Question

1. you are testing chocolate chip cookies to estimate the mean number of chips per cookie. you sample 25 cookies and you find a sample mean of 10 chips per cookie. assume (sigma = 2). find a 95% confidence interval and interpret.

Explanation:

Step1: Find the z - score for 95% confidence level

For a 95% confidence level, the significance level \(\alpha=1 - 0.95=0.05\). Then \(\alpha/2=0.025\). Looking up in the standard normal table, \(z_{\alpha/2}=z_{0.025} = 1.96\)

Step2: Calculate the standard error

The formula for the standard error of the mean is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\). Given \(\sigma = 2\) and \(n = 25\), then \(\sigma_{\bar{x}}=\frac{2}{\sqrt{25}}=\frac{2}{5}=0.4\)

Step3: Calculate the margin of error

The margin of error \(E=z_{\alpha/2}\times\sigma_{\bar{x}}\). Substituting \(z_{\alpha/2}=1.96\) and \(\sigma_{\bar{x}} = 0.4\), we get \(E=1.96\times0.4 = 0.784\)

Step4: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is given by \(\bar{x}-E<\mu<\bar{x} + E\). Given \(\bar{x}=10\), then \(10 - 0.784<\mu<10+0.784\)

Answer:

The 95% confidence interval is \((9.216, 10.784)\). This means that we are 95% confident that the true mean number of chips per cookie lies between \(9.216\) and \(10.784\)