Question

A simple random sample of​ front-seat occupants involved in car crashes is obtained. Among 2902 occupants not wearing seat​ belts, 30 were killed. Among 7866 occupants wearing seat​ belts, 20 were killed. Use a 0.05 significance level to test the claim that seat belts are effective in reducing fatalities. Complete parts​ (a) through​ (c) below. a. Test the claim using a hypothesis test. Consider the first sample to be the sample of occupants not wearing seat belts and the second sample to be the sample of occupants wearing seat belts. What are the null and alternative hypotheses for the hypothesis? test?

Answer 1

Null hypothesis:
`p_(SB) \geq p_(NSB)`

Alternative hypothesis:
`p_(SB) < \mu_(NSB)`

The p value is a very low value and using the significance given
`\alpha` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of people death using seat belts is significant lower than the proportion of deaths no using seat belts .

Explanation:

1) Data given and notation

`X_(NSB)=30` represent the number of people killed not using seat belts

`X_(SB)=20` represent the number of people killed using seat belts

`n_(NSB)=2902` sample of people not wearing seat belts

`n_(SB)=7866` sample of people wearing seat belts

`p_(NSB)=(30)/(2902)=0.0103` represent the proportion of people killed not using seat belts

`p_(SB)=(20)/(7866)=0.00254` represent the proportion of people killed using seat belts

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that seat belts are effective in reducing fatalities (If using seat belts reduce the proportion of deaths we need to see that the proportion of death using seat belts is lower than not using seat belts) , the system of hypothesis would be:

Null hypothesis:
`p_(SB) \geq p_(NSB)`

Alternative hypothesis:
`p_(SB) < \mu_(NSB)`

We need to apply a z test to compare proportions, and the statistic is given by:

`z=\frac{p_(SB)-p_(NSB)}{\sqrt{\hat p (1-\hat p)((1)/(n_(SB))+(1)/(n_(NSB)))}}` (1)

Where
`\hat p=(X_(SB)+X_(NSB))/(n_(SB)+n_(NSB))=(20+30)/(7866+2902)=0.00464`

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:

`z=\frac{0.00254-0.0103}{\sqrt{0.00464(1-0.00464)((1)/(7866)+(1)/(2902))}}=-5.26`

4) Statistical decision

Using the significance level provided
`\alpha=0.05`, the next step would be calculate the p value for this test.

Since is a one side lower test the p value would be:

`p_v =P(Z<-5.26)=7.2x10^(-8)`

So the p value is a very low value and using the significance given
`\alpha` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of people death using seat belts is significant lower than the proportion of deaths no using seat belts .