Question

Suppose we are testing people to see if the rate of use of seat belts has changed from a previous value of 88%. Suppose that in our random sample of 500 people, we see that 450 have the seat belt fastened. (a) About how many of 500 would we expect to be using their seat belts if the proportion who use seat belts is unchanged?

(b) We observe 450 people out of a random sample of 500 using their seatbelt. The p-value is 0.167. Explain the meaning of the p-value.

Answer 1

a) We would expect to see 500*0.88=440

b)
`z=\frac{0.9 -0.88}{\sqrt{(0.88(1-0.88))/(500)}}=1.376`

`p_v =2*P(Z>1.376)=0.167`

So the p value obtained was a very high value and using the significance level assumed
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion is not significant different from 0.9.

The p value is a criterion to decide if we reject or not the null hypothesis, when
`p_v <\alpha` we reject the null hypothesis in other case we FAIL to reject the null hypothesis. And represent the "probability of obtaining the observed results of a test, assuming that the null hypothesis is correct".

Explanation:

Data given and notation

n=500 represent the random sample taken

X=450 represent the people that have the seat belt fastened

`\hat p=(450)/(500)=0.9` estimated proportion of people that have the seat belt fastened

`p_o=0.88` is the value that we want to test

`\alpha` represent the significance level

z would represent the statistic (variable of interest)

Alternative hypothesis:
`p \\eq 0.88`

When we conduct a proportion test we need to use the z statisitc, 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`.

Calculate the statistic

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

`z=\frac{0.9 -0.88}{\sqrt{(0.88(1-0.88))/(500)}}=1.376`

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 bilateral test the p value would be:

`p_v =2*P(Z>1.376)=0.167`

So the p value obtained was a very high value and using the significance level assumed
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion is not significant different from 0.9.

The p value is a criterion to decide if we reject or not the null hypothesis, when
`p_v <\alpha` we reject the null hypothesis in other case we FAIL to reject the null hypothesis. And represent the "probability of obtaining the observed results of a test, assuming that the null hypothesis is correct".