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?

Respuesta :

Answer:

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:

[tex]CI=\bar x_{d}\pm t_{\alpha/2, (n-1)}\times \frac{s_{d}}{\sqrt{n}}[/tex]

Here,

[tex]\bar x_{d}[/tex] = sample mean of the difference,

[tex]s_{d}[/tex] = sample standard deviation of the difference,

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

[tex]t_{\alpha/2, (n-1)}[/tex] = critical value of t

(n - 1) = degrees of freedom

The information provided is:

n = 5

Confidence level = 95%

The critical value of t is:

[tex]t_{\alpha/2, (n-1)}=t_{0.05/2, (5-1)}[/tex]

               [tex]=t_{0.025, 4}[/tex]

               [tex]=2.776[/tex]

*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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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 :

  • -3, - 3, - 4, - 8, 0

Using a calculator :

  • Mean difference, Xd = -3.6
  • Standard deviation of difference, Sd = 2.88

The test statistics :

  • Test statistic = Xd ÷ (Sd/n)

Test statistic = -3.6 ÷ (2.88/√5)

Test statistic = -2.79

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

  • df = n - 1 ; 5 - 1 = 4

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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