Question

An e-commerce research company claims that 60% or more graduate students have bought merchandise on-line at their site. A consumer group is suspicious of the claim and thinks that the proportion is lower than 60%. A random sample of 80 graduate students show that only 44 students have ever done so. Is there enough evidence to show that the true proportion is lower than 60%

Answer 1

Answer:

$z=\frac{0.55 -0.6}{\sqrt{\frac{0.6(1-0.6)}{80}}}=-0.913$  

$p_v =P(z$  

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ 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 graduate students show that only 44 students have ever done so is not significantly lower than 0.6

Step-by-step explanation:

Data given and notation

n=80 represent the random sample taken

X=44 represent the students that have bought merchandise on-line at their site

$\hat p=\frac{44}{80}=0.55$ estimated proportion of graduate students show that only 44 students have ever done so

$p_o=0.6$ 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 the true proportion of interest is lower than 0.6 or 60%, so then the system of hypothesis are.:  

Null hypothesis:$p \geq 0.6$  

Alternative hypothesis:$p < 0.6$  

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:  

$z=\frac{0.55 -0.6}{\sqrt{\frac{0.6(1-0.6)}{80}}}=-0.913$  

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 left tailed test the p value would be:  

$p_v =P(z$  

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ 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 graduate students show that only 44 students have ever done so is not significantly lower than 0.6