Question

An insurance company states that 90% of its claims are settled within a month. A consumer group selected a random sample of 75 of the company’s claims to test this statement and found that only 55 of the claims were settled within a month.
(a) (12 points) At a 5% level of significance, does the consumer group have sufficient evidence to support their contention that fewer than 90% of the claims are settled within a month? Find also the P-value

Answer 1

Answer:

$z=\frac{0.733 -0.9}{\sqrt{\frac{0.9(1-0.9)}{75}}}=-4.82$  

$p_v =P(Z$  

The p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ 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  claims that were settled within a month is significantly less than 0.9 or 90%.

Step-by-step explanation:

1) Data given and notation

n=75 represent the random sample taken

X=55 represent the number of claims that were settled within a month.

$\hat p=\frac{55}{75}=0.733$ estimated proportion of claims that were settled within a month.

$p_o=0.9$ 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 the claim that the true proportion is less than 0.9 or 90%.:  

Null hypothesis:$p\geq 0.9$  

Alternative hypothesis:$p$  

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

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

$z=\frac{0.733 -0.9}{\sqrt{\frac{0.9(1-0.9)}{75}}}=-4.82$  

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 left tailed test the p value would be:  

$p_v =P(Z$  

The p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ 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  claims that were settled within a month is significantly less than 0.9 or 90%.