Question

Daniel is paying $600 for his auto insurance, and he is wondering if he is overpaying compared to his friends. He sent an email to all his friends in his contact list, and 9 of them replied with their paid amount. Suppose the 9 friends who replied are a random sample, and the paid amount for auto insurance has approximately a normal distribution. What can you conclude on the study?
564 578 478 507 621 564 489 612 538

Answer 1

Answer:

$t=\frac{550.11-600}{\frac{51.185}{\sqrt{9}}}=-2.924$    

$p_v =P(t_{(8)}$  

If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is significantly lower than 600 so then Daniel is not overpaying for this case

Step-by-step explanation:

Data given and notation  

Data: 564 578 478 507 621 564 489 612 538

We can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}[tex][tex] s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=550.11$ represent the mean height for the sample  

$s=51.185$ represent the sample standard deviation for the sample  

$n=9$ sample size  

$\mu_o =600$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

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

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 600 or no, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 600$  

Alternative hypothesis:$\mu < 600$  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$  (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".  

Calculate the statistic

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

$t=\frac{550.11-600}{\frac{51.185}{\sqrt{9}}}=-2.924$    

P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=9-1=8$  

Since is a one side test the p value would be:  

$p_v =P(t_{(8)}$  

Conclusion  

If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is significantly lower than 600 so then Daniel is not overpaying for this case