Answer:
[tex]t=\frac{67-60}{\frac{12}{\sqrt{16}}}=2.33[/tex] Â Â
Step-by-step explanation:
Data given and notation Â
[tex]\bar X=67[/tex] represent the sample mean
[tex]s=12[/tex] represent the sample standard deviation
[tex]n=16[/tex] sample size Â
[tex]\mu_o =60[/tex] represent the value that we want to test
t would represent the statistic (variable of interest) Â
State the null and alternative hypotheses. Â
We need to conduct a hypothesis in order to check if the true mean is higher than 60, the system of hypothesis would be: Â
Null hypothesis:[tex]\mu \leq 60[/tex] Â
Alternative hypothesis:[tex]\mu > 60[/tex] Â
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: Â
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] Â (1) Â
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". Â
Calculate the statistic
We can replace in formula (1) the info given like this: Â
[tex]t=\frac{67-60}{\frac{12}{\sqrt{16}}}=2.33[/tex] Â Â