g. Interpret the confidence interval in the context of the problem. h. If someone hypothesized that the majority of daytime Highline College students work, would you accept or reject this hypothesis?
g. Interpret the confidence interval in the context of the problem. h. If someone hypothesized that the majority of daytime Highline College students work, would you accept or reject this hypothesis?
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
![### Transcription of Statistical Analysis
---
#### e. Compute the Margin of Error (high – low)/2
\[
0.5926 - 0.2646 / 2 = 0.164
\]
#### f. Check two important assumptions:
1. \(\hat{p} n > 10\)
2. \(n(1 - \hat{p}) > 10\)
If one of these is not true, then your results cannot be trusted.
\[
np = 35 \times 0.5926 = 20.741 > 10
\]
\[
n(1-p) = 35(1-0.5926) = 14.259 > 10
\]
#### g. Interpret the confidence interval in the context of the problem.
*Text goes here.*
#### h. If someone hypothesized that the majority of daytime Highline College students work, would you accept or reject this hypothesis?
*Text goes here.*
---
### Explanation of Graphs and Diagrams
There are no graphs or diagrams included in this section.
---](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F521a46c7-b5be-4646-b544-bbde7c597bf9%2F9e2a8b03-340f-45e6-9c09-effdbe56a3ef%2Fogyj9z_processed.png&w=3840&q=75)
Transcribed Image Text:### Transcription of Statistical Analysis
---
#### e. Compute the Margin of Error (high – low)/2
\[
0.5926 - 0.2646 / 2 = 0.164
\]
#### f. Check two important assumptions:
1. \(\hat{p} n > 10\)
2. \(n(1 - \hat{p}) > 10\)
If one of these is not true, then your results cannot be trusted.
\[
np = 35 \times 0.5926 = 20.741 > 10
\]
\[
n(1-p) = 35(1-0.5926) = 14.259 > 10
\]
#### g. Interpret the confidence interval in the context of the problem.
*Text goes here.*
#### h. If someone hypothesized that the majority of daytime Highline College students work, would you accept or reject this hypothesis?
*Text goes here.*
---
### Explanation of Graphs and Diagrams
There are no graphs or diagrams included in this section.
---
![2. The data below shows hours worked per week by a sample of daytime Math& 146 students. I am assuming that these students represent a random sample of daytime Highline College Students.
a. **Clearly state the population for this problem.**
The population is Math& 146 daytime students at Highline College.
b. **Clearly state the sample for this problem.**
The sample is 35 Math& 146 daytime students at Highline College.
c. **Compute the sample proportion \( \hat{p} \) of students in the sample who worked 20 or more hours per week.**
\[
\text{x} = 15 \\
p = x/n = 15/35 = 0.4286
\]
d. **Compute the 95% confidence interval for the population proportion of all daytime Highline students who work 20 or more hours per week.**
\[
\hat{X} \pm Z(S \div \sqrt{n}) \\
0.4286 \pm 1.96 \sqrt{0.4286(1-0.4286)/35} = 0.5926 \\
0.4286 + 1.96 \sqrt{0.4286(1-0.4286)/35} = 0.5926 \\
0.4286 - 1.96 \sqrt{0.4286(1-0.4286)/35} = 0.2646 \\
\text{95% confidence interval is (0.5926, 0.2646)}
\]
e. **Compute the Margin of Error (high - low)/2**
\[
0.5926 - 0.2646/2 = 0.164
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F521a46c7-b5be-4646-b544-bbde7c597bf9%2F9e2a8b03-340f-45e6-9c09-effdbe56a3ef%2Fafsj55_processed.png&w=3840&q=75)
Transcribed Image Text:2. The data below shows hours worked per week by a sample of daytime Math& 146 students. I am assuming that these students represent a random sample of daytime Highline College Students.
a. **Clearly state the population for this problem.**
The population is Math& 146 daytime students at Highline College.
b. **Clearly state the sample for this problem.**
The sample is 35 Math& 146 daytime students at Highline College.
c. **Compute the sample proportion \( \hat{p} \) of students in the sample who worked 20 or more hours per week.**
\[
\text{x} = 15 \\
p = x/n = 15/35 = 0.4286
\]
d. **Compute the 95% confidence interval for the population proportion of all daytime Highline students who work 20 or more hours per week.**
\[
\hat{X} \pm Z(S \div \sqrt{n}) \\
0.4286 \pm 1.96 \sqrt{0.4286(1-0.4286)/35} = 0.5926 \\
0.4286 + 1.96 \sqrt{0.4286(1-0.4286)/35} = 0.5926 \\
0.4286 - 1.96 \sqrt{0.4286(1-0.4286)/35} = 0.2646 \\
\text{95% confidence interval is (0.5926, 0.2646)}
\]
e. **Compute the Margin of Error (high - low)/2**
\[
0.5926 - 0.2646/2 = 0.164
\]
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