Dole Pineapple Inc. is concerned that the 16-ounce can of sliced pineapple is being overfilled. Assume the standard deviation of the process is .03 ounces. The quality-control department took a random sample of 50 cans and found that the arithmetic mean weight was 16.05 ounces. At the 5 percent level of significance, can we conclude that the mean weight is greater than 16 ounces?

1. Write the null and alternative hypothesis

2. What is the significance level?

3. Solve for z value

4. Write the decision rule

5. Accept or Reject: Be Very Specific what you plan to do and why.

1. the null and alternative hypothesis are:

2. The significance level is α=0.05.

3. Test statistic z=11.785

4. Critical value approach: The decision rule is that if the test statistic is above zc=1.96, the null hypothesis is rejected.

5. Reject. There is  enough evidence to support the claim that the mean weight is greater than 16 ounces.  The machine should be calibrated to correct the deviation.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the mean weight is greater than 16 ounces.

Then, the null and alternative hypothesis are:

The significance level is 0.05.

The sample has a size n=50.

The sample mean is M=16.05.

The standard deviation of the population is known and has a value of σ=0.03.

We can calculate the standard error as:

Then, we can calculate the z-statistic as:

This test is a right-tailed test, so the P-value for this test is calculated as:

As the P-value (0) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is  enough evidence to support the claim that the mean weight is greater than 16 ounces.

Critical value approach:

The critical value for a level of significance is α=0.05 is z=1.96.

The decision rule is that if the test statistic is above zc=1.96, the null hypothesis is rejected.

As the test statistic z=11.785 is above the critical value zc=1.96, then the null hypothesis is rejected.

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