Answer:
The 95% confidence interval for the mean score, , of all students taking the test is
[tex]28.37< L\ 30.63[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is [tex]n = 59[/tex]
The mean score is [tex]\= x = 29.5[/tex]
The standard deviation [tex]\sigma = 5.2[/tex]
Generally the standard deviation of mean is mathematically represented as
[tex]\sigma _{\= x} = \frac{\sigma }{\sqrt{n} }[/tex]
substituting values
[tex]\sigma _{\= x} = \frac{5.2 }{\sqrt{59} }[/tex]
[tex]\sigma _{\= x} = 0.677[/tex]
The degree of freedom is mathematically represented as
[tex]df = n - 1[/tex]
substituting values
[tex]df = 59 -1[/tex]
[tex]df = 58[/tex]
Given that the confidence interval is 95% then the level of significance is mathematically represented as
[tex]\alpha = 100 -95[/tex]
[tex]\alpha =[/tex]5%
[tex]\alpha = 0.05[/tex]
Now the critical value at this significance level and degree of freedom is
[tex]t_{df , \alpha } = t_{58, 0.05 } = 1.672[/tex]
Obtained from the critical value table
So the the 95% confidence interval for the mean score, , of all students taking the test is mathematically represented as
[tex]\= x - t*(\sigma_{\= x}) < L\ \= x + t*(\sigma_{\= x})[/tex]
substituting value
[tex](29.5 - 1.672* 0.677) < L\ (29.5 + 1.672* 0.677)[/tex]
[tex]28.37< L\ 30.63[/tex]
