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?

So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] 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

[tex]\hat p=\frac{47}{500}=0.094[/tex] estimated proportion of potatoes that have blemishes

[tex]p_o=0.08[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

[tex]p_v[/tex] 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:[tex]p \leq 0.08[/tex]

Alternative hypothesis:[tex]p > 0.08[/tex]

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

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)

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

Calculate the statistic

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

[tex]z=\frac{0.094 -0.08}{\sqrt{\frac{0.08(1-0.08)}{500}}}=1.154[/tex]

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 [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

[tex]p_v =P(z>1.154)=0.124[/tex]

So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] 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%