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A veterinarian investigating possible causes of enteroliths in horses suspects that feeding alfalfa may be to blame. She wishes to estimate the proportion of horses with enteroliths who are fed at least two flakes of alfalfa per day. In a sample of 62 horses with enteroliths, she finds 42 are fed two or more flakes of alfalfa.
A 95% confidence interval is given by:

a. (0.524, 0.83).
b. (0.561, 0.794).
c. (0.601, 0.754).
d. (0.576, 0.744).

Respuesta :

JeanaShupp

Answer: b. (0.561, 0.794).

Step-by-step explanation:

We know that the confidence interval for population proportion (p) is given by:-

[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

, where [tex]\hat{p}[/tex]= Sample proportion.

n = Sample size.

z*  = Critical z-value.

Let p = proportion of horses with enteroliths who are fed at least two flakes of alfalfa per day.

As per given ,  n = 62

[tex]\hat{p}=\dfrac{42}{62}=0.67742[/tex]

z-value for 95% confidence = z*=1.96

A 95% confidence interval is given by:

[tex]0.67742\pm (1.96)\sqrt{\dfrac{0.67742(1-0.67742)}{62}}\\\\ =0.67742\pm 0.11636\\\\ =(0.67742-0.11636,0.67742+0.11636 )\\\\=(0.56106, 0.79378)\approx(0.561,0.794 )[/tex]

Thus , the required 95% confidence interval is (0.561, 0.794).

Hence, the correct option is b. (0.561, 0.794)..