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Len [333]
3 years ago
10

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).
Mathematics
1 answer:
SpyIntel [72]3 years ago
8 0

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] ± t_{n-1;1-\alpha /2} * \frac{S}{\sqrt{20} }

t_{n-1;1-\alpha /2} = t_{19;0.975}= 2.093

[15.3 ± 2.093 * \frac{6.8}{\sqrt{20} }]

[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!

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If P=$10, MPL= $25 and APL=$ 40, then the value of marginal product of labor is $250

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