Respuesta :
Part A:
For a hypothesis test, the significant level is given by α. From the question, we were told that α = 0.01.
Therefore, the significant level is 0.01.
Our previous knowledge is that a mean arsenic level of μ = 8.0 parts per billion (ppb) is considered safe for agricultural use, we know want to know from a sample of 36 tests whether the mean level of arsenic in a particular well is less than 8 ppb.
Therefore, the null and the alternative hypothesis are:
[tex]H_0:\mu=8\ pbb \\ \\ H_a:\mu\ \textless \ 8\ pbb[/tex]
Part B:
The sampling distribution to be used is the student's t, since the sample size is large and σ is unknown.
The test statistics is given by:
[tex]t= \frac{\bar{x}-\mu}{s/\sqrt{n}} \\ \\ = \frac{6.8-8}{2.9/\sqrt{36}} \\ \\ = \frac{-1.2}{2.9/6} = \frac{-1.2}{0.4833} \\ \\ =-2.483[/tex]
Part C:
The p-value for a t-distribution test statistics of -2.483 with a degree of freedom of 35 is given by 0.0090
Thus 0.005 < p-value < 0.025
Part D:
Since the p-value is less that the significant level, we reject the null hypothesis and conclude the data are statistically significant.
i.e.
at the α = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant.
Part E:
Therefore, we conclude that there is sufficient evidence at the 0.01 level to conclude that the mean level of arsenic in the well is less than 8 ppb.
For a hypothesis test, the significant level is given by α. From the question, we were told that α = 0.01.
Therefore, the significant level is 0.01.
Our previous knowledge is that a mean arsenic level of μ = 8.0 parts per billion (ppb) is considered safe for agricultural use, we know want to know from a sample of 36 tests whether the mean level of arsenic in a particular well is less than 8 ppb.
Therefore, the null and the alternative hypothesis are:
[tex]H_0:\mu=8\ pbb \\ \\ H_a:\mu\ \textless \ 8\ pbb[/tex]
Part B:
The sampling distribution to be used is the student's t, since the sample size is large and σ is unknown.
The test statistics is given by:
[tex]t= \frac{\bar{x}-\mu}{s/\sqrt{n}} \\ \\ = \frac{6.8-8}{2.9/\sqrt{36}} \\ \\ = \frac{-1.2}{2.9/6} = \frac{-1.2}{0.4833} \\ \\ =-2.483[/tex]
Part C:
The p-value for a t-distribution test statistics of -2.483 with a degree of freedom of 35 is given by 0.0090
Thus 0.005 < p-value < 0.025
Part D:
Since the p-value is less that the significant level, we reject the null hypothesis and conclude the data are statistically significant.
i.e.
at the α = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant.
Part E:
Therefore, we conclude that there is sufficient evidence at the 0.01 level to conclude that the mean level of arsenic in the well is less than 8 ppb.