Based on a sample of n = 28, the least-squares method was used to develop the prediction line ₁ = 8 + 3X₁. In addition, Syx = 2.3, X = 7, and Σ (x₁ - x)² = 28. Complete parts (a) through (c) below. i=1 a. Construct a 90% confidence interval estimate of the population mean response for X= 5. SHY|X=5³ (Round to two decimal places as needed.) b. Construct a 90% prediction interval of an individual response for X = 5. ≤YX=5³ (Round to two decimal places as needed.) c. Compare the results of (a) and (b) with those for X= 7, which are 28.26 ≤ Hy|x=7 ≤29.74 and 25.01 ≤Yx=732.99. Which intervals are wider? Why? the mean X. The intervals for X= ▼are wider because X = 7 is
Based on a sample of n = 28, the least-squares method was used to develop the prediction line ₁ = 8 + 3X₁. In addition, Syx = 2.3, X = 7, and Σ (x₁ - x)² = 28. Complete parts (a) through (c) below. i=1 a. Construct a 90% confidence interval estimate of the population mean response for X= 5. SHY|X=5³ (Round to two decimal places as needed.) b. Construct a 90% prediction interval of an individual response for X = 5. ≤YX=5³ (Round to two decimal places as needed.) c. Compare the results of (a) and (b) with those for X= 7, which are 28.26 ≤ Hy|x=7 ≤29.74 and 25.01 ≤Yx=732.99. Which intervals are wider? Why? the mean X. The intervals for X= ▼are wider because X = 7 is
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![Based on a sample of n = 28, the least-squares method was used to develop the prediction line \( \hat{Y}_i = 8 + 3X_i \). In addition, \( S_{YX} = 2.3 \), \( \bar{X} = 7 \), and
\[
\sum_{i=1}^{n} \left( X_i - \bar{X} \right)^2 = 28
\].
Complete parts (a) through (c) below.
a. **Construct a 90% confidence interval estimate of the population mean response for X = 5.**
\[ \underline{\hspace{1cm}} \le \mu_{Y|X=5} \le \underline{\hspace{1cm}} \]
(Round to two decimal places as needed.)
b. **Construct a 90% prediction interval of an individual response for X = 5.**
\[ \underline{\hspace{1cm}} \le Y_{X=5} \le \underline{\hspace{1cm}} \]
(Round to two decimal places as needed.)
c. **Compare the results of (a) and (b) with those for X = 7, which are \( 28.26 \le \mu_{Y|X=7} \le 29.74 \) and \( 25.01 \le Y_{X=7} \le 32.99 \). Which intervals are wider? Why?**
The intervals for X = \(\begin{array}{c} \text{____} \\ \text{(dropdown)} \end{array}\) are wider because X = 7 is \(\begin{array}{c} \text{____} \\ \text{(dropdown)} \end{array}\) the mean \( \bar{X} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc91d435c-8a20-4fa9-b80f-5aaeb9e13244%2F904b4c42-5f41-4b1a-a180-78c80fb37245%2Fwawn8ut_processed.png&w=3840&q=75)
Transcribed Image Text:Based on a sample of n = 28, the least-squares method was used to develop the prediction line \( \hat{Y}_i = 8 + 3X_i \). In addition, \( S_{YX} = 2.3 \), \( \bar{X} = 7 \), and
\[
\sum_{i=1}^{n} \left( X_i - \bar{X} \right)^2 = 28
\].
Complete parts (a) through (c) below.
a. **Construct a 90% confidence interval estimate of the population mean response for X = 5.**
\[ \underline{\hspace{1cm}} \le \mu_{Y|X=5} \le \underline{\hspace{1cm}} \]
(Round to two decimal places as needed.)
b. **Construct a 90% prediction interval of an individual response for X = 5.**
\[ \underline{\hspace{1cm}} \le Y_{X=5} \le \underline{\hspace{1cm}} \]
(Round to two decimal places as needed.)
c. **Compare the results of (a) and (b) with those for X = 7, which are \( 28.26 \le \mu_{Y|X=7} \le 29.74 \) and \( 25.01 \le Y_{X=7} \le 32.99 \). Which intervals are wider? Why?**
The intervals for X = \(\begin{array}{c} \text{____} \\ \text{(dropdown)} \end{array}\) are wider because X = 7 is \(\begin{array}{c} \text{____} \\ \text{(dropdown)} \end{array}\) the mean \( \bar{X} \).
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