Question

Last year it was found that on average it took students 20 minutes to fill out the forms required for graduation. This year the department has changed the form and asked graduating student to report how much time it took them to complete the forms. Of the students 22 replied with their time, the average time that they reported was 18.5 minutes, and the sample standard deviation was 5.2. Can we conclude that the new forms take less time to complete than the older forms?

Answer 1

Answer:

$t=\frac{18.5-20}{\frac{5.2}{\sqrt{22}}}=-1.35$  

$p_v =P(t_{21}$  

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean for the time is not significantly less than 20 minutes at 5% of significance.  

Step-by-step explanation:

Data given and notation  

$\bar X=18.5$ represent the sample mean  

$s=5.2$ represent the standard deviation for the sample

$n=22$ sample size  

$\mu_o =20$ 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 to be tested

We need to conduct a hypothesis in order to determine if the mean for the new forms take less time to complete than the older forms, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 20$  

Alternative hypothesis:$\mu < 20$  

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:  

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

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

$t=\frac{18.5-20}{\frac{5.2}{\sqrt{22}}}=-1.35$  

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

$df=n-1=22-1=21$

What do you conclude?

Compute the p-value

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

$p_v =P(t_{21}$  

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean for the time is not significantly less than 20 minutes at 5% of significance.