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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