Question

Let p equal the proportion of drivers who use a seat belt in a country that does not have a mandatory seat belt law. It was claimed that p = 0.14. An advertising campaign was conducted to increase this proportion. Two months after the campaign, y = 104 out of a random sample of n = 590 drivers were wearing their seat belts . Was the campaign successful? (a) Define the null and alternative hypotheses. (b) Define a rejection region with an α = 0.01 significance level. (c) Determine the approximate p-value and state your conclusion.

Answer 1

Answer:

a)Null hypothesis:$p\leq 0.14$  

Alternative hypothesis:$p > 0.14$  

b) For this case we are conducting a right tailed test so then we need to look in the normal standard distribution a quantile that accumulates 0.01 of the are in the left and we got;

$z_{crit} = 2.33$

So then the rejection region would be $(2.33 , \infty)$

c) The significance level provided $\alpha=0.01$. The next step would be calculate the p value for this test.  

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

$p_v =P(z>2.52)=0.006$  

And since the p value is lower than the significance level then we can reject the null hypothesis. So then we can conclude that the true proportion of interest is higher than 0.14 at 1% of significance.

Step-by-step explanation:

Data given and notation

n=590 represent the random sample taken

X=104 represent the drivers were wearing their seat belts

We can estimate the sample proportion like this:

$\hat p=\frac{104}{590}=0.176$ estimated proportion of  drivers were wearing their seat belts

$p_o=0.14$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

a) System of hypothesis

We need to conduct a hypothesis in order to test the claim that the true proportion of drivers were wearing their seat belts is higher than 0.14 or no, so the system of hypothesis are.:  

Null hypothesis:$p\leq 0.14$  

Alternative hypothesis:$p > 0.14$  

Part b

For this case we are conducting a right tailed test so then we need to look in the normal standard distribution a quantile that accumulates 0.01 of the are in the left and we got;

$z_{crit} = 2.33$

So then the rejection region would be $(2.33 , \infty)$

Part c

The statistic is given by:

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

Calculate the statistic  

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

$z=\frac{0.176 -0.14}{\sqrt{\frac{0.14(1-0.14)}{590}}}=2.52$  

Statistical decision  

The significance level provided $\alpha=0.01$. The next step would be calculate the p value for this test.  

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

$p_v =P(z>2.52)=0.006$  

And since the p value is lower than the significance level then we can reject the null hypothesis. So then we can conclude that the true proportion of interest is higher than 0.14 at 1% of significance.

Answer 2

Answer:

See explanation

Step-by-step explanation:

Solution:-

- Let the proportion of drivers who use a seat belt in a country that does not have a mandatory seat belt law = p.

- A claim was made that p = 0.14. We will state our hypothesis:

                Null hypothesis: p = 0.14

- Two months after the campaign, y = 104 out of a random sample of n = 590 drivers were wearing their seat belts. The sample proportion can be determined as:

               Sample proportion ( p^ ) = y / n = 104 / 590 = 0.176

- The alternate hypothesis will be defined by a population proportion that supports an increase. So we state the hypothesis:

               Alternate hypothesis: p > 0.14

- The rejection is defined by the significance level ( α = 0.01 ). The rejection region is defined by upper tail of standard normal.

- The Z-critical value that limits the rejection region is defined as:

                            P ( Z < Z-critical ) = 1 - 0.01 = 0.99

                            Z-critical = 2.33

- All values over Z-critical are rejected.

- Determine the test statistics by first determining the population standard deviation ( σ ):

- Estimate σ using the given formula:

                      σ = $\sqrt{\frac{p*(1-p)}{n} } = \sqrt{\frac{0.14*(1-0.14)}{590} }= 0.01428$

- The Z-test statistics is now evaluated:

                     Z-test = ( p^ - p ) / σ

                     Z-test = ( 0.1763 - 0.14 ) / 0.01428

                    Z-test = 2.542

- The Z-test is compared whether it lies in the list of values from rejection region.

                     2.542 > 2.33

                     Z-test > Z-critical

Hence,

                     Null hypothesis is rejected

- The claim made over the effectiveness of campaign is statistically correct.