Question

Based on information from a large insurance company, 68% of all damage liability claims are made by single people under the age of 25. A random sample of 53 claims showed that 41 were made by single people under the age of 25. Does this indicate that the insurance claims of single people under the age of 25 is higher than the national percent reported by the large insurance company? State the null and alternate hypothesis then give the test statistic and your conclusion.
a)z = 2.326; reject Hoat the 5% significance level
b)z = 1.826; reject Hoat the 5% significance level
c)z = -2.326; reject Hoat the 5% significance level
d)z = 2.326; fail to reject Hoat the 5% significance level
e)z = -1.826; fail to reject Hoat the 5% significance level

Answer 1

Answer:

z=1.461, fail to reject the null hypothesis since $p_v>\alpha$ at 5% of singificance.

Step-by-step explanation:

1) Data given and notation

n=53 represent the random sample taken

X=41 represent the adults with damage liability claims by single people under the age of 25.

$\hat p=\frac{41}{53}=0.774$ estimated proportion of adults with damage liability claims by single people under the age of 25.

$p_o=0.68$ is the value that we want to test

$\alpha=0.05$ 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)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test that the insurance claims of single people under the age of 25 is higher than the national percent reported by the large insurance company (68%).:  

Null hypothesis:$p\leq 0.68$  

Alternative hypothesis:$p > 0.68$  

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

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

$z=\frac{0.774 -0.68}{\sqrt{\frac{0.68(1-0.68)}{53}}}=1.461$

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 provided $\alpha=0.05$. The next step would be calculate the p value for this test.  

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

$p_v =P(z>1.461)=0.072$  

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we don't have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults with damage liability claims by single people under the age of 25 is not significantly higher from 0.68 or 68% .