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

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

`p_v =P(Z<-4.82)=7.18x10^(-7)`

The p value obtained was a very low value and using 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 claims that were settled within a month is significantly less than 0.9 or 90%.

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=(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 <0.9`

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{0.733 -0.9}{\sqrt{(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<-4.82)=7.18x10^(-7)`

The p value obtained was a very low value and using 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 claims that were settled within a month is significantly less than 0.9 or 90%.