A veterinarian collects data on the number of times race horses are raced during their careers. The veterinarian finds that the
average number of races a horse enters is ¯ x = 15.3 , with a standard deviation of s = 6.8 in a sample of n = 20 horses. A 95% confidence for μ , the average number of times a horse races, is given by___________. a. (12.32, 18.28).
b. (10.95, 19.65).
c. (12.12, 18.48).
d. (11.38, 19.22).
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] ±
[15.3 ± 2.093 * ]
[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].
The next term in the sequence will be increased by 27.
The difference between 15 and 34 is 19
The difference between 34 and 55 is 21
The difference between 55 and 78 is 23
The difference between 78 and 103 is 25
Following this pattern, the next number will be 130