Question

A potato-chip producer has just received a truckload of potatoes from their main supplier. If the producer finds convincing evidence that more than 8% of the potatoes in the shipment have blemishes, the truck will be sent away to get another load from the supplier. A supervisor selects a random sample of 500 potatoes from the truck. An inspection reveals that 47 of the potatoes have blemishes. Carry out a significance test at the α = 0.05 significance level. What should the producer conclude?

Answer 1

Answer:

$z=\frac{0.094 -0.08}{\sqrt{\frac{0.08(1-0.08)}{500}}}=1.154$  

$p_v =P(z>1.154)=0.124$  

So the p value obtained was a very high value 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 potatoes that have blemishes is not significantly higher than 0.08 or 8%

Step-by-step explanation:

Data given and notation

n=500 represent the random sample taken

X=47 represent the potatoes that have blemishes

$\hat p=\frac{47}{500}=0.094$ estimated proportion of potatoes that have blemishes

$p_o=0.08$ 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 the true proportion i higher than 8% or 0.08.:  

Null hypothesis:$p \leq 0.08$  

Alternative hypothesis:$p > 0.08$  

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:  

$z=\frac{0.094 -0.08}{\sqrt{\frac{0.08(1-0.08)}{500}}}=1.154$  

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 right tailed test the p value would be:  

$p_v =P(z>1.154)=0.124$  

So the p value obtained was a very high value 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 potatoes that have blemishes is not significantly higher than 0.08 or 8%