Question

Suppose that 500 parts are tested in manufacturing and 10 are rejected.

Answer 1

Answer:

a) $z=\frac{0.02 -0.03}{\sqrt{\frac{0.03(1-0.03)}{500}}}=-1.31$  

$p_v =P(Z$  

If we compare the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of rejected items is less than 0.03.  

b) We can use a 95 percent upper confidence interval.

On this case we want a interval on this form : $(-\infty,\hat p +z_{\alpha}\sqrt{\frac{\hat p (1-\hat p)}{n}})$

So the critical value would be on this case $Z_{\alpha}=1.64$ and we can use the following excel code to find it: "=NORM.INV(1-0.05,0,1)"

We found the upper limit like this:

$0.02+1.64\sqrt{\frac{0.02 (1-0.02)}{500}}=0.03026$

And the interval would be: $(-\infty,0.03026)$

And since our value (0.02) is contained in the interval We fail to reject the hypothesis that p=0.03

Step-by-step explanation:

Part a

Data given and notation  

n=500 represent the random sample taken

X=10 represent the number of objects rejected

$\hat p=\frac{10}{500}=0.02$ estimated proportion of objects rejected

$p_o=0.03$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that 70% of adults say that it is morally wrong to not report all income on tax returns.:  

Null hypothesis:$p=0.03$  

Alternative hypothesis:$p < 0.03$  

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 $p_o$.

Calculate the statistic  

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

$z=\frac{0.02 -0.03}{\sqrt{\frac{0.03(1-0.03)}{500}}}=-1.31$  

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 $\alpha=0.05$. The next step would be calculate the p value for this test.  

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

$p_v =P(Z$  

If we compare the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of rejected items is less than 0.03.  

Part b

We can use a 95 percent upper confidence interval.

On this case we want a interval on this form : $(-\infty,\hat p +z_{\alpha}\sqrt{\frac{\hat p (1-\hat p)}{n}})$

So the critical value would be on this case $Z_{\alpha}=1.64$ and we can use the following excel code to find it: "=NORM.INV(1-0.05,0,1)"

We found the upper limit like this:

$0.02+1.64\sqrt{\frac{0.02 (1-0.02)}{500}}=0.03026$

And the interval would be: $(-\infty,0.03026)$

And since our value (0.02) is contained in the interval We fail to reject the hypothesis that p=0.03