Question 1193206: The National Bank, like most other large banks, found that using automatic teller machines (ATMs) reduces the cost of routine bank transactions. National installed an ATM in the corporate offices of the Fun Toy Company. The ATM is for the exclusive use of Fun's 571 employees. After several months of operation, a sample of 100 employees revealed the following use of the ATM machine by Fun employees in a month:
Number of Times
ATM Used Frequency
0 20
1 30
2 20
3 5
4 5
5 20
a. What is the estimate of the proportion of employees who do not use the ATM in a month?
b-1. For estimate of the proportion of employees who do not use the ATM in a month, develop a 90% confidence interval. (Round the final answers to 3 decimal places.)
b-2. Can National be sure that at least 40% of the employees of Fun Toy Company will use the ATM:
c. How many transactions does the average Fun employee make per month? (Round the final answer to 2 decimal places.)
d. Develop a 90% confidence interval for the mean number of transactions per month. (Round the final answers to 3 decimal places.)
90% confidence interval for the mean number of transaction per month is
and
.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! **a. Estimate of the proportion of employees who do not use the ATM in a month**
* **Calculate the number of employees who do not use the ATM:**
* Number of employees who do not use the ATM = 20 employees (from the table)
* **Estimate the proportion:**
* Proportion = (Number of employees who do not use the ATM) / (Total number of employees in the sample)
* Proportion = 20 / 100 = 0.20
**Therefore, the estimate of the proportion of employees who do not use the ATM in a month is 0.20.**
**b-1. 90% Confidence Interval for the Proportion**
* **Find the sample proportion (p̂):**
* p̂ = 0.20
* **Find the standard error (SE):**
* SE = √[(p̂ * (1 - p̂)) / n]
* SE = √[(0.20 * 0.80) / 100]
* SE = √(0.16 / 100)
* SE = √0.0016
* SE = 0.04
* **Find the critical value (zα/2) for a 90% confidence level:**
* For a 90% confidence level, α = 0.10
* zα/2 = z0.05
* Using a standard normal distribution table, we find z0.05 ≈ 1.645
* **Calculate the margin of error (E):**
* E = zα/2 * SE
* E = 1.645 * 0.04
* E ≈ 0.0658
* **Calculate the confidence interval:**
* Lower limit = p̂ - E = 0.20 - 0.0658 = 0.1342
* Upper limit = p̂ + E = 0.20 + 0.0658 = 0.2658
**The 90% confidence interval for the proportion of employees who do not use the ATM in a month is (0.134, 0.266).**
**b-2. Can National be sure that at least 40% of the employees of Fun Toy Company will use the ATM?**
* No, National cannot be sure that at least 40% of the employees will use the ATM.
* The lower limit of the 90% confidence interval for the proportion of employees who do not use the ATM is 0.134.
* This means that there is a possibility that the true proportion of employees who do not use the ATM could be as high as 13.4%.
* If 13.4% of employees do not use the ATM, then only 86.6% would use it, which is below 90%.
**c. How many transactions does the average Fun employee make per month?**
* **Calculate the mean number of transactions:**
* Mean = Σ(x * f(x)) / Σf(x)
* Where:
* x represents the number of ATM uses
* f(x) represents the frequency of each number of uses
* Mean = (0 * 20 + 1 * 30 + 2 * 20 + 3 * 5 + 4 * 5 + 5 * 20) / 100
* Mean = (0 + 30 + 40 + 15 + 20 + 100) / 100
* Mean = 205 / 100
* Mean = 2.05
**The average Fun employee makes 2.05 transactions per month.**
**d. 90% Confidence Interval for the Mean Number of Transactions**
* **Calculate the sample standard deviation (s):**
* This requires calculating the variance first:
* Variance (s²) = Σ[(x - Mean)² * f(x)] / (n - 1)
* Variance (s²) = [(0 - 2.05)² * 20 + (1 - 2.05)² * 30 + ... + (5 - 2.05)² * 20] / (100 - 1)
* Variance (s²) ≈ 2.3025
* Standard deviation (s) = √s²
* Standard deviation (s) = √2.3025
* Standard deviation (s) ≈ 1.517
* **Calculate the standard error of the mean (SEM):**
* SEM = s / √n
* SEM = 1.517 / √100
* SEM = 1.517 / 10
* SEM = 0.1517
* **Find the critical value (tα/2) for a 90% confidence level with 99 degrees of freedom (n - 1):**
* Using a t-distribution table, we find tα/2 ≈ 1.660
* **Calculate the margin of error (E):**
* E = tα/2 * SEM
* E = 1.660 * 0.1517
* E ≈ 0.2518
* **Calculate the confidence interval:**
* Lower limit = Mean - E = 2.05 - 0.2518 = 1.798
* Upper limit = Mean + E = 2.05 + 0.2518 = 2.302
**The 90% confidence interval for the mean number of transactions per month is (1.798, 2.302).**
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