Answer:
The p-value of the test is 0.1515.
Step-by-step explanation:
The hypothesis for the test can be defined as follows:
<em>H</em>₀: The mean level of arsenic is 80 ppb, i.e. <em>μ</em> = 80.
<em>Hₐ</em>: The mean level of arsenic is greater than 80 ppb, i.e. <em>μ</em> > 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,
.
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](https://tex.z-dn.net/?f=%5Cbar%20x%3D%5Cfarc%7B57%2B64%2B70%2B82%2B84%2B123%7D%7B6%7D%3D80%5C%5C%5C%5Cs%3D%5Csqrt%7B%5Cfrac%7B1%7D%7B6-1%7D%5Ctimes%20%5B%2857-80%29%5E%7B2%7D%2B%2864-80%29%5E%7B2%7D%2B%2870-80%29%5E%7B2%7D%2B...%2B%28123-80%29%5E%7B2%7D%5D%7D%3D23.47)
Compute the test statistic value as follows:

Thus, the test statistic value is 1.148.
Compute the p-value of the test as follows:


*Use a t-table.
Thus, the p-value of the test is 0.1515.
Decision rule:
If the <em>p</em>-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.