Q. 26 Refer to Problem 24.聽The follow... [FREE SOLUTION]

Chapter 9: Q. 26 (page 393)

Refer to Problem 24.The following table provides last year's cheese consumption, in pounds, for 35 randomly selected Americans.
Part (a): 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 $蟽 = 6.9 l b$ . Use a z-test.
Part (b): Given the conclusion in part (a), if an error has been made, what type must it be? Explain your answer.

Short Answer

Expert verified

Part (a): The data provide sufficient evidence to conclude that last year's mean cheese consumption for all Americans has increased over the 2010 mean.

Part (b): There are two types of error.

Type I error is error committed in rejecting a null hypothesis $(H_{0})$ , when it is true.

Type I error is error committed in rejecting a null hypothesis $(H_{0})$ , when it is false.

Step by step solution

Part (a) Step 1. Given information.

Consider the given question,

Part (a) Step 2. State the null and alternative hypothesis.

The null hypothesis is given below,

$H_{0} : 渭 = 33 lb$

There is no evidence that last year's mean cheese consumption for Americans has increased over the 2010mean.

The alternative hypothesis is given below,

$H_{a} : 渭 > 33 l b$

Last year's mean cheese consumption for Americans has increased over the 2010mean.

Part (a) Step 3. Compute the confidence interval.

On using the MINITAB procedure,

Choose Stat>Basic Statistics>1-Sample Z.
In Samples in Column, enter the column of Consumption.
In Standard deviation, enter 6.9.
In Perform hypothesis test, enter the test mean as 33.
Check Options, enter Confidence level as 90.
Choose greater than in alternative.
Click OK in all dialogue boxes.

Hence, from the MINITAB output, the value of test statistics is 1.54 and the P-value is 0.061.

Critical value is given below,

From "Table II Areas under the standard normal curve" the required value of $z_{\frac{a}{2}}$ with 95% confidence level is $卤 1.96$ .

P-value is given below,

From the MINITAB output, the P-value is 0.061.

Part (a) Step 4. Use the critical value and P-value approach.

Using the rejection rule on the critical value approach,

If the value of test statistic falls in the rejection region, reject the null hypothesis $H_{0}$ .

Here, the value of test statistics fall sin the rejection region. Therefore, the null hypothesis is rejected at 10% level.

It can be concluded that the test results are statistically significant at 10% level of significance.

Using the P-value approach,

If $P 鈮� 伪$ , then reject the null hypothesis.

The P-value is 0.061, which is greater than the level of significance, that is $P (= 0.061) < 伪 (= 0.1)$ .

Therefore, the null hypothesis is rejected at 5% level.

It can be concluded that the test results are statistically significant at 10% level of significance.

On interpreting, we can say that the data provide sufficient evidence to conclude that last year's mean cheese consumption for all Americans has increased over the 2010mean.

Part (b) Step 1. Identify the type of error.

Consider the type I error,

This is the error committed in rejecting a null hypothesis $(H_{0})$ , when it is true.

Consider the type II error,

This is the error committed in rejecting a null hypothesis $(H_{0})$ , when it is false.

In this situation, the Type I error would be appropriate because the null hypothesis is rejected in part (a).

The conclusion is that last year's mean cheese consumption for all Americans has increased over the 2010 mean but in reality the last year's mean cheese consumption for all Americans does not increased over the2010mean.