Question

A study by Consumer Reports showed that 64% of supermarket shoppers believe supermarket brands to be as good as national name brands. To investigate whether this result applies to its own product, the manufacturer of a national name-brand ketchup asked a sample of shoppers whether they believed that supermarket ketchup was as good as the national brand ketchup. A random sample of 350 shoppers showed that 217 thought the supermarket brand was as good as the national brand. Should we conduct a one or two tail test in this situation?

Answer 1

Answer:

$z=\frac{0.62 -0.64}{\sqrt{\frac{0.64(1-0.64)}{350}}}=-0.780$  

$p_v =2*P(z$  

So the p value obtained was a very high 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 FAIL reject the null hypothesis.

Step-by-step explanation:

She need to conduct a two tailed hypothesis test since she want to check if the standar dof 0.64 for the proportion is satisfied.

Data given and notation

n=350 represent the random sample taken

X=217 represent the shoppers who thought the supermarket brand was as good as the national brand

$\hat p=\frac{217}{350}=0.62$ estimated proportion of shoppers who thought the supermarket brand was as good as the national brand

$p_o=0.64$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95 (assumed)

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 shoppers who thought the supermarket brand was as good as the national brand is 0.64.:  

Null hypothesis:$p=0.64$  

Alternative hypothesis:$p \neq 0.64$  

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:  

$z=\frac{0.62 -0.64}{\sqrt{\frac{0.64(1-0.64)}{350}}}=-0.780$  

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 bilateral test the p value would be:  

$p_v =2*P(z$  

So the p value obtained was a very high 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 FAIL reject the null hypothesis.