Question

A restaurant chain is measuring the levels of arsenic in chicken from its suppliers. The question is whether there is evidence that the mean level of arsenic is greater than 80 ppb, so we are testing vs , where represents the average level of arsenic in all chicken from a certain supplier. It takes money and time to test for arsenic so samples are often small. A sample of chickens from one supplier is tested, and the resulting sample mean is . Subtracting 11 from the sample data to move the mean down to the null mean of results in the following data: .
57, 64, 70, 82, 84, 123
Use StatKey or other technology to create the randomization distribution for this test. Find the p-value.

Answer 1

Answer:

The p-value of the test is 0.1515.

Step-by-step explanation:

The hypothesis for the test can be defined as follows:

H₀: The mean level of arsenic is 80 ppb, i.e. μ = 80.

Hₐ: The mean level of arsenic is greater than 80 ppb, i.e. μ > 80.

As the population standard deviation is not known we will use a t-test for single mean.

It is provided that the sample mean was, $\bar X=91$.

The adjusted sample provided is:

S = {57, 64, 70, 82, 84, 123}

Compute the sample standard deviation as follows:

$\bar x=\farc{57+64+70+82+84+123}{6}=80\\\\s=\sqrt{\frac{1}{6-1}\times [(57-80)^{2}+(64-80)^{2}+(70-80)^{2}+...+(123-80)^{2}]}=23.47$

Compute the test statistic value as follows:

 $t=\frac{\bar X-\mu}{\s/\sqrt{n}}=\frac{91-80}{23.47/\sqrt{6}}=1.148$

Thus, the test statistic value is 1.148.

Compute the p-value of the test as follows:

 $p-\text{value}=P(t_{n-1}$

                $=P(t_{6-1}$

*Use a t-table.

Thus, the p-value of the test is 0.1515.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

p-value = 0.1515 > α = 0.05

The null hypothesis will not be rejected at 5% level of significance.

Thus, concluding that the mean level of arsenic in chicken from the suppliers is 80 ppb.