Answer to Question #245372 in Statistics and Probability for nieks

Question

A consumer group is investigating a producer of diet meals to examine if its prepackaged meals actually contain the advertised 6 ounces of protein in each package. Based on the following data, is there any evidence that the meals do not contain the advertised amount of protein? Run the appropriate test at a 5% level of significance.

5.1

4.9

6.0

5.1

5.7

5.5

4.9

6.1

6.0

5.8

5.2

4.8

4.7

4.2

4.9

5.5

5.6

5.8

6.0

6.1

Answer 1

$$\bar{x}=\dfrac{1}{n}\displaystyle\sum_{i=1}^nx_i=\dfrac{107.9}{20}=5.395$$

$$s^2=\dfrac{1}{n-1}\displaystyle\sum_{i=1}^n(x_i-\bar{x})^2=\dfrac{5.7895}{20-1}=0.3047$$

$$s=\sqrt{s^2}=0.552$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu=6$

$H_1:\mu\not=6$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.05, df=n-1=20-1$ degrees of freedom, and the critical value for a two-tailed test is $t_c = 2.093024.$

The rejection region for this two-tailed test is$R = \{t: |t| > 2.093024\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{5.395-6}{0.552/\sqrt{20}}=-4.9015$$

Since it is observed that $|t| = 4.902 > 2.093024=t_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for two-tailed $\alpha=0.05, df=19$ degrees of freedom, $t=-4.9015,$ is $p= 0.000099,$ and since $p= 0.000099<0.5=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 6, at the $\alpha = 0.05$ significance level.