Answer:
The value is [tex]C = 85\%[/tex]
Step-by-step explanation:
From the question we are told that
The standard deviation is [tex]\sigma = 21.73[/tex]
The sample size is n = 14
The mean score is [tex]\= x = 74.57[/tex]
The confidence interval is 65.7425 to 83.3975
Generally the degree of freedom is mathematically represented as
[tex]df = n -1[/tex]
=> [tex]df = 14 -1[/tex]
=> [tex]df = 13[/tex]
Generally the margin of error is mathematically evaluated as
[tex]E = \frac{83.3975 -65.7425 }{2}[/tex]
=> [tex]E = 8.828[/tex]
Generally this margin of error is mathematically represented as
[tex]E = t_{\frac{\alpha }{2} , df} * \frac{\sigma }{\sqrt{n} }[/tex]
=> [tex]8.828 = t_{\frac{\alpha }{2} , df} * \frac{21.73 }{\sqrt{14} }[/tex]
=> [tex]t_{\frac{\alpha }{2} , df} = 1.520[/tex]
From the t distribution table the probability of [tex]( t> 1.520 )[/tex] at a degree of freedom of 13 is
[tex]\frac{\alpha }{2} = P( t > 1.52 ) = 0.076[/tex]
=> [tex]\alpha = 2 * 0.076[/tex]
=> [tex]\alpha = 1.52[/tex]
So the confidence level is evaluated as
[tex]C = 1- \alpha[/tex]
=> [tex]C = 1- 0.152[/tex]
=> [tex]C = 0.85[/tex]
Converting to percentage
=> [tex]C = 0.85 * 100[/tex]
=> [tex]C = 85\%[/tex]
