Question

. You are interested in the amount of time teenagers spend weekly working at part-time jobs. A random sample of 15 teenagers was drawn and each reported the amount of time spent at part-time jobs (in minutes), with the following results: sample mean is 147.3 and sample standard deviation is 50. Assume that the population is normally distributed. Is there evidence that the mean amount of time teenagers spend weekly working at part-time jobs is more than 120 minutes? Use α = 0.05 and the critical value approach to hypothesis testing.

Answer 1

Answer:

$t=\frac{147.3-120}{\frac{50}{\sqrt{15}}}=2.115$    

Now we can calculate the degrees of freedom

$df=n-1=15-1=14$  

If we find a critical value in the t distribution with 14 degrees of freedom who accumulates 0.05 of the area in the right we got $t_{critc}= 1.761$

Since the calculated value is higher than the critical value we have enough evidence to conclude that the true mean is significantly higher than 120 minutes for the average time of part time jobs

Step-by-step explanation:

Information given

$\bar X=147.3$ represent the sample mean for the amount of time spent at part time jobs

$s=50$ represent the sample standard deviation

$n=15$ sample size  

$\mu_o =120$ represent the value to check

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value for the test

System of hypothesis

We want to analyze if the true mean for the amount of time spent at part time jobs is higher than 120, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 120$  

Alternative hypothesis:$\mu > 120$  

Since we don't know the population deviation the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

Replacing the info given we got:

$t=\frac{147.3-120}{\frac{50}{\sqrt{15}}}=2.115$    

Now we can calculate the degrees of freedom

$df=n-1=15-1=14$  

If we find a critical value in the t distribution with 14 degrees of freedom who accumulates 0.05 of the area in the right we got $t_{critc}= 1.761$

Since the calculated value is higher than the critical value we have enough evidence to conclude that the true mean is significantly higher than 120 minutes for the average time of part time jobs