Question

A veterinarian pregnancy tests cows each year for various ranches. In a random sample of 200 cows, 11 were found to be “open” (not pregnant).

• Construct a 98 percent confidence interval for the true proportion of “open” cows.
• Write a sentence using the confidence interval found in part a.
• Show if the sample proportion may be assumed to be normally distributed.

Answer 1

solution Here n=200, x=11, p = 1 = 0.055 here np=11>10 n(1-D = $189 >10 so, we can use the nounal distubution- At 98% confident tal-2-33 0.055 Jn the 984. c- I for p is 055(1-0.055 P + 2 p q = 0.055 + 2-33) 200 = 0.055+ 0.03756 =(0.017uu, 0.09256) we are 98% confident that the true of open is lies between 0.01741 and 8.09256 cours