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

Answer:

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

$p_v =P(z>3.913)=0.000046$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ 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%

Step-by-step 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

$\hat p=\frac{135}{500}=0.27$ 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:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (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 required we can replace in formula (1) like this:  

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

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 $p_v$ 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%