Chapter 9, Problem 1CRE

A random sample of 40 quarters minted after 1964 are weighed. The mean weight is 5.639 grams and the standard deviation is 0.062 gram. Assume that we want to use a 0.05 significance level to test the claim that the population of quarters has a mean weight of 5.670 grams, which is the manufacturing specification.

1. a. What is the null hypothesis?
2. b. What is the alternative hypothesis?
3. c. What is the value of the standard score for the sample mean of 5.639 grams?
4. d. What is the P-value?
5. e. What do you conclude? (Be sure to address the original claim that the mean weight is 5.670 grams.)
6. f. Describe a type I error for this test.
7. g. Describe a type II error for this test.

To determine

Identify the null hypothesis.

Explanation of Solution

The given information is that the mean weight for a sample of 40 quarters is 5.639 grams and the standard deviation is 0.062 gram.

Here, the claim is that the population of quarters has a mean weight of 5.670 grams. This can be written as $\mu =5.670\text{ grams}$.

Therefore, the null hypothesis is $H_0:\mu =5.670\text{ grams}$.

To determine

Identify the alternative hypothesis.

Explanation of Solution

Here, the claim is that the population of quarters has a mean weight of 5.670 grams. This can be written as $\mu =5.670\text{ grams}$. The complement of the claim is $\mu \ne 5.670\text{ grams}$.

Therefore, the alternative hypothesis is $H_a:\mu \ne 5.670\text{ grams}$.

To determine

Find the value of the standard score for the sample mean of 5.639 grams.

Answer to Problem 1CRE

The value of the standard score for the sample mean of 5.639 grams is –3.16.

Explanation of Solution

Calculation:

Assume that $\sigma =s$ because the sample size is large.

The formula for standard score is,

$z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}$

Substitute $\overline{x}=5.639$, $\mu =5.670$, $\sigma =0.062$ and $n=40$ in the standard score.

$\begin{array}{c}z=\frac{5.639-5.670}{\frac{0.062}{\sqrt{40}}}\\ =\frac{-0.031}{\left(\frac{0.062}{6.3246}\right)}\\ =\frac{-0.031}{0.0098}\\ =-3.16\end{array}$

Thus, the value of the standard score is –3.16.

To determine

Find the P-value.

Answer to Problem 1CRE

The P-value is 0.001578.

Explanation of Solution

Calculation:

P-value:

Software procedure:

Step-by-step software procedure to obtain P-value using EXCEL is as follows:

• Open an EXCEL file.
• In cell A1, enter the formula “=NORM.S.DIST(–3.16,1)”.
• Output using Excel software is given below:

From the output, the P-value for two tailed test is $2\times 0.000789=0.001578$.

To determine

State the conclusion.

Answer to Problem 1CRE

There is enough evidence to reject the claim that the population of quarters has a mean weight of 5.670 grams.

Explanation of Solution

Decision rule:

If $P\text{-value}\le \alpha$, then reject the null hypothesis.

Conclusion:

Here, the P-value is less than the level of significance.

That is, $P\text{-value}\left(=0.001578\right)<\alpha \left(=0.05\right)$

Therefore, the null hypothesis is rejected.

Hence, there is enough evidence to reject the claim that the population of quarters has a mean weight of 5.670 grams.

To determine

Describe a type I error.

Explanation of Solution

Type I error: Reject the null hypothesis, when it is actually true.

A Type I error occurs when the hypotheses test indicates that the mean weight differs from 5.670 grams. However, in reality, the mean weight is 5.670 grams.

To determine

Describe a type II error.

Explanation of Solution

Type II error: Fail to reject the null hypothesis, when it is false.

A Type II error occurs when the hypotheses test indicates that the mean weight is 5.670 grams. However, in reality, the mean weight differs from 5.670 grams.