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

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

`p_v =P(z<-0.913)=0.181`

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

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=(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{(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{(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<-0.913)=0.181`

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