Mathematics, 10.03.2020 19:04 emzaa

The null and alternative hypotheses for a population proportion, as well as the sample results, are given. Use StatKey or other technology to generate a randomization distribution and calculate a p-value. StatKey tip: Use "Test for a Single Proportion" and then "Edit Data" to enter the sample information.

Hypotheses: H0: p = 0.5 vs Ha: p<⁢0.5 ;
Sample data: p^ = 38/100 = 0.38 with n = 100.

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Answer from: tyrique86

$z=\frac{0.38 -0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}=-2.4$

p_v =P(z

And we can use the following excel code to find it:

"=NORM.DIST(-2.4,0,1,TRUE)"

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

$\hat p=0.38$ estimated proportion of interesst

p_o=0.5 is the value that we want to test

$\alpha$ represent the significance level

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 proportion is lower than 0.5:

Null hypothesis: $p \geq 0.5$

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:

$z=\frac{0.38 -0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}=-2.4$

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 next step would be calculate the p value for this test.

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

p_v =P(z

And we can use the following excel code to find it:

"=NORM.DIST(-2.4,0,1,TRUE)"

Answer from: Quest

answer: 1,380

step-by-step explanation:

1: you turn %15 into a decimal by moving the decimal point two places to the left.

2: and then you multiply .15 by 1,200 , which will get you 180.00

3: from there you will add 180 to 1,200

4: which will get you 1,380

hope it *im kind of slow so idk if my answer is right*

Answer from: Quest

.19875994384

step-by-step explanation:

1699/ 8548

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