eat your cereal: boxes of cereal are labeled as containing 14 ounces. following are the weights, in ounces…

Explanation:

Step1: Calculate sample mean $\bar{x}$

First, sum all the data values: $13.01 + 14.96+13.10 + 13.11+13.09+13.01+13.14+14.96+13.04+13.03+13.10+13.11 = 156.66$. Then divide by the sample size $n = 12$. So, $\bar{x}=\frac{156.66}{12}=13.055$.

Step2: Calculate sample standard - deviation $s$

The formula for the sample standard - deviation is $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$. First, calculate $(x_{i}-\bar{x})^{2}$ for each data point: $(13.01 - 13.055)^{2}=(-0.045)^{2}=0.002025$ $(14.96 - 13.055)^{2}=(1.905)^{2}=3.639025$ $(13.10 - 13.055)^{2}=(0.045)^{2}=0.002025$ $(13.11 - 13.055)^{2}=(0.055)^{2}=0.003025$ $(13.09 - 13.055)^{2}=(0.035)^{2}=0.001225$ $(13.01 - 13.055)^{2}=(-0.045)^{2}=0.002025$ $(13.14 - 13.055)^{2}=(0.085)^{2}=0.007225$ $(14.96 - 13.055)^{2}=(1.905)^{2}=3.639025$ $(13.04 - 13.055)^{2}=(-0.015)^{2}=0.000225$ $(13.03 - 13.055)^{2}=(-0.025)^{2}=0.000625$ $(13.10 - 13.055)^{2}=(0.045)^{2}=0.002025$ $(13.11 - 13.055)^{2}=(0.055)^{2}=0.003025$

The sum $\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=0.002025+3.639025+0.002025+0.003025+0.001225+0.002025+0.007225+3.639025+0.000225+0.000625+0.002025+0.003025 = 7.293$. Then $s=\sqrt{\frac{7.293}{12 - 1}}=\sqrt{\frac{7.293}{11}}\approx0.812$.

Step3: Determine the critical value $t_{\alpha/2}$

The confidence level is $90%$, so $\alpha=1 - 0.90 = 0.10$ and $\alpha/2=0.05$. The degrees of freedom $df=n - 1=12 - 1 = 11$. Looking up in the $t$-distribution table, $t_{0.05,11}=1.796$.

Step4: Calculate the margin of error $E$

The formula for the margin of error for a confidence interval of the mean when the population standard - deviation $\sigma$ is unknown is $E = t_{\alpha/2}\frac{s}{\sqrt{n}}$. Substitute the values: $E = 1.796\times\frac{0.812}{\sqrt{12}}\approx1.796\times\frac{0.812}{3.464}\approx0.421$.

Step5: Construct the confidence interval

The confidence interval for the population mean $\mu$ is $\bar{x}-E<\mu<\bar{x} + E$. Substitute the values: $13.055-0.421<\mu<13.055 + 0.421$, which gives $12.634<\mu<13.476$.

Answer:

$12.634<\mu<13.476$