Question

Motor vehicle accidents are rare events, but it is always better to have fewer. Vehicle speed may be related to the number of accidents, and in general it is believed that slower speeds will lead to fewer accidents. On a stretch of highway 42, the average number of accidents per day was 0.23. A month ago, the speed limit was changed from 60 miles per hour to 50 miles per hour. In the one-month period following the change in the speed limit, three accidents were observed. Explain in detail how you would test whether this is a significant decrease (3pts). Include a description of your null and alternative hypotheses.

Answer 1

Null hypothesis:
`p\geq 0.23`

Alternative hypothesis:
`p < 0.23`

`p_v =P(z<-1.692)=0.045`

If we compare the p value obtained and the significance level given
`\alpha=0.05` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of accidents per day is significantly less than 0.23.

Explanation:

1) Data given and notation n

n=30 represent the random sample taken (assuming 1 month=30days)

X=3 represent the number of accidents in the month selected

`\hat p=(3)/(30)=0.1` estimated proportion of accidents

`p_o=0.23` is the value that we want to test

`\alpha` represent the significance level

z would represent the statistic (variable of interest)

`p_v` represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion of accidents per day is less than 0.23.:

Null hypothesis:
`p\geq 0.23`

Alternative hypothesis:
`p < 0.23`

When we conduct a proportion test we need to use the z statistic, and the is given by:

`z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}` (1)

The One-Sample Proportion Test is used to assess whether a population proportion
`\hat p` is significantly different from a hypothesized value
`p_o`.

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

`z=\frac{0.1 -0.23}{\sqrt{(0.23(1-0.23))/(30)}}=-1.692`

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level assumed is
`\alpha=0.05`. The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

`p_v =P(z<-1.692)=0.045`

If we compare the p value obtained and the significance level given
`\alpha=0.05` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of accidents per day is significantly less than 0.23.