An insurance company analyst wants to test the hypothesis that the average amount claimed by its policy holders over one year is $800 against the alternative that is less than $800. A random sample of 169 policy holders reveals that the sample mean claim is $739.98 and the sample standard deviation is $312.70. Perform this test at a 5% error level.

If we compare the p value and a significance level assumed we see that so we can conclude that we can reject the null hypothesis, and the actual true mean is significantly lower than 800 at 5% of significance.  

Step-by-step explanation:

Data given and notation

represent the sample mean

represent the standard deviation for the sample

sample size  

represent the value that we want to test  

represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

represent the p value for the test (variable of interest)  

State the null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the mean is lower than 800, the system of hypothesis would be:  

Null hypothesis:  

Alternative hypothesis:  

Compute the test statistic  

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

(1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

We can replace in formula (1) the info given like this:  

Now we need to find the degrees of freedom for the t distirbution given by:

What do you conclude?  

Use the p-value approach

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

If we compare the p value and a significance level assumed we see that so we can conclude that we can reject the null hypothesis, and the actual true mean is significantly lower than 800 at 5% of significance.  

70

step-by-step explanation:

2*5=10

4*15=60

60+10=

70