An investigative bureau uses a laboratory method to match the lead in a bullet found at a crime scene with unexpended lead cartridges found in the possession of a suspect. The value of this evidence depends on the chance of a false positive positive that is the probability that the bureau finds a match given that the lead at the crime scene and the lead in the possession of the suspect are actually from two differant melts or sources. To estimate the false positive rate the bureau collected 1851 bullets that the agency was confident all came from differant melts. The using its established ctireria the bureau examined every possible pair of bullets and found 658 matches. Use this info to to compute the chance of a false positive.

Answers

Answer 1

Answer:

Answer:

Step-by-step explanation:

Given that, we have 1851 bullets that we KNOW are NOT MATCHES of one another. One by one they examine two bullets at a time.

So, there are 1851 bullets but each time we choose 2.

We have, N choose K = N! / K! (N-k)!

Here, N = 1851 and K = 2

Therefore, 1851 choose 2 = 1851! / 2! (1851-2)!

= 1851! / 2! * 1849!

= 1712175 Possible Combinations

Out of these 653 are false positive.

The chance of getting false positive is = 658 / 1712175

= 0.000384

= 0.0384 %

Therefore, The correct option is

The chance of false positive is 0.0384% Because this probability is sufficiently small (< or = 1%) There is high confidence in the agency's forensic evidence.

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Answer:

there isn't even a y here

Step-by-step explanation:

Did you not write the while question?

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Answer:

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Explanation:

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The amount of chlorine needed to treat a swimming pool is directly proportional to the volume of the poolWhat is the constant of proportionality for this relationship

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Paired t‐Test for Mean Comparison with Dependent Samples To study the effects of an advertising campaign at a supply chain, several stores are randomly selected with the following observed before‐ and after‐advertising monthly sales revenues: Store number 1 2 3 4 5
Old sales revenue (mil. $) 5.2 6.5 7.2 5.7 7.6
New sales revenue (mil. $) 6.4 7.8 6.8 6.5 8.2
Let μ₁ and μ₂ be the means of old and new sales revenues, both in millions of dollars per month.
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(b)Sketch and find the p‐value of the test. Would you reject H0 if α = 0.01?

Answers

Answer:

a) $t=(\bar d -0)/((s_d)/(√(n)))=(0.7 -0)/((0.678)/(√(5)))=2.308$

$p_v =P(t_((4))>2.308) =0.0411$

So the p values is lower than the significance level given 0.05, so then we can conclude that we reject the null hypothesis.

b) The p value is illustrated on the figure attached.

If we select $\alpha=0.01$ we see that $p_v >\alpha$ so then we have enough evidence to FAIL to reject the null hypothesis.

Step-by-step explanation:

Part a

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

Let put some notation

1=test value old , 2 = test value new

1: 5.2 6.5 7.2 5.7 7.6

2: 6.4 7.8 6.8 6.5 8.2

The system of hypothesis for this case are:

Null hypothesis: $\mu_2- \mu_1 \leq 0$

Alternative hypothesis: $\mu_2 -\mu_1 >0$

The first step is calculate the difference $d_i=y_i-x_i$ and we obtain this:

d: 1.2, 1.3, -0.4, 0.8, 0.6

The second step is calculate the mean difference

$\bar d= (\sum_(i=1)^n d_i)/(n)=0.7$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =(\sum_(i=1)^n (d_i -\bar d)^2)/(n-1) =0.678$

The 4 step is calculate the statistic given by :

$t=(\bar d -0)/((s_d)/(√(n)))=(0.7 -0)/((0.678)/(√(5)))=2.308$

The next step is calculate the degrees of freedom given by:

$df=n-1=5-1=4$

Now we can calculate the p value, since we have a right tailed test the p value is given by:

$p_v =P(t_((4))>2.308) =0.0411$

So the p values is lower than the significance level given 0.05, so then we can conclude that we reject the null hypothesis.

Part b

The p value is illustrated on the figure attached.

If we select $\alpha=0.01$ we see that $p_v >\alpha$ so then we have enough evidence to FAIL to reject the null hypothesis.

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Answer:

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Step-by-step explanation:

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