Question

An insurance company states that at least 87% of its claims are settled within 30 days. A consumer group selected a random sample of 64 of the company's claims to test this statement. They found that 53 of the claims were settled within 30 days. Does the consumer group have evidence to disbelieve the insurance company's claim?

Answer 1

`z=\frac{0.828 -0.87}{\sqrt{(0.87(1-0.87))/(64)}}=-0.999`

`p_v =P(z<-0.999)=0.159`

So the p value obtained was a very high 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 FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of he claims were settled within 30 days is not significantly lower than 0.87, and the claim makes sense .

Explanation:

Data given and notation

n=64 represent the random sample taken

X=53 represent the claims were settled within 30 days

`\hat p=(53)/(64)=0.828` estimated proportion of the claims were settled within 30 days

`p_o=0.87` is the value that we want to test

`\alpha=0.05` represent the significance level (assumed)

Confidence=95% or 0.95 (assumed)

z would represent the statistic (variable of interest)

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

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is at leat 0.87 or 87%:

Null hypothesis:
`p\geq 0.87`

Alternative hypothesis:
`p < 0.87`

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`.

Calculate the statistic

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

`z=\frac{0.828 -0.87}{\sqrt{(0.87(1-0.87))/(64)}}=-0.999`

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 left tailed test the p value would be:

`p_v =P(z<-0.999)=0.159`

So the p value obtained was a very high 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 FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of he claims were settled within 30 days is not significantly lower than 0.87, and the claim makes sense .