Mathematics, 18.12.2019 18:31 dmarte11092001

We are conducting a hypothesis test to determine if fewer than 80% of st 311 hybrid students complete all modules before class. our hypotheses are h0 : p = 0.8 and ha : p < 0.8. we looked at 110 randomly selected students and found that 97 of these students had completed the modules before class. what is the appropriate conclusion for this test?

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Answer from: jazzhands16

$z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124$

Null hypothesis: $p\geq 0.8$

Alternative hypothesis: p < 0.8

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

p_v =P(Z

So the p value obtained was a very high value and using the significance level assumed $\alpha=0.05$ we have $p_v\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis: $p\leq 0.8$

Alternative hypothesis: p 0.8

p_v =P(Z2.124)=0.013

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have p_v so we can conclude that we have enough evidence to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation

n=110 represent the random sample taken

X=97 represent the students who completed the modules before class

$\hat p=\frac{97}{110}=0.882$ estimated proportion of students who completed the modules before class

p_o=0.8 is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v{/tex} represent the p value (variable of interest) 2) Concepts and formulas to use We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.8 or 80%: Null hypothesis:[tex]p\geq 0.8$

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 .

3) Calculate the statistic

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

$z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124$

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 assumed $\alpha=0.05$ . 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

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis: $p\leq 0.8$

Alternative hypothesis: p 0.8

p_v =P(Z2.124)=0.013

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have p_v so we can conclude that we have enough evidence to reject the null hypothesis.

Answer from: Quest

answer: b

step-by-step explanation:

How is the interquartile range calculated?

Answer from: Quest

this is not the answer this is just a question. is this middle school math?

step-by-step explanation:

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