Question

g we found that 15 customers from a random sample of 60 would be willing to buy a software upgrade that costs 100. If the upgrade is to be profitable you will need to sell it to more than 20% of your customers. Do the sample data give good evidence that more than 20% are willing to buy

Answer 1

`z=\frac{0.25 -0.2}{\sqrt{(0.2(1-0.2))/(100)}}=1.25`

`p_v =P(Z>1.25)=0.106`

So the p value obtained was a very high value and using the significance level assumed
`\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 then we don't have enough evidence to conclude that the true proportion is higher than 0.2 .

Explanation:

1) Data given and notation

n=60 represent the random sample taken

X=15 represent the customers who would be willing to buy a software upgrade that costs 100

`\hat p=(15)/(60)=0.25` estimated proportion of customers who would be willing to buy a software upgrade that costs 100

`p_o=0.2` 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)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportions is higher than 20%:

Null hypothesis:
`p\leq 0.2`

Alternative hypothesis:
`p > 0.2`

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`.

3) Calculate the statistic

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

`z=\frac{0.25 -0.2}{\sqrt{(0.2(1-0.2))/(100)}}=1.25`

4) 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 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>1.25)=0.106`

So the p value obtained was a very high value and using the significance level assumed
`\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 then we don't have enough evidence to conclude that the true proportion is higher than 0.2 .