Question

A coffee shop owner claims that more than 80% of coffee drinkers think that taste of a shops coffee is very important in determining where they purchase their coffee. In a random sample of 36 coffee drinkers, 31 think that the taste of a shops coffee is very important in determining where they purchase their coffee.
1. At α = 0.10, is there enough evidence to support the owners claim?

Answer 1

Answer:

$z=\frac{0.861 -0.8}{\sqrt{\frac{0.8(1-0.8)}{36}}}=0.915$  

$p_v =P(z>0.915)=0.18$  

If we compare the p value obtained and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% of significance the proportion of coffee drinkers  who  think that the taste of a shops coffee is very important in determining where they purchase their coffee is not significantly higher than 0.8 or 80%.

Step-by-step explanation:

Data given and notation

n=36 represent the random sample taken

X=31 represent the coffee drinkers  who  think that the taste of a shops coffee is very important in determining where they purchase their coffee.

$\hat p=\frac{31}{36}=0.861$ estimated proportion of coffee drinkers  who  think that the taste of a shops coffee is very important in determining where they purchase their coffee.

$p_o=0.8$ is the value that we want to test

$\alpha=0.1$ represent the significance level

Confidence=90% or 0.90

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 true proportion is higher than 0.8:  

Null hypothesis:$p\leq 0.8$  

Alternative hypothesis:$p > 0.8$  

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:  

$z=\frac{0.861 -0.8}{\sqrt{\frac{0.8(1-0.8)}{36}}}=0.915$  

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 $\alpha=0.1$. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

$p_v =P(z>0.915)=0.18$  

If we compare the p value obtained and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% of significance the proportion of coffee drinkers  who  think that the taste of a shops coffee is very important in determining where they purchase their coffee is not significantly higher than 0.8 or 80%.

Answer 2

Answer:

Step-by-step explanation:

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