Business, 01.04.2020 20:27, AdoNice

Since many people have trouble programming their dvd players, an electronics company has developed what it hopes will be easier instructions. the goal is to have at least 9595% of customers succeed at being able to program their dvd players. the company tests the new system on 100100 people, 9292 of whom were successful. is this strong evidence that the new system fails to meet the company's goal? a student's test of this hypothesis is provided. how many mistakes can you find?

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answered: atran6955

$z=\frac{0.92-0.95}{\sqrt{\frac{0.95(1-0.95)}{100}}}=-1.376$

p_v =P(z

So the p value obtained was a very low 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 of interest is not significantly lower than 0.95, so then the program works since we fail to reject the null hypothesis.

Explanation:

Data given and notation

n=100 represent the random sample taken

X=92 represent the successful

$\hat p=\frac{92}{100}=0.92$ estimated proportion of successes

p_o=0.95 is the value that we want to test

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 true proportion is lower than 0.95 or no.:

Null hypothesis: $p\geq 0.95$

Alternative hypothesis: p < 0.95

When we conduct a proportion test we need to use the z statisitc, 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 required we can replace in formula (1) like this:

$z=\frac{0.92-0.95}{\sqrt{\frac{0.95(1-0.95)}{100}}}=-1.376$

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

answered: Guest

b is the correct answer

answered: Guest

Iam pretty sure a because if u are exiting you need to know the exit plan

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