Mathematics, 29.01.2020 05:50 lilkobe6982

Acoffee 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?

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Answer from: naruto63

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

p_v =P(z0.915)=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(z0.915)=0.18

Answer from: Quest

false

step-by-step explanation:

Answer from: Quest

step-by-step explanation:

because that's the answer

The stem-and-leaf plot above lists the number of ingredients found in different industrial cleaners.

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