Hii! hopefully you can answer this :)

An insurance company states that it settles 85% of all life insurance claims within 30 days. A consumer group asks the state ins

BlackZzzverrR [31]

Using the z-distribution, it is found that since the p-value of the test is less than 0.05, we reject the null hypothesis and find that there is enough evidence to conclude that the true proportion of all life insurance claims made to this company that are settled within 30 days is less than 85%.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, it is tested if there is not enough evidence that the proportion is below 85%, that is:

$H_0: p \geq 0.85$

At the alternative hypothesis, it is tested if there is enough evidence that the proportion is below 85%, that is:

$H_0: p < 0.85$

<h3>What is the test statistic?</h3>

The test statistic is given by:

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

In which:

$\overline{p}$ is the sample proportion.

p is the proportion tested at the null hypothesis.

n is the sample size.

For this problem, the parameters are:

$p = 0.85, n = 250, \overline{p} = \frac{203}{250} = 0.812$

Hence the test statistic is:

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

$z = \frac{0.812 - 0.85}{\sqrt{\frac{0.85(0.15)}{250}}}$

z = -1.68

<h3>What is the p-value and the conclusion</h3>

Using a z-distribution calculator, with a left-tailed test, as we are testing if the proportion is less than a value, and z = -1.68, the p-value is of 0.0465.

Since the p-value of the test is less than 0.05, we reject the null hypothesis and find that there is enough evidence to conclude that the true proportion of all life insurance claims made to this company that are settled within 30 days is less than 85%.

More can be learned about the z-distribution at brainly.com/question/16313918

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4 0

1 year ago

I GIVE BRAINLILEST<br> //////////////////////////////////////

elena-s [515]

Answer:

answer is c makes sense because you just add 1.25 x 20

3 0

2 years ago

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti

melomori [17]

Answer:

a) 0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

b) Capacity of 252.6 cubic feet per second

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second).

This means that $m = 100, \mu = \frac{1}{100} = 0.01$

(a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

We have that:

$P(X > x) = e^{-\mu x}$

This is P(X > 190). So

$P(X > 190) = e^{-0.01*190} = 0.1496$

0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

(b) What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.08?

This is x for which:

$P(X > x) = 0.08$