The weights of four randomly and independently selected bags of tomatoes labeled 5 pounds were found to be 5.2, 4.9, 5.2, and 5. Assume Normality.A. Using a two-sided alternative hypothesis, with a significance level of 0.05. B.find the test statistics and p value .

Respuesta :

The Solution:

Step 1:

We shall find the mean and standard deviation.

[tex]\text{ Mean =}\frac{5.2+4.9+5.2+5}{4}=\frac{20.3}{4}=5.075[/tex][tex]\begin{gathered} \text{ Standard deviation =}\sigma=\sqrt[]{\frac{(4.9-5.075)^2+2(5.2-5.075)^2+(5-5.075)^2}{4}} \\ \\ =\text{ }\sqrt[]{\frac{(0.0306+0.03125+0.005625}{4}}=\text{ }\sqrt[]{0.01686875}\text{ =0.12988} \end{gathered}[/tex][tex]Z=\frac{5-5.075}{\frac{0.12988}{\sqrt[]{4}}}=-1.15491[/tex]

From the Zscore statistics table, the p-value is

[tex]\begin{gathered} p=0.2485 \\ \alpha=0.05\text{ (given)} \end{gathered}[/tex]

Since:

[tex]\begin{gathered} p=0.2485>\alpha,\text{ we conclude that the test fail to reject the null hypothesis.} \\ \end{gathered}[/tex]