Question

part b
a cereal makers container machine is designed to fill boxes so that the mean weight of cereal in the boxes is 18 ounces.

mean standard deviation standard error mean min q1 median q3 max
weight 17.920 0.200 0.0395 17.55 17.75 17.90 18.00 18.25

question 2 (33 points)
conclusion & interpretation:
we are ___ % confident that the
conclusion & interpretation:
we are _ % confident that the interval ( _, ___ ) ounces contains the true mean weight of cereal boxes filled by this machine.
round your answer to the second decimal place. use a comma between the answers. do not put your answer in parenthesis.

since the claimed mean of 18 ounces (is/is not) in this confidence interval...

since the claimed mean of 18 ounces (is/is not) in this confidence interval, we have evidence that the mean (is/is not) 18 ounces.

Explanation:

Step1: Assume a 95% confidence level

For a normal - distribution, a common confidence level is 95%. The critical value $z$ for a 95% confidence interval is approximately $z = 1.96$.

Step2: Calculate the confidence interval

The formula for a confidence interval for the population mean $\mu$ when the population standard - deviation $\sigma$ is known (or when the sample size $n$ is large and we use the sample standard - deviation $s$ as an estimate) is $\bar{x}\pm z\times SE$, where $\bar{x}$ is the sample mean, $z$ is the critical value, and $SE$ is the standard error of the mean. Given $\bar{x}=17.920$, $SE = 0.0385$, and $z = 1.96$.
The lower limit of the confidence interval is $L=\bar{x}-z\times SE=17.920 - 1.96\times0.0385=17.920-0.0755 = 17.8445\approx17.84$.
The upper limit of the confidence interval is $U=\bar{x}+z\times SE=17.920 + 1.96\times0.0385=17.920 + 0.0755=17.9955\approx17.99$.

Step3: Check if the claimed mean is in the interval

The claimed mean is $\mu = 18$ ounces. Since $18$ is not in the interval $(17.84,17.99)$.

Step4: Make a conclusion

We have evidence that the mean is not 18 ounces.

Answer:

95
17.84,17.99
is not
is not