The branch manager of an outlet (Store 1) of a nationwide chain of pet supply stores wants to study characteristics of her customers. In particular, she decides to focus on two variables: the amount of money spent by customers and whether the customers own only one dog, only one cat, or more than one dog and/or cat. The results from a sample of 70 customers are follows:
- Amount of money spent: X = $21.34, S=$9.22
- 37 customers own only a dog
- 26 customers own only a cat
- 7 customers own more than one dog and/or cat
a) Construct a 95% confidence interval estimate of the population mean amount in the pet supply.
b) Construct a 90% confidence interval estimate of the population proportion of customers who own only a cat
The branch manager of another outlet (Store 2) wishes to conduct a similar survey in the store. The manager does not have access to the information generated by the manager of Store 1. Answer the following questions:
c) What sample size is needed to have 95% confidence of estimating the population mean amount spent in his store within +/-$1.50 if the standard deviation is $10
d) What sample size is needed to have 90% confidence of estimating the population proportion of customers who own only a cat to within +/-0.045?
e) Based on your answers for c and d, how large a sample should the manager take?
Note*The manager at store 2 only has enough money budgeted to do one sample analysis
- Amount of money spent: X = $21.34, S=$9.22
- 37 customers own only a dog
- 26 customers own only a cat
- 7 customers own more than one dog and/or cat
a) Construct a 95% confidence interval estimate of the population mean amount in the pet supply.
b) Construct a 90% confidence interval estimate of the population proportion of customers who own only a cat
The branch manager of another outlet (Store 2) wishes to conduct a similar survey in the store. The manager does not have access to the information generated by the manager of Store 1. Answer the following questions:
c) What sample size is needed to have 95% confidence of estimating the population mean amount spent in his store within +/-$1.50 if the standard deviation is $10
d) What sample size is needed to have 90% confidence of estimating the population proportion of customers who own only a cat to within +/-0.045?
e) Based on your answers for c and d, how large a sample should the manager take?
Note: The manager at store 2 only has enough money budgeted to do one sample analysis.
Incomplete post
For complete post, please see Statistics help with Statistics help Part 2 and Part 3
a) To construct a 95% confidence interval estimate of the population mean amount spent, we can use the formula:
Confidence interval = X ± (Z * (S / √(n)))
Where:
- X is the sample mean
- Z is the Z-score corresponding to the desired confidence level (in this case, 95%, which corresponds to a Z-score of 1.96)
- S is the sample standard deviation
- n is the sample size
Given that X = $21.34, S = $9.22, and the sample size is 70, we can plug these values into the formula:
Confidence interval = $21.34 ± (1.96 * ($9.22 / √(70)))
Calculating the expression inside the parentheses, we get:
$21.34 ± (1.96 * $1.105)
Using the Z-score value, we can find the range of the confidence interval:
Lower limit = $21.34 - $2.1668 ≈ $19.17
Upper limit = $21.34 + $2.1668 ≈ $23.51
Therefore, the 95% confidence interval estimate of the population mean amount spent in the pet supply store is approximately $19.17 to $23.51.
b) To construct a 90% confidence interval estimate of the population proportion of customers who own only a cat, we can use the formula:
Confidence interval = p̂ ± (Z * √((p̂ * (1 - p̂)) / n))
Where:
- p̂ is the sample proportion (owning only a cat)
- Z is the Z-score corresponding to the desired confidence level (in this case, 90%, which corresponds to a Z-score of 1.645)
- n is the sample size
Given that 26 customers own only a cat out of a sample size of 70, we can calculate the sample proportion:
p̂ = 26 / 70 ≈ 0.371
Plugging these values into the formula, we get:
Confidence interval = 0.371 ± (1.645 * √((0.371 * (1 - 0.371)) / 70))
Calculating the expression inside the square root, we have:
√((0.371 * (1 - 0.371)) / 70) ≈ 0.0857
Using the Z-score value, we can find the range of the confidence interval:
Lower limit = 0.371 - (1.645 * 0.0857) ≈ 0.218
Upper limit = 0.371 + (1.645 * 0.0857) ≈ 0.524
Therefore, the 90% confidence interval estimate of the population proportion of customers who own only a cat is approximately 0.218 to 0.524.
c) To determine the sample size needed to estimate the population mean amount spent within +/-$1.50 with 95% confidence, we can use the formula:
Sample size (n) = (Z^2 * (S^2)) / (E^2)
Where:
- Z is the Z-score corresponding to the desired confidence level (in this case, 95%, which corresponds to a Z-score of 1.96)
- S is the estimated standard deviation
- E is the desired margin of error
Given that the standard deviation is $10 and the desired margin of error is +/-$1.50, we can plug these values into the formula:
n = (1.96^2 * 10^2) / (1.5^2)
Simplifying the expression, we get:
n ≈ 6.086
Since the sample size cannot be fractional, we need a minimum sample size of 7 to achieve a margin of error within +/-$1.50.
d) To determine the sample size needed to estimate the population proportion of customers who own only a cat within +/-0.045 with 90% confidence, we can use the formula:
Sample size (n) = (Z^2 * (p̂ * (1 - p̂))) / (E^2)
Where:
- Z is the Z-score corresponding to the desired confidence level (in this case, 90%, which corresponds to a Z-score of 1.645)
- p̂ is the estimated proportion (owning only a cat)
- E is the desired margin of error
Given that the estimated proportion is p̂ = 0.371 and the desired margin of error is +/-0.045, we can plug these values into the formula:
n = (1.645^2 * (0.371 * (1 - 0.371))) / (0.045^2)
Simplifying the expression, we get:
n ≈ 120.515
Since the sample size cannot be fractional, we need a minimum sample size of 121 to achieve a margin of error within +/-0.045.
e) Since the manager at Store 2 only has enough budget for one sample analysis, they should choose the larger sample size requirement. In this case, the larger sample size is 121 (from question d). Therefore, the manager should take a sample size of 121 to achieve both the desired confidence level and margin of error in estimating the population proportion of customers who own only a cat.