Chapter 23, Problem 32E

(a)

To determine

To explain do these data satisfy assumptions for inference.

Answer to Problem 32E

These data satisfy assumptions for inference.

Explanation of Solution

In the question it is given that, some students checked six bags of Doritos marked with a net weight of $28.3$ grams. They then carefully weighed the content of each bag and recorded it as:

  $\{29.2,28.5,28.7,28.9,29.1,29.5\}$

Thus, now we will check if these data satisfy assumptions for inference as:

Random condition: It is satisfied assuming that the six bags were randomly selected.

Independent condition: It is satisfied since the sample of six bags of Doritos is less than $10\%$ of the population of all bags of Doritos.

Normal condition: It is satisfied because the histogram shows no outlier and the normal quantile plot is nearly linear.

Thus, all the conditions are met.

(b)

To determine

To find the mean and standard deviation of the weight.

Answer to Problem 32E

The mean is $28.9833$ .

The standard deviation is $0.3601$ .

Explanation of Solution

In the question it is given that, some students checked six bags of Doritos marked with a net weight of $28.3$ grams. They then carefully weighed the content of each bag and recorded it as:

  $\{29.2,28.5,28.7,28.9,29.1,29.5\}$

The mean of this data is:

  $\begin{array}{l}\overline{x}=\frac{29.2+28.5+29.1+28.7+28.9+29.5}{6}\\ =28.9833\end{array}$

The standard deviation of the data is:

  $\begin{array}{l}s=\sqrt{\frac{{(29.2-28.9833)}^2+\mathrm{....}+{(29.5-28.9833)}^2}{6-1}}\\ =0.3601\end{array}$

(c)

To determine

To create a $95\%$ confidence interval for the mean weight of such bags of chips.

Answer to Problem 32E

The confidence interval is: $(28.6053,29.3613)$ .

Explanation of Solution

In the question it is given that, some students checked six bags of Doritos marked with a net weight of $28.3$ grams. They then carefully weighed the content of each bag and recorded it as:

  $\{29.2,28.5,28.7,28.9,29.1,29.5\}$

  $n=6$

  $\begin{array}{l}\overline{x}=28.9833\\ s=0.3601\end{array}$

The degrees of freedom is:

  $\begin{array}{l}df=n-1\\ =6-1\\ =5\end{array}$

The t -value is:

  $t_{\alpha /2}=2.571$

The confidence interval is:

  $\begin{array}{l}\overline{x}-t_{\alpha /2}\times \frac{s}{\sqrt{n}}\\ =28.9833-2.571\times \frac{0.3601}{\sqrt{6}}\\ =28.6053\\ \overline{x}+t_{\alpha /2}\times \frac{s}{\sqrt{n}}\\ =28.9833+2.571\times \frac{0.3601}{\sqrt{6}}\\ =29.3613\end{array}$

Thus, the confidence interval is: $(28.6053,29.3613)$ .

(d)

To determine

To explain in context what your interval means.

Explanation of Solution

In the question it is given that, some students checked six bags of Doritos marked with a net weight of $28.3$ grams. They then carefully weighed the content of each bag and recorded it as:

  $\{29.2,28.5,28.7,28.9,29.1,29.5\}$

  $n=6$

  $\begin{array}{l}\overline{x}=28.9833\\ s=0.3601\end{array}$

The confidence interval is: $(28.6053,29.3613)$ . Thus, we are $95\%$ confident that the true net weight of the bags of Doritos marked with a net weight of $28.3$ grams is between $28.6053\text{Grams and }29.3613\text{Grams}$ .

(e)

To determine

To comment on the company’s stated net weight of $28.3$ grams.

Explanation of Solution

In the question it is given that, some students checked six bags of Doritos marked with a net weight of $28.3$ grams. They then carefully weighed the content of each bag and recorded it as:

  $\{29.2,28.5,28.7,28.9,29.1,29.5\}$

  $n=6$

  $\begin{array}{l}\overline{x}=28.9833\\ s=0.3601\end{array}$

The confidence interval is: $(28.6053,29.3613)$ . Thus, the company's stated net weight of $28.3$ grams appears to be incorrect and the actual net weight appears to be more $28.3$ grams, because the confidence interval does not contain $28.3$ and only contains values higher than $28.3$ grams.