Question

1. the manager at tim hortons is interested in the distribution of change that is given in coins to each customer at his store. he assumes the average change given would be $0.50. he runs a simulation using this assumption 100 times each with a sample size of 40. the results are shown below:

a. a random sample of 40 customers yielded a sample mean of $0.65. could the manager conclude that the mean of all change given is more than $0.50 at his store based on the simulation? support your conclusion with a 95% confidence interval.
the manager can conclude that the mean of all change given is more than the 0.50 since the 95% interval is (0.63726 - 0.66274 and 0.50 is not in this interval
b. if a random sample of 40 customers yielded a sample mean of $0.52, could the manager make the same conclusion as in part a? explain.
c. state a value for a sample mean that would cause the manager to think that the average change given is less than $0.50. explain. there are multiple correct answers.

Explanation:

Step1: Identify the formula for confidence - interval

For a large - sample (n = 40, n>30), the confidence - interval for the population mean $\mu$ is given by $\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$. Here, we assume the standard deviation $\sigma$ is estimated from the simulation results. Let's assume the standard deviation $\sigma$ is known or estimated as $\sigma = 0.041$ (from the problem context), the sample size $n = 40$, and for a 95% confidence interval, $z_{\alpha/2}=1.96$.

Step2: Calculate the margin of error for part a

The margin of error $E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=1.96\times\frac{0.041}{\sqrt{40}}\approx1.96\times\frac{0.041}{6.3246}\approx1.96\times0.0065 = 0.01274$. The sample mean $\bar{x}=0.65$. The 95% confidence interval is $\bar{x}\pm E=0.65\pm0.01274=(0.63726, 0.66274)$. Since $0.50$ is not in this interval, the manager can conclude that the mean of all change given is more than $0.50$.

Step3: Calculate the confidence - interval for part b

The sample mean $\bar{x}=0.52$. The margin of error $E = 0.01274$ (same as above). The 95% confidence interval is $0.52\pm0.01274=(0.50726, 0.53274)$. Since $0.50$ is in this interval, the manager cannot conclude that the mean of all change given is more than $0.50$.

Step4: Determine a sample mean for part c

A sample mean $\bar{x}<0.50 - E$ would cause the manager to think that the average change given is less than $0.50$. Since $E = 0.01274$, a sample mean $\bar{x}=0.48$ (for example) would work. Because if $\bar{x}=0.48$, the 95% confidence interval is $0.48\pm0.01274=(0.46726, 0.49274)$, and $0.50$ is not in this interval, so the manager would think the population mean is less than $0.50$.

Answer:

a. The manager can conclude that the mean of all change given is more than $0.50$ since the 95% confidence interval is $(0.63726, 0.66274)$ and $0.50$ is not in this interval.
b. No, the manager cannot make the same conclusion. The 95% confidence interval for a sample mean of $0.52$ is $(0.50726, 0.53274)$ and $0.50$ is in this interval.
c. A sample mean of $0.48$ (for example). Since the 95% confidence interval for $\bar{x}=0.48$ is $(0.46726, 0.49274)$ and $0.50$ is not in this interval, the manager would think the average change given is less than $0.50$.