Based on a sample of 60 textbooks at the store, you find an average of 75.97 and a standard deviation of 25.4. The point estimate is: (to 3 decimals) The 90% confidence interval (use z*) is: (to 3 decimals) Based on this we: to
Based on a sample of 60 textbooks at the store, you find an average of 75.97 and a standard deviation of 25.4. The point estimate is: (to 3 decimals) The 90% confidence interval (use z*) is: (to 3 decimals) Based on this we: to
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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![### Testing a Claim about the Average Textbook Price
Suppose that BC's financial aid office allocates a textbook stipend by claiming that the average textbook at the BC bookstore costs $66.77. You want to test this claim.
#### Formulating Hypotheses
The null and alternative hypothesis in symbols would be:
- \( H_0 : \mu = 66.77 \)
- \( H_1 : \mu \neq 66.77 \)
(The fourth option from the top is correct in this case.)
#### Null Hypothesis in Words
The null hypothesis in words would be:
- The average price of all textbooks from the store is $66.77.
#### Data Observed
Based on a sample of 60 textbooks at the store, you find:
- Average price: $75.97
- Standard deviation: $25.4
#### Calculations
**Point Estimate**
The point estimate is the sample mean:
\[ 75.97 \]
**90% Confidence Interval**
To find the 90% confidence interval for the mean, use the formula for the confidence interval for a population mean (\(\mu\)) when the standard deviation (\(s\)) is known. The formula is:
\[
\text{CI} = \bar{x} \pm z^* \left( \frac{s}{\sqrt{n}} \right)
\]
Where:
- \( \bar{x} \) is the sample mean
- \( z^* \) is the z-value corresponding to the 90% confidence level (which is approximately 1.645 for a two-tailed test)
- \( s \) is the sample standard deviation
- \( n \) is the sample size
Substitute the values:
\[ \bar{x} = 75.97 \]
\[ s = 25.4 \]
\[ n = 60 \]
\[ z^* = 1.645 \]
\[
\text{CI} = 75.97 \pm 1.645 \left( \frac{25.4}{\sqrt{60}} \right)
\]
First, calculate the standard error:
\[
\frac{25.4}{\sqrt{60}} \approx 3.28
\]
Next, multiply by \( z^* \):
\[
1.645 \times 3.28 \approx 5.39
\]
So, the confidence](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc7e5ddd4-9774-49f5-a3a5-694b08ae0d81%2F432cec3e-2758-4297-8dd7-a305306f9a36%2Fu60npdh_processed.png&w=3840&q=75)
Transcribed Image Text:### Testing a Claim about the Average Textbook Price
Suppose that BC's financial aid office allocates a textbook stipend by claiming that the average textbook at the BC bookstore costs $66.77. You want to test this claim.
#### Formulating Hypotheses
The null and alternative hypothesis in symbols would be:
- \( H_0 : \mu = 66.77 \)
- \( H_1 : \mu \neq 66.77 \)
(The fourth option from the top is correct in this case.)
#### Null Hypothesis in Words
The null hypothesis in words would be:
- The average price of all textbooks from the store is $66.77.
#### Data Observed
Based on a sample of 60 textbooks at the store, you find:
- Average price: $75.97
- Standard deviation: $25.4
#### Calculations
**Point Estimate**
The point estimate is the sample mean:
\[ 75.97 \]
**90% Confidence Interval**
To find the 90% confidence interval for the mean, use the formula for the confidence interval for a population mean (\(\mu\)) when the standard deviation (\(s\)) is known. The formula is:
\[
\text{CI} = \bar{x} \pm z^* \left( \frac{s}{\sqrt{n}} \right)
\]
Where:
- \( \bar{x} \) is the sample mean
- \( z^* \) is the z-value corresponding to the 90% confidence level (which is approximately 1.645 for a two-tailed test)
- \( s \) is the sample standard deviation
- \( n \) is the sample size
Substitute the values:
\[ \bar{x} = 75.97 \]
\[ s = 25.4 \]
\[ n = 60 \]
\[ z^* = 1.645 \]
\[
\text{CI} = 75.97 \pm 1.645 \left( \frac{25.4}{\sqrt{60}} \right)
\]
First, calculate the standard error:
\[
\frac{25.4}{\sqrt{60}} \approx 3.28
\]
Next, multiply by \( z^* \):
\[
1.645 \times 3.28 \approx 5.39
\]
So, the confidence
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