Mathematics, 21.05.2020 02:02 kate5895

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

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Answer from: tornadowolfcosplay

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 = proportion of deaths when drivers were not wearing seat belts.

p_2 = proportion of deaths when drivers were wearing seat belts.

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 Two-sample z test for proportions;

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, the test statistics = $\frac{(0.011-0.002)-(0)}{\sqrt{\frac{0.011(1-0.011)}{2823}+\frac{0.002(1-0.002)}{7765} } }$

= 4.438

The value of z test statistics is 4.438.

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

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 we reject our null hypothesis.

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

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

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

= 0.00397

Answer from: Quest

step-by-step explanation:

Answer from: Quest

oof cheaters cheat because they’re lazy that’s why i cheat xd

: )

step-by-step explanation:

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