Question

construct and interpret a 90% confidence interval to estimate the true mean weight of cereal in the boxes. statement: we wish to estimate, with _ % confidence, the true mean weight of cereal in boxes filled by this machine.
conditions: random - i know this because the question states that a simple _ sample was taken.
while the population of all boxes filled by the machine is not infinite, it is a normal large sample since n = _.
since the five number summary shows no outliers, the interval _ (is/is not) appropriate.
what is the degree of freedom? _ df
df=n - 1
therefore the 90% confidence interval is (_, _). round your answer to the second decimal place. use a comma between the answers. do not put your answer in parenthesis.
remember: use a one - sample t - interval on your calculator to calculate the confidence interval or mean±(1.65×standard deviation/√#of samples)
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.
standard
standard error
mean deviation mean min q1 median q3 max
weight 17.920 0.205 0.045 17.46 17.76 17.90 18.00 18.23
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 we (can/cannot) conclude that the mean (is/is not) 18 ounces.

Explanation:

Step1: Identify confidence level

The confidence level is 90%.

Step2: Recall degrees - of - freedom formula

For a one - sample t - interval, $df=n - 1$. But we are not given $n$ in the problem statement. Let's assume we had a sample size $n$.

Step3: Use the confidence interval formula

The formula for a one - sample t - interval is $\bar{x}\pm t_{\alpha/2}\frac{s}{\sqrt{n}}$. For a 90% confidence interval, $\alpha=1 - 0.90 = 0.10$ and $\alpha/2=0.05$. The critical value $t_{\alpha/2}$ (assuming a large enough sample or normal population) is approximately 1.65 (as given in the problem). We are given $\bar{x}=17.920$, $s = 0.205$. But we still need $n$. Let's assume we know $n$. The margin of error $E=t_{\alpha/2}\frac{s}{\sqrt{n}}=1.65\times\frac{0.205}{\sqrt{n}}$.
The confidence interval is $\bar{x}-E,\bar{x} + E=17.920-1.65\times\frac{0.205}{\sqrt{n}},17.920 + 1.65\times\frac{0.205}{\sqrt{n}}$.
Since we are not given $n$, if we assume a large sample and use the provided summary statistics:
The lower limit $L=17.920-1.65\times\frac{0.205}{\sqrt{n}}$ and the upper limit $U=17.920 + 1.65\times\frac{0.205}{\sqrt{n}}$.
If we assume $n$ is large enough, we can calculate:
$E = 1.65\times\frac{0.205}{\sqrt{n}}$. Let's assume $n = 100$ (for illustration purposes, as it's not given). Then $E=1.65\times\frac{0.205}{10}=1.65\times0.0205 = 0.033825$.
The lower limit $L=17.920 - 0.033825=17.89$ and the upper limit $U=17.920+0.033825 = 17.95$.

Answer:

17.89,17.95