Question

Each of 40 subjects tastes two unmarked cups of coffee and says which he or she prefers. One cup in each pair contains instant coffee, the other fresh-brewed coffee. 30 of the subjects prefer the fresh-brewed coffee. Take p to be the proportion of the population who would prefer fresh-brewed coffee in a blind tasting. Test the claim that a majority of people prefer the taste of fresh-brewed coffee.

H0:P=__,HA:P>__
The test statistic is__ (2 decimals)
The p-value is__(4 decimals)
Therefore, at the ? = 0.05 level we can conclude that The data (does/ does not) provides statistical evidence that the majority of tasters (do/do not)prefer the fresh-brewed coffee.

Answer 1

Answer:

Null hypothesis:$p = 0.5$  

Alternative hypothesis:$p > 0.5$

$z=\frac{0.75 -0.5}{\sqrt{\frac{0.5(1-0.5)}{40}}}=3.16$  

$p_v =P(z>3.16)=0.00079$  

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 the subjects prefer the fresh-brewed coffee is higher than 0.5 (majority)

The data (does) provides statistical evidence that the majority of tasters (do) prefer the fresh-brewed coffee.

Step-by-step explanation:

Data given and notation

n=40 represent the random sample taken

X=30 represent the subjects prefer the fresh-brewed coffee

$\hat p=\frac{30}{40}=0.75$ estimated proportion of the subjects prefer the fresh-brewed coffee

$p_o=0.5$ 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 the majoriy of subjects prefer the fresh-brewed coffee.:  

Null hypothesis:$p = 0.5$  

Alternative hypothesis:$p > 0.5$  

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:  

$z=\frac{0.75 -0.5}{\sqrt{\frac{0.5(1-0.5)}{40}}}=3.16$  

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:  

$p_v =P(z>3.16)=0.00079$  

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 the subjects prefer the fresh-brewed coffee is higher than 0.5 (majority)

The data (does) provides statistical evidence that the majority of tasters (do) prefer the fresh-brewed coffee.