Question

Every road has one at some point - construction zones that have much lower speed limits. To see if drivers obey these lower speed limits, a police officer uses a radar gun to measure the speed (in miles per hours, or mph) of a random sample of 10 drivers in a 25 mph construction zone. Here are the data: 27; 33; 32; 21; 30; 30; 29; 25; 27; 34. Is there convincing evidence that the average speed of drivers in this construction zone is greater than the posted speed limit?

Answer 1

Answer:

$t=\frac{28.8-25}{\frac{3.938}{\sqrt{10}}}=3.05$    

$p_v =P(t_{(9)}>3.05)=0.0069$  

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, so we can conclude that the true mean is higher than 25 mph at 1% of signficance.  

Step-by-step explanation:

Data given and notation  

Data: 27; 33; 32; 21; 30; 30; 29; 25; 27; 34

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

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

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

$\bar X=28.8$ represent the mean height for the sample  

$s=3.938$ represent the sample standard deviation for the sample  

$n=10$ sample size  

$\mu_o =25$ 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 25mph, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 25$  

Alternative hypothesis:$\mu > 25$  

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:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (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:  

$t=\frac{28.8-25}{\frac{3.938}{\sqrt{10}}}=3.05$    

P-value

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

$df=n-1=10-1=9$  

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

$p_v =P(t_{(9)}>3.05)=0.0069$  

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, so we can conclude that the true mean is higher than 25 mph at 1% of signficance.