical value of t

(n - 1) = degrees of freedom

The information provided is:

n = 5

Confidence level = 95%

The critical value of t is:

*Use a t-table for the critical value of t.

Thus, the t-value used for the 95% confidence interval of paired data is 2.776.

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step-by-step explanation:

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