Answer :
Answer:
99% confidence interval for the true mean Perry score of all undergraduate engineering students is [3.022 , 3.518].
Step-by-step explanation:
We are given that Intellectual development (Perry) scores were determined for 21 students in a first-year, project-based design course.
The average Perry score for the 21 students was 3.27 and the standard deviation was 0.40.
Firstly, the pivotal quantity for 99% confidence interval for the true mean is given by;
P.Q. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample average Perry score for the 21 students = 3.27
s = sample standard deviation = 0.40
n = sample of students = 21
[tex]\mu[/tex] = population mean Perry score
Here for constructing 99% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.
So, 99% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-2.845 < [tex]t_2_0[/tex] < 2.845) = 0.99 {As the critical value of t at 20 degree
of freedom are -2.845 & 2.845 with P = 0.5%}
P(-2.845 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.845) = 0.99
P( [tex]-2.845 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.845 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.99
P( [tex]\bar X-2.845 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X +2.845 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.99
99% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.845 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.845 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]3.27-2.845 \times {\frac{0.40}{\sqrt{21} } }[/tex] , [tex]3.27+2.845 \times {\frac{0.40}{\sqrt{21} } }[/tex] ]
= [3.022 , 3.518]
Therefore, 99% confidence interval for the true mean Perry score of all undergraduate engineering students is [3.022 , 3.518].
Interpretation of this confidence interval is that we are 99% confident that the true mean Perry score of all undergraduate engineering students will lie between 3.022 and 3.518.