Respuesta :
Answer:
(a) H₀: p₁ ≤ p₂ vs. Hₐ: p₁ > p₂.
(b) The point estimate of the proportion of wells drilled in 2005 that were dry is 0.202.
(c) The point estimate of the proportion of wells drilled in 2012 that were dry is 0.111.
(d) The p-value of the test is 0.017.
The wells drilled in 2005 were more likely to be dry than wells drilled in 2012.
Step-by-step explanation:
The data provided for the wells drilled in 2005 and 2012 are as follows:
2005 2012
Wells drilled 119 162
Dry Wells 24 18
(a)
The hypothesis to test whether the wells drilled in 2005 were more likely to be dry than wells drilled in 2012 are defined as follows:
H₀: The wells drilled in 2005 were not more likely to be dry than wells drilled in 2012, i.e. p₁ ≤ p₂.
Hₐ: The wells drilled in 2005 were more likely to be dry than wells drilled in 2012, i.e. p₁ > p₂.
(b)
A point estimate of a parameter (population) is a distinct value used for the estimation the parameter (population). For instance, the sample mean [tex]\bar x[/tex] is a point-estimate of the population mean μ.
Similarly the point estimate of population proportion p is, [tex]\hat p[/tex].
It is computed using the formula:
[tex]\hat p=\frac{X}{n}[/tex]
Compute the point estimate of the proportion of wells drilled in 2005 that were dry as follows:
[tex]\hat p_{1}=\frac{X_{1}}{n_{1}}=\frac{24}{119}=0.202[/tex]
Thus, the point estimate of the proportion of wells drilled in 2005 that were dry is 0.202.
(c)
Compute the point estimate of the proportion of wells drilled in 2012 that were dry as follows:
[tex]\hat p_{2}=\frac{X_{2}}{n_{2}}=\frac{18}{162}=0.111[/tex]
Thus, the point estimate of the proportion of wells drilled in 2012 that were dry is 0.111.
(d)
A z - test for two proportions will be used to perform the test.
The test statistic is defined as:
[tex]z=\frac{\hat p_{1}-\hat p_{2}}{\sqrt{\hat P(1-\hat P)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex]
Compute the value of standard error as follows:
[tex]SE=\sqrt{\hat P(1-\hat P)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}\\\hat P=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{24+18}{119+162}=0.15\\SE=\sqrt{0.15(1-0.15)(\frac{1}{119}+\frac{1}{162})}\\=0.043[/tex]
Compute the test statistic value as follows:
[tex]z=\frac{\hat p_{1}-\hat p_{2}}{\sqrt{\hat P(1-\hat P)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}=\frac{0.202-0.111}{0.043}=2.12[/tex]
Compute the p-value of the test as follows:
[tex]p-value=P(Z>2.12)=1-P(Z<2.12)=1-0.983=0.017[/tex]
*Use a z-table for the probability.
Thus, the p-value of the test is 0.017.
The significance level of the test is α = 0.025.
The p-value = 0.017 < α = 0.025.
As the p-value is less than the significance level the null hypothesis will be rejected at 2.5% level of significance.
Conclusion:
As the null hypothesis is rejected it can b concluded that the wells drilled in 2005 were more likely to be dry than wells drilled in 2012.
