Question

A large manufacturer that sells consumer products on-line wishes to publicize its customer satisfaction in an advertisement. Specifically, it wants to state that over 90% of the manufacturer's customers would tell a friend to buy a product from the manufacturer. The manufacturer selects a random sample of 400 customers from its database, contacts them via email and asks them the question "Would you tell a friend to buy a product from us?" 372 say Yes, and 28 say No. Is thos enough evidence for this manufacturer to state that more than 90% of its costumers would tell their friend to buy a product from the manufacturer?

Answer 1

Answer:

$z=\frac{0.93 -0.9}{\sqrt{\frac{0.9(1-0.9)}{400}}}=2$  

$p_v =P(z>2)=0.0228$  

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who say yes is significantly higher than 0.9 or 90%.

Step-by-step explanation:

Data given and notation

n=400 represent the random sample taken

X=372 represent the number of people who say yes

$\hat p=\frac{372}{400}=0.93$ estimated proportion of people who say yes

$p_o=0.9$ is the value that we want to test

$\alpha$ represent the significance level

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 that more than 90% of its costumers would tell their friend to buy a product from the manufacturer.:  

Null hypothesis:$p\leq 0.9$  

Alternative hypothesis:$p > 0.9$  

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:  

$z=\frac{0.93 -0.9}{\sqrt{\frac{0.9(1-0.9)}{400}}}=2$  

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

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

$p_v =P(z>2)=0.0228$  

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who say yes is significantly higher than 0.9 or 90%.