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just olya [345]
3 years ago
9

Forty percent of prison inmates were unemployed when they entered prison. If 5 inmates are randomly selected. Find these possibi

lities.
a. Exactly 3 were unemployed.
b. At most 4 were unemployed.
c. At least 3 were unemployed.
d. Fewer than 2 were unemployed.
Mathematics
2 answers:
melamori03 [73]3 years ago
8 0
D. Fewer than 1 were unemployed
irinina [24]3 years ago
6 0
D, I think because 40 % of 100 is 4 then you divide them into in half and get every 2/5 inmates were unemployed before they entered prison
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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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