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.

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.

Explanation:

Step1: Identify the formula for confidence interval

For a large - sample (n = 40, usually n>30 is considered large), the confidence interval for the population mean $\mu$ is given by $\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z_{\alpha/2}$ is the z - score, $\sigma$ is the standard deviation, and n is the sample size. For a 95% confidence interval, $\alpha=1 - 0.95 = 0.05$, so $\alpha/2=0.025$ and $z_{\alpha/2}=1.96$. We are given that the mean of the simulation distribution $\mu_{sim}=0.499$ and the standard deviation of the simulation distribution $\sigma_{sim}=0.041$.

Step2: Calculate the confidence interval for part a

Given $\bar{x}=0.65$, $n = 40$, $z_{\alpha/2}=1.96$, and $\sigma = 0.041$.
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 confidence interval is $\bar{x}\pm E=0.65\pm0.01274=(0.63726,0.66274)$. Since the hypothesized mean $\mu_0 = 0.50$ is not in the confidence interval $(0.63726,0.66274)$, the manager can conclude that the mean of all change given is more than $0.50$ at the 95% confidence level.

Step3: Calculate the confidence interval for part b

Given $\bar{x}=0.52$, $n = 40$, $z_{\alpha/2}=1.96$, and $\sigma = 0.041$.
The margin of error $E=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=1.96\times\frac{0.041}{\sqrt{40}}\approx0.01274$.
The confidence interval is $\bar{x}\pm E=0.52\pm0.01274=(0.50726,0.53274)$. Since the hypothesized mean $\mu_0 = 0.50$ is in the confidence interval $(0.50726,0.53274)$, the manager cannot conclude that the mean of all change given is more than $0.50$ at the 95% confidence level.

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. The manager cannot conclude that the mean of all change given is more than $0.50$ since the 95% confidence interval is $(0.50726,0.53274)$ and $0.50$ is in this interval.