1. From a sample of 230 pet owners, 104 indicate that they buy their pet food in brick-and-mortar stores. a. Write the value of the sample statistic to three decimal places, using correct notation. Show your calculation here. b. Define the parameter this sample statistic estimates. I c. An estimate of the standard error for this statistic is 0.03117. Find a 95% confidence interval to three decimal places for the parameter defined here. Show all calculations. d. Interpret the 95% confidence interval in context. e. Which of the following best explains the meaning of 95% confidence in this question? Check only one. The proportion of 95% of all pet owners who buy their pet food in brick-and-mortar stores is in the interval. The proportion of 95% of the pet owners in this sample who buy their pet food in brick-and- mortar stores is in the interval. There is a 95% chance that the proportion of all not own ub bir a
1. From a sample of 230 pet owners, 104 indicate that they buy their pet food in brick-and-mortar stores. a. Write the value of the sample statistic to three decimal places, using correct notation. Show your calculation here. b. Define the parameter this sample statistic estimates. I c. An estimate of the standard error for this statistic is 0.03117. Find a 95% confidence interval to three decimal places for the parameter defined here. Show all calculations. d. Interpret the 95% confidence interval in context. e. Which of the following best explains the meaning of 95% confidence in this question? Check only one. The proportion of 95% of all pet owners who buy their pet food in brick-and-mortar stores is in the interval. The proportion of 95% of the pet owners in this sample who buy their pet food in brick-and- mortar stores is in the interval. There is a 95% chance that the proportion of all not own ub bir a
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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![### Statistical Analysis of Pet Owners’ Shopping Preferences
#### Question 1:
From a sample of 250 pet owners, 104 indicate that they buy their pet food in brick-and-mortar stores.
**a. Calculate the value of the sample statistic:**
To find the sample statistic (proportion), use the formula:
\[ \hat{p} = \frac{x}{n} \]
Where:
- \( \hat{p} \) is the sample proportion,
- \( x \) is the number of successes (pet owners who buy pet food in brick-and-mortar stores),
- \( n \) is the total sample size.
Plugging in the numbers:
\[ \hat{p} = \frac{104}{250} = 0.416 \]
**b. Define the parameter this sample statistic estimates:**
The parameter this sample statistic estimates is the proportion of all pet owners who buy their pet food in brick-and-mortar stores.
**c. Calculate a 95% confidence interval for the parameter:**
Given:
- Sample proportion (\( \hat{p} \)) = 0.416
- Standard error (\( SE \)) = 0.03117
The formula for the confidence interval is:
\[ \hat{p} \pm z \cdot SE \]
For a 95% confidence interval, \( z \) (the critical value) is approximately 1.96.
\[ CI = 0.416 \pm (1.96 \cdot 0.03117) \]
Calculate the margin of error:
\[ MOE = 1.96 \cdot 0.03117 = 0.0611 \]
Thus, the 95% confidence interval is:
\[ (0.416 - 0.0611, 0.416 + 0.0611) = (0.3549, 0.4771) \]
**d. Interpret the 95% confidence interval in context:**
We are 95% confident that the true proportion of all pet owners who buy their pet food in brick-and-mortar stores lies between 35.49% and 47.71%.
**e. Meaning of 95% confidence interval:**
Which of the following best explains the meaning of 95% confidence in this question? Check only one.
- [ ] The proportion of 95% of all pet owners who buy their pet food in brick-and](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe88a8cea-da71-40ab-aaed-a0e0afa0772e%2F4031745e-763c-46c3-9022-c4807a0b56cc%2Ffqppww_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Statistical Analysis of Pet Owners’ Shopping Preferences
#### Question 1:
From a sample of 250 pet owners, 104 indicate that they buy their pet food in brick-and-mortar stores.
**a. Calculate the value of the sample statistic:**
To find the sample statistic (proportion), use the formula:
\[ \hat{p} = \frac{x}{n} \]
Where:
- \( \hat{p} \) is the sample proportion,
- \( x \) is the number of successes (pet owners who buy pet food in brick-and-mortar stores),
- \( n \) is the total sample size.
Plugging in the numbers:
\[ \hat{p} = \frac{104}{250} = 0.416 \]
**b. Define the parameter this sample statistic estimates:**
The parameter this sample statistic estimates is the proportion of all pet owners who buy their pet food in brick-and-mortar stores.
**c. Calculate a 95% confidence interval for the parameter:**
Given:
- Sample proportion (\( \hat{p} \)) = 0.416
- Standard error (\( SE \)) = 0.03117
The formula for the confidence interval is:
\[ \hat{p} \pm z \cdot SE \]
For a 95% confidence interval, \( z \) (the critical value) is approximately 1.96.
\[ CI = 0.416 \pm (1.96 \cdot 0.03117) \]
Calculate the margin of error:
\[ MOE = 1.96 \cdot 0.03117 = 0.0611 \]
Thus, the 95% confidence interval is:
\[ (0.416 - 0.0611, 0.416 + 0.0611) = (0.3549, 0.4771) \]
**d. Interpret the 95% confidence interval in context:**
We are 95% confident that the true proportion of all pet owners who buy their pet food in brick-and-mortar stores lies between 35.49% and 47.71%.
**e. Meaning of 95% confidence interval:**
Which of the following best explains the meaning of 95% confidence in this question? Check only one.
- [ ] The proportion of 95% of all pet owners who buy their pet food in brick-and
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