Mathematics, 07.04.2020 00:25 shadow56

Intellectual development of engineering students. Refer to the Journal of Engineering Education (Jan. 2005) study of the intellectual development of undergraduate engineering students, Exercise 1.19 (p. 10). Intellectual development (Perry) scores were determined for 21 students in a first-year, project-based design course. (Recall that a Perry score of 1 indicates the lowest level of intellectual development, and a Perry score of 5 indicates the highest level.) The average Perry score for the 21 students was 3.27 and the standard deviation was .40. Apply the confidence interval method of this section to estimate the true mean Perry score of all undergraduate engineering students with 99% confidence. Interpret the results.

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Answer from: anonymousanon

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. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ t_n_-_1

where, $\bar X$ = sample average Perry score for the 21 students = 3.27

s = sample standard deviation = 0.40

n = sample of students = 21

$\mu$ = 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, $\mu$ is ;

P(-2.845 < t_2_0 < 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 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.845) = 0.99

P( $-2.845 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.845 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-2.845 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X +2.845 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X-2.845 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.845 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $3.27-2.845 \times {\frac{0.40}{\sqrt{21} } }$ , $3.27+2.845 \times {\frac{0.40}{\sqrt{21} } }$ ]

= [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.

Answer from: Quest

add all the numbers and check the ones place.

the answer is 486. as the ones place is more than 4, the estimation will be the next 10 ie, 490

Answer from: Quest

the answer is true.

step-by-step explanation:

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