Respuesta :
Answer:
Step-by-step explanation:
We would use the t- distribution.
From the information given,
Mean, μ = 2950
Standard deviation, σ = 115
number of sample, n = 25
Degree of freedom, (df) = 25 - 1 = 24
Alpha level,α = (1 - confidence level)/2
α = (1 - 0.98)/2 = 0.01
We will look at the t distribution table for values corresponding to (df) = 24 and α = 0.01
The corresponding z score is 2.492
We will apply the formula
Confidence interval
= mean ± z ×standard deviation/√n
It becomes
2950 ± 2.492 × 115/√25
= 2950 ± 2.492 × 23
= 2950 ± 57.316
The lower end of the confidence interval is 2950 - 57.316 =2892.68
The upper end of the confidence interval is 2950 + 57.316 = 3007.32
The solution is correct.
The confidence intervals express the distribution of sampling in a survey.
The 98% confidence interval is [tex]\mathbf{(2892.684,3007.316)}[/tex]
The given parameters are:
[tex]\mathbf{\mu = 2800}[/tex] --- the population mean
[tex]\mathbf{\bar x = 2950}[/tex] --- the sample mean
[tex]\mathbf{\sigma = 115}[/tex] -- the standard deviation
[tex]\mathbf{n = 25}[/tex] --- the sample size
Calculate the degrees of freedom
[tex]\mathbf{df = n - 1}[/tex]
[tex]\mathbf{df = 25 - 1}[/tex]
[tex]\mathbf{df = 24}[/tex]
Calculate the significance level
[tex]\mathbf{\alpha /2=\frac{1 - 98\%}2 = 0.01}[/tex]
The critical value at [tex]\mathbf{\alpha/2 = 0.01}[/tex] and df = 24 is:
[tex]\mathbf{t_{\alpha/2} = 2.492}[/tex]
Calculate the margin of error
[tex]\mathbf{E = t_{\alpha/2} \frac{\sigma}{\sqrt n}}[/tex]
So, we have:
[tex]\mathbf{E = 2.492 \times \frac{115}{\sqrt{25}}}[/tex]
[tex]\mathbf{E = 2.492 \times \frac{115}{5}}[/tex]
[tex]\mathbf{E = 2.492 \times 23}[/tex]
[tex]\mathbf{E = 57.316}[/tex]
The confidence interval is then calculated as:
[tex]\mathbf{CI =\bar x \pm E}[/tex]
So, we have:
[tex]\mathbf{CI = 2950 \pm 57.316}}[/tex]
Split
[tex]\mathbf{CI = (2950 - 57.316,2950 + 57.316)}[/tex]
[tex]\mathbf{CI = (2892.684,3007.316)}[/tex]
So, the 98% confidence interval is [tex]\mathbf{(2892.684,3007.316)}[/tex]
Read more about confidence intervals and margin of errors at:
https://brainly.com/question/19426829
