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

Null hypothesis:
`p = 0.5`

Alternative hypothesis:
`p > 0.5`

`z=\frac{0.75 -0.5}{\sqrt{(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.

Explanation:

Data given and notation

n=40 represent the random sample taken

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

`\hat p=(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{(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{(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.