Hello! Any help is appreciated (:

Test the hypothesis using the P value approach. Be sure the verify the requirements of the test.

Andreas93 [3]

Answer:

$p_v =2*P(z$

If we compare the p 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 to reject the null hypothesis, and we can said that at 5% of significance the proportion is not significantly different from 0.77.

Step-by-step explanation:

1) Data given and notation

n=500 represent the random sample taken

X=380 represent the number of people with some characteristic

$\hat p=\frac{380}{500}=0.76$ estimated proportion of adults that said that it is morally wrong to not report all income on tax returns

$p_o=0.76$ 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)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is 0.7 .:

Null hypothesis: $p=0.77$

Alternative hypothesis: $p \neq 0.77$

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 $p_o$ .

<em>Check for the assumptions that he sample must satisfy in order to apply the test </em>

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

$np_o =500*0.77=385>10$

$n(1-p_o)=384*(1-0.77)=115>10$

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.76-0.77}{\sqrt{\frac{0.77(1-0.77)}{500}}}=-0.531$

4) 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:

$p_v =2*P(z$

If we compare the p 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 to reject the null hypothesis, and we can said that at 5% of significance the proportion is not significantly different from 0.77.

6 0

4 years ago

A kennel is moving 160 dogs to a new facility each dog has its own crate the facility manager rents 17 trucks each truck holds 9

mina [271]

This will only be able to hold 153 dogs in total. He will need 1 additional truck to hold the last 7 dogs.

4 0

3 years ago

Help number 2 ASAP thank you

emmasim [6.3K]

Answer:

yes

Step-by-step explanation: