Question

Cheese Consumption. Refer to Problem $24$. The following table provides last year's cheese consumption: in pounds, 35 randomly selected Americans.

$$\begin{gathered} \begin{array}{lllllll}46& 29& 33& 38& 42& 40& 34\\ 33& 32& 36& 28& 47& 26& 42\\ 36& 32& 45& 24& 39& 28& 33\\ 44& 33& 26& 37& 27& 31& 36\end{array} \\ \begin{array}{lllllll}37& 37& 36& 22& 44& 36& 29\end{array} \end{gathered}$$

1. At the $10\%$significance level, do the data provide sufficient evidence to conclude that last year's mean cheese consumption for all Americans has increased over the 2010 mean? Assume that $\sigma =6.9\mathrm{lb}$. Use a z-test. (Note: The sum of the data is $1218\mathrm{lb}$.)
2. Given the conclusion in part (a), if an error has been made, what type must it be? Explain your answer.

Short answer

(a) There is sufficient data to support the assumption that total American cheese consumption increased last year compared to $2010$.

(b) Type I is error

Step-by-step solution

Part (a) Step 1: Given information

Given in the question that, last year's cheese consumption, in pounds, for 35 randomly selected Americans

$\begin{array}{lllllll}46& 29& 33& 38& 42& 40& 34\\ 33& 32& 36& 28& 47& 26& 42\\ 36& 32& 45& 24& 39& 28& 33\\ 44& 33& 26& 37& 27& 31& 36\\ 37& 37& 36& 22& 44& 36& 29\end{array}$

Part (a) Step 2: Explanation

The given information states that the mean has increased from 33. The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population mean is equal to the value mentioned in the claim. If the null hypothesis is the claim, then the alternative hypothesis states the opposite of the null hypothesis.

$$\begin{gathered} H_0:\mu =33 \\ {H}_0:\mu >33 \end{gathered}$$

The mean is calculated by dividing the total of all values by the number of values:

$$\begin{gathered} \overline{x}=\frac{{\displaystyle \sum _{}}xi}{n} \\ =\frac{1218}{35} \\ =34.8 \end{gathered}$$

The sample mean's sampling distribution has a mean $\mu$and a standard deviation of $\frac{\sigma }{\sqrt{n}}$. The sample mean is subtracted from the population mean and divided by the standard deviation to get the $z$value:

$$\begin{gathered} z=\frac{\overline{x}-\mu }{{\displaystyle \raisebox{1ex}{$\sigma $}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}} \\ =\frac{34.8-33}{6.9\sqrt{35}} \\ \approx 1.54 \end{gathered}$$

The test is left-tailed if the alternative hypothesis $H_a$contains$<$.
The test is right-tailed if the alternative hypothesis $H_a$contains$>$.
The test is two-tailed if the alternative hypothesis $H_a$contains$\ne$.
If the null hypothesis is true, the $P$-value is the probability of getting a value that is more extreme or equal to the standardized test statistic $z$.

Part (a) Step 3: P-value

Here, the test is two-tailed, and the $P$-value represents the likelihood that the $z$-score is greater than $z=1.54$.

$P(z>1.54)$

The probability $P(z>1.54)$is shown in the table's row beginning with " $1.5$" and column beginning with ".04 ".

$$\begin{gathered} P=P(z>1.54) \\ =1-P(z<1.54) \\ =1-0.9382 \\ =0.0618 \end{gathered}$$

The null hypothesis is rejected if the$P$-value is less than the significance level $a$.

$0.0618<0.10$,
Therefore, Reject$H_0$

There is sufficient data to support the assumption that total American cheese consumption increased last year compared to 2010.

Part (b) Step 1: Given information

In order to define the type of error, we must first describe the error type (a).

Explanation

Type I error is When the null hypothesis $H_0$is true, reject the null hypothesislocalid="1654764107290" $H_0$.
Type Il error is When the null hypothesis localid="1654764112725" $H_0$is untrue, the null hypothesislocalid="1654764119120" $H_0$ is not rejected.
They could have made a Type I error because they rejected the null hypothesis localid="1654764123114" $H_0$in part (a).