Chapter 10, Problem 1SR

Heinz, a manufacturer of ketchup, uses a particular machine to dispense 16 ounces of its ketchup into containers. From many years of experience with the particular dispensing machine, Heinz knows the amount of product in each container follows a normal distribution with a mean of 16 ounces and a standard deviation of 0.15 ounce. A sample of 50 containers filled last hour revealed the mean amount per container was 16.017 ounces. Does this evidence suggest that the mean amount dispensed is different from 16 ounces? Use the .05 significance level.

1. (a) State the null hypothesis and the alternate hypothesis.
2. (b) What is the probability of a Type I error?
3. (c) Give the formula for the test statistic.
4. (d) State the decision rule.
5. (e) Determine the value of the test statistic.
6. (f) What is your decision regarding the null hypothesis?
7. (g) Interpret, in a single sentence, the result of the statistical test.

To determine

State the hypotheses.

Answer to Problem 1SR

The null hypothesis is $\underset{\_}{H_0:\mu =16}$.

The alternative hypothesis is $\underset{\_}{H_1:\mu \ne 16}$.

Explanation of Solution

Here, the claim is that there is evidence that the mean amount dispensed is different from 16 ounces. This defines the alternative hypothesis.

Let $\mu$ denotes the mean amount dispensed.

The hypotheses are given below:

Null hypothesis:

$H_0:\mu =16$

Alternative hypothesis:

$H_1:\mu \ne 16$

To determine

Write the probability of a Type I error

Explanation of Solution

Type I error:

Probability of rejecting $H_0$, when it is true.

Here, the null hypothesis is rejected. But in actual the mean amount dispensed is 16 ounces.

To determine

Write the formula for the test statistic.

Explanation of Solution

The formula for the test statistic is given below:

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

Where, $\overline{x}$ is sample mean, $\mu$ is population mean, $\sigma$ is population standard deviation, and $n$ is sample size.

To determine

Write the decision rule.

Explanation of Solution

Step-by-step procedure to obtain the critical value using MINITAB:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose Probability and Both Tail for the region of the curve to shade.
• Enter the Probability value as 0.05.
• Click OK.

Output using the MINITAB software is given below:

From the output, the critical value is ±1.96.

Decision rule:

If $z\text{-value}<-1.96$, then reject the null hypothesis. or

If $z\text{-value}>1.96$, then reject the null hypothesis.

To determine

Find the value of test statistic.

Answer to Problem 1SR

The value of test statistic is 0.8.

Explanation of Solution

Step by step procedure to obtain the test statistic using MINITAB software is given below:

• Choose Stat > Basic Statistics > 1-Sample Z.
• In Summarized data, enter the sample size as 50 and mean as 16.017.
• In Standard deviation, enter 0.15.
• Check Options, enter Confidence level as 95.
• In Perform Hypothesis test, enter 16 Under Hypothesized mean.
• Choose Mean ≠ Hypothesized mean in alternative.
• Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

From the MINITAB output, the value of test statistic is 0.8.

To determine

Find the decision.

Answer to Problem 1SR

The decision is that fail to reject the null hypothesis.

Explanation of Solution

Decision:

Here, the computed z-value is 0.8.

The computed z-value lies between ±1.96.

From the decision rule, fail to reject the null hypothesis.

To determine

Write the single sentence for the result of the test.

Explanation of Solution

The null hypothesis is not rejected. Hence, it can be concluded that there is no evidence that the mean amount dispensed is different from 16 ounces.