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madreJ [45]
3 years ago
11

Consider a drug testing company that provides a test for marijuana usage. Among 317 tested​ subjects, results from 25 subjects w

ere wrong​ (either a false positive or a false​negative). Use a 0.05 significance level to test the claim that less than 10 percent of the test results are wrong. Identify the null and alternative hypotheses for this test. Choose the correct answer below. A. Upper H 0H0​: pequals=0.10.1 Upper H 1H1​: pless than<0.10.1 B. Upper H 0H0​: pless than<0.10.1 Upper H 1H1​: pequals=0.10.1 C. Upper H 0H0​: pequals=0.10.1 Upper H 1H1​: pgreater than>0.10.1 D. Upper H 0H0​: pequals=0.10.1 Upper H 1H1​: pnot equals≠0.10.1 Identify the test statistic for this hypothesis test. The test statistic for this hypothesis test is nothing. ​(Round to two decimal places as​ needed.) Identify the​P-value for this hypothesis test. The​ P-value for this hypothesis test is nothing. ​(Round to three decimal places as​ needed.) Identify the conclusion for this hypothesis test. A. Fail to rejectFail to reject Upper H 0H0. There is notis not sufficient evidence to warrant support of the claim that less than 1010 percent of the test results are wrong. B. RejectReject Upper H 0H0. There is notis not sufficient evidence to warrant support of the claim that less than 1010 percent of the test results are wrong. C. RejectReject Upper H 0H0. There isis sufficient evidence to warrant support of the claim that less than 1010 percent of the test results are wrong. D. Fail to rejectFail to reject Upper H 0H0. There isis sufficient evidence to warrant support of the claim that less than 1010 percent of the test results are wrong.
Mathematics
1 answer:
svet-max [94.6K]3 years ago
5 0

Answer:

We conclude that we fail to reject H_0 as there is not sufficient evidence to warrant support of the claim that less than 10 percent of the test results are wrong.

Step-by-step explanation:

We are given that among 317 tested​ subjects, results from 25 subjects were wrong.

We have to test the claim that less than 10 percent of the test results are wrong.

<u><em>Let p = proportion of subjects that were wrong.</em></u>

So, Null Hypothesis, H_0 : p = 10%     {means that 10 percent of the test results are wrong}

Alternate Hypothesis, H_A : p < 10%     {means that less than 10 percent of the test results are wrong}

The test statistics that would be used here <u>One-sample z proportion</u> <u>statistics</u>;

                      T.S. =  \frac{\hat p-p}{\sqrt{\frac{\hat p (1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of test results that were wrong = \frac{25}{317} = 0.08

           n = sample of tested subjects = 317

So, <u><em>test statistics</em></u>  =  \frac{0.08-0.10}{\sqrt{\frac{0.08 (1-0.08)}{317} } }

                              =  -1.31

The value of z test statistics is -1.31.

Now, the P-value of the test statistics is given by the following formula;

                P-value = P(Z < -1.31) = 1 - P(Z \leq 1.31)

                              = 1 - 0.9049 = <u>0.095</u>

<u><em>Now, at 0.05 significance level the z table gives critical value of -1.645 for left-tailed test.</em></u><em> Since our test statistics is more than the critical value of z as -1.31 > -1.645, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which </em><em><u>we fail to reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that we fail to reject H_0 as there is not sufficient evidence to warrant support of the claim that less than 10 percent of the test results are wrong.

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melamori03 [73]

Answer:

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Step-by-step explanation:

Given:

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4 0
2 years ago
In the past, the average age of employees of a large corporation has been 40 years. Recently, the company has been hiring older
Viktor [21]

Answer:

p_v =P(t_{(63)}>2.5)=0.0075  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean age is significantly higher than 45 years at 5% of significance.  

Step-by-step explanation:

1) Data given and notation  

\bar X=45 represent the mean height for the sample  

s=16 represent the sample standard deviation for the sample  

n=64 sample size  

\mu_o =40 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean age is higher than 40 years, the system of hypothesis would be:  

Null hypothesis:\mu \leq 40  

Alternative hypothesis:\mu > 40  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{45-40}{\frac{16}{\sqrt{64}}}=2.5    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=64-1=63  

Since is a one right tailed test the p value would be:  

p_v =P(t_{(63)}>2.5)=0.0075  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean age is significantly higher than 45 years at 5% of significance.  

6 0
3 years ago
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Volgvan
2.3 is smaller than 2.6: we can see it just by looking at the digits from left to right.

now, where does 2\frac{4}{5} stand?

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\frac{4*2}{5*2}=  \frac{8}{10}

do that means that 2\frac{4}{5}=2.8

which is bigger than 2.6

so the final order is

2.3 2.6 2\frac{4}{5}




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