Question

When testing gas pumps for​ accuracy, fuel-quality enforcement specialists tested pumps and found that 1294 of them were not pumping accurately​ (within 3.3 oz when 5 gal is​ pumped), and 5705 pumps were accurate. Use a 0.01 significance level to test the claim of an industry representative that less than​ 20% of the pumps are inaccurate. Use the​ P-value method and use the normal distribution as an approximation to the binomial distribution. Identify the null hypothesis and alternative hypothesis.A. H0​: p=0.2H1​:p>0.2B. H0​: p=0.2H1​: ≠0.2C. H0​: p≠0.2H1​: p=0.2D.H0​: p>0.2H1​: p=0.2E.H0​: p=0.2H1​: p<0.2F. H0​: P<0.2H1​: P=0.2The Test static is z=The P-Value is=Because the P-Value is (greater than/less than) the significance level (fail to reject/reject) the null hypothesis. There is (sufficient/insufficient) evidence support the claim that less than 20% of the pumps are inaccurate.

Answer 1

Answer:

E.H0​: p=0.2H1​: p<0.2

$z=\frac{0.185 -0.2}{\sqrt{\frac{0.2(1-0.2)}{6999}}}=-3.137$  

$p_v =P(z$  

Because the P-Value is (less than) the significance level (reject) the null hypothesis. There is (sufficient) evidence support the claim that less than 20% of the pumps are inaccurate.

Step-by-step explanation:

1) Data given and notation

n=1294+5705=6999 represent the random sample taken

X=1294 represent the number of pumps that were not pumping accurately

$\hat p=\frac{1294}{6999}=0.185$ estimated proportion of pumps that were not pumping accurately

$p_o=0.2$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that that less than​ 20% of the pumps are inaccurate:  

Null hypothesis:$p\geq 0.2$  

Alternative hypothesis:$p < 0.2$  

E.H0​: p=0.2H1​: p<0.2

When we conduct a proportion test we need to use the z statistic, and the is given by:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

$z=\frac{0.185 -0.2}{\sqrt{\frac{0.2(1-0.2)}{6999}}}=-3.137$  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided $\alpha=0.01$. The next step would be calculate the p value for this test.  

Since is a one left tailed test the p value would be:  

$p_v =P(z$  

Because the P-Value is (less than) the significance level (reject) the null hypothesis. There is (sufficient) evidence support the claim that less than 20% of the pumps are inaccurate.