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Imagine that researchers appropriately conduct a paired-samples t test with 25 pairs of related scores. The mean difference between those 25 pairs of scores was 5.00. The standard deviation for the 25 difference scores was 7.50. Which of the following is the appropriate 95% confidence interval for these data (when reporting three decimal places)?

a. [2.768, 7.332]
b. [-10.480, 20.480]
c. [2.434, 7.567]
d. [1.904, 8.096]
e. None of the Above

Respuesta :

Answer:

The correct option is;

d. [1.904, 8.096]

Step-by-step explanation:

The confidence interval is given as

[tex]CI=\bar{x}\pm z\frac{s}{\sqrt{n}}[/tex]

Where:

[tex]\bar {x}[/tex] = Mean = 5

s = Standard deviation = 7.5

n = Number in sample = 25

z = z value at 95% =

As we have a small sample size and an unknown population standrd deviation, we get from a two tailed confidence level

z = from the students at 95% two tailed confidence level = 2.0639

[tex]CI=5\pm 2.0639\frac{7.5}{\sqrt{25}}[/tex]

Which gives the confidence interval as;

CI = (1.90415, 8.0958)

The critical z = 2.0639  

fichoh

Uisng the confidence interval relation, the appropriate interval for the experiment conducted would be [1.904, 8.096]

Recall :

  • [tex]x ± Z* \frac{s}{\sqrt{n}}[/tex]
  • Sample size, n = 25
  • Z* at 95%, df = 25 - 1 ; 2.063

Plugging the values into the relation :

[tex] 5 ± 2.063(\frac{7.50}{\sqrt{25}}[/tex]

[tex] 5 ± 2.063(\frac{7.50}{5}[/tex]

[tex] 5 ± 3.0945[/tex]

Lower boundary = 5 - 3.0945 = 1.904

Upper boundary = 5 + 3.0945 = 8.096

Hence, the confidence interval is [1.904, 8.096]

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