Answer:
c. (12.12, 18.48)
Step-by-step explanation:
Hello!
The study variable is X: number of times a racehorse is raced during its career.
The average number is X[bar]= 15.3 and the standard deviation is S= 6.8 obtained from a sample of n=20 horses.
To estimate the population mean you need that the variable has a normal distribution, in this case, we have no information about its distribution so I'll assume that it has a normal distribution. With n=20 the most accurate statistic to use for the estimation is a Students-t for one sample, the formula for the interval is:
X[bar] ± [tex]t_{n-1;1-\alpha /2} * \frac{S}{\sqrt{20} }[/tex]
[tex]t_{n-1;1-\alpha /2} = t_{19;0.975}= 2.093[/tex]
[15.3 ± 2.093 * [tex]\frac{6.8}{\sqrt{20} }[/tex]]
[12.12; 18.48]
Using a significance level of 95% you'd expect that the true average of times racehorses are raced during their career is included in the interval [12.12; 18.48].
I hope it helps!