Answer:
The limit of the (228.533, 234.735)
Step-by-step explanation:
The values are missing in the given question. Therefore, in order to attempt this question, we will make assumptions.
So, let's assume that:
sample size of the journal article that was reported = 5
which is applied for determining a 95% CI
Also, assuming that the resulting interval = (229.764, 233.504)
However, if we are to use a 99% CI which we deemed to be more appropriate;
Then, our objective is to find the limits of this particular interval in question.
To do that:
We need to first find the sample mean at 95% CI by using the formula:
[tex]\Big ( \bar {x} - t_{\alpha/2, df} \ \dfrac{s}{\sqrt{n}}}, \bar x + t_{\alpha/2, df} \ \dfrac{s}{\sqrt{n}}} \Big) = (229.764,233.504)[/tex]
Since; df = n - 1
df = 5 - 1
df = 4
Then;
[tex]\Big ( \bar {x} - t_{\alpha/2, 4} \ \dfrac{s}{\sqrt{n}}}, \bar x + t_{\alpha/2, 4} \ \dfrac{s}{\sqrt{n}}} \Big) = (233.504+229.764)[/tex]
[tex]2 \bar {x} = (463.268)[/tex]
[tex]\bar {x} =\dfrac{463.268}{2 }[/tex]
[tex]\bar {x} =231.634[/tex]
Sample mean = 231.634
Using the same formula to determine the standard deviation, we have:
[tex]\Big ( \bar {x} - t_{\alpha/2, df} \ \dfrac{s}{\sqrt{n}}}, \bar x + t_{\alpha/2, df} \ \dfrac{s}{\sqrt{n}}} \Big) = (229.764,233.504)[/tex]
[tex]\Big ( \bar {x} - t_{\alpha/2, 4} \ \dfrac{s}{\sqrt{n}}}, \bar x + t_{\alpha/2, 4} \ \dfrac{s}{\sqrt{n}}} \Big) = (233.504-229.764)[/tex]
[tex]\Big ( (\bar x + t_{\alpha/2, 4} \ \dfrac{s}{\sqrt{n}}} )- (\bar {x} - t_{\alpha/2, 4} \ \dfrac{s}{\sqrt{n}}}) \Big) = (233.504-229.764)[/tex]
[tex]\Big ( (\bar x + t_{0.05/2, 4} \ \dfrac{s}{\sqrt{4}}} )- (\bar {x} - t_{0.05/2, 4} \ \dfrac{s}{\sqrt{5}}}) \Big) = (233.504-229.764)[/tex]
At t = 0.025 and df = 4; = 2.776
[tex]2\times 2.776 \dfrac{s}{\sqrt{5}}= 3.74[/tex]
[tex]5.552 \dfrac{s}{\sqrt{5}}= 3.74[/tex]
[tex]\dfrac{s}{\sqrt{5}}= \dfrac{ 3.74}{5.552}[/tex]
[tex]\dfrac{s}{\sqrt{5}}= 0.6736[/tex]
[tex]s = 0.6736 \times \sqrt{5}[/tex]
s = 1.5063
The 99% CI is:
[tex]\implies \Big(\bar x \pm t_{\alpha/2,4} \dfrac{s}{\sqrt{n}} \Big)[/tex]
At t =0.005 and df =4; = 4.604
[tex]\implies \Big(231.634 \pm 4.604 \dfrac{1.5063}{\sqrt{5}} \Big)[/tex]
[tex]\implies \Big(231.634 \pm 4.604 (0.67364) \Big)[/tex]
[tex]\implies \Big(231.634 \pm 3.1014 \Big)[/tex]
[tex]\implies \Big((231.634 - 3.1014), (231.634 + 3.1014) \Big)[/tex]
[tex]\implies \Big( 228.533, 234.735 \Big)[/tex]
