Question

A cellphone provider has the business objective of wanting to estimate the proportion of subscribers who would upgrade to a new cellphone with improved features if it were made available at a substantially reduced cost. Data are collected from a random sample of 500 subscribers. The results indicate that 135 of the subscribers would upgrade to a new cellphone at a reduced cost. At the 0.05 level of significance, is there evidence that more than 20% of the customers would upgrade to a new cellphone at a reduced cost

Answer 1

`z=\frac{0.27 -0.2}{\sqrt{(0.2(1-0.2))/(500)}}=3.913`

`p_v =P(z>3.913)=0.000046`

So the p value obtained was a very low value and using the significance level given
`\alpha=0.05` we have
`p_v<\alpha` 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 subscribers that would upgrade to a new cellphone at a reduced cost is significantly higher than 0.2 or 20%

Explanation:

Data given and notation

n=500 represent the random sample taken

X=135 represent the subscribers that would upgrade to a new cellphone at a reduced cost

`\hat p=(135)/(500)=0.27` estimated proportion of subscribers that would upgrade to a new cellphone at a reduced cost

`p_o=0.2` is the value that we want to test

`\alpha=0.05` represent the significance level

Confidence=95% or 0.95

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 true proportion is higher than 0.2 or not.:

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`.

Calculate the statistic

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

`z=\frac{0.27 -0.2}{\sqrt{(0.2(1-0.2))/(500)}}=3.913`

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 provided
`\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>3.913)=0.000046`

So the p value obtained was a very low value and using the significance level given
`\alpha=0.05` we have
`p_v<\alpha` 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 subscribers that would upgrade to a new cellphone at a reduced cost is significantly higher than 0.2 or 20%