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spayn [35]
3 years ago
11

2. The National Safety Council routinely analyzes the benefit of seat belt use on driver safety. Their data showed that among 28

23 drivers not wearing seat belts, 31 died as a result of injuries, and among 7765 drivers wearing seat belts 16 were killed. Test the hypothesis (at 95% confidence) that there is no difference in the proportion of deaths between the 2 groups. What do you conclude? Calculate the margin of error (E) at 95%.
Mathematics
1 answer:
JulijaS [17]3 years ago
5 0

Answer:

We conclude that there is difference in the proportion of deaths between the 2 groups.

Step-by-step explanation:

We are given that among 2823 drivers not wearing seat belts, 31 died as a result of injuries, and among 7765 drivers wearing seat belts 16 were killed.

Let p_1 = <u><em>proportion of deaths when drivers were not wearing seat belts.</em></u>

p_2 = <u><em>proportion of deaths when drivers were wearing seat belts.</em></u>

So, Null Hypothesis, H_0 : p_1=p_2      {means that there is no difference in the proportion of deaths between the 2 groups}

Alternate Hypothesis, H_A : p_1\neq p_2     {means that there is difference in the proportion of deaths between the 2 groups}

The test statistics that would be used here <u>Two-sample z test for proportions;</u>

                          T.S. =  \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }  ~ N(0,1)

where, \hat p_1 = sample proportion of deaths when drivers were not wearing seat belts = \frac{31}{2823} = 0.011

\hat p_2 = sample proportion of deaths when drivers were wearing seat belts = \frac{16}{7765} = 0.002

n_1 = sample of drivers not wearing seat belts = 2823

n_2 = sample of drivers wearing seat belts = 7765

So, <u><em>the test statistics</em></u>  =  \frac{(0.011-0.002)-(0)}{\sqrt{\frac{0.011(1-0.011)}{2823}+\frac{0.002(1-0.002)}{7765} } }

                                       =  4.438

The value of z test statistics is 4.438.

<u>Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.</u>

Since our test statistic doesn't lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that there is difference in the proportion of deaths between the 2 groups.

Also, <u>Margin of error</u> (E) =  1.96 \times \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} }

                                        =  1.96 \times \sqrt{\frac{0.011(1-0.011)}{2823}+\frac{0.002(1-0.002)}{7765} }

                                        =  <u>0.00397</u>

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Step-by-step explanation:

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

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Let X the random variable of interest, on this case we now that:  

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The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

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Part a

We want this probability:

P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)

And we can use the probability mass function and we got:

P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115  

P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576  

P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369  

And adding we got:

P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061

Part b

We want this probability:

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And using the probability mass function we got:

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Part c

We want this probability:

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We can use the complement rule and we got:

P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]

P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115  

P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576  

P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369

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And replacing we got:

P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886

Part d

The expected value is given by:

E(X) = np

And replacing we got:

E(X) = 20*0.2= 4

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