The Wall Street Journal reported that the age at first startup for 55% of entrep

Question

The Wall Street Journal reported that the age at first startup for 55% of entrepreneurs was 29 years or less and the age at first startup for 45% of entrepreneurs was 30 years or more. a. Suppose a sample of 200 entrepreneurs will be taken to learn about the most impor- 32, tant qualities of entrepreneurs. Show the sampling distribution of p where P is the sample proportion of entrepreneurs whose first startup was at 29 years of age or less. What is the probability that the sample proportion in part (a) will be within ±.05 of its population proportion? Suppose a sample of 200 entrepreneurs will be taken to learn about the most important qualities of entrepreneurs. Show the sampling distribution of p where p is now the sample proportion of entrepreneurs whose first startup was at 30 years of age or more. What is the probability that the sample proportion in part (c) will be within ±05 of its population proportion? Is the probability different in parts (b) and (d)? Why? Answer part (b) for a sample of size 400. Is the probability smaller? Why? b. c. d. e. f.

Explanation / Answer

Here Pr(enterprenuers age when they start startup is less then 29) = 0.55

Pr(enterprenuers age when they start startup is more then 29) = 0.45

(a) Here sample size = 200

Here estimated proportion = p0 = 0.55

standard deviation of the sample proportion = sqrt(0.55 * 0.45/200) = 0.0352

(b) Here we have to calculate that sample proportion is under +-0.5 of population proportion.

Pr(0.50 < p < 0.60 ; 0.55; 0.0352) = Pr(0.60 < p ; 0.55; 0.0352) - Pr(0.50 < p ; 0.55; 0.0352) = (Z2) - (Z1)

Z2 = (0.60 - 0.55)/ 0.0352 = 1.42

Z1 = (0.50 - 0.55)/ 0.0352 = -1.42

(Z2) - (Z1) = 0.9222 - 0.0778 = 0.8444

Here is the cumulative standard normal distribution.

(c) Here the mean proportion of that sampling distribution where age is more than 30 years = 0.45

and, standard deviation of the sample proportion = sqrt(0.55 * 0.45/200) = 0.0352

(d) Here

Here we have to calculate that sample proportion is under +-0.5 of population proportion.

whereis, population proportion p0= 0.45

Pr(0.40 < p < 0.40 ; 0.45; 0.0352) = Pr(0.50 < p ; 0.45; 0.0352) - Pr(0.40 < p ; 0.45; 0.0352) = (Z2) - (Z1)

Z2 = (0.50 - 0.45)/ 0.0352 = 1.42

Z1 = (0.40 - 0.45)/ 0.0352 = -1.42

(Z2) - (Z1) = 0.9222 - 0.0778 = 0.8444

Here is the cumulative standard normal distribution.

(e) Now, the probability is not different in parts(b) and (d). It is because standard error of the proportion is same for both.

(f) Size = 400.

Here we have to calculate that sample proportion is under +-0.5 of population proportion when size is 400

standard error of the proportion se = sqrt(0.55 * 0.45/400) = 0.0249

Pr(0.50 < p < 0.60 ; 0.55; 0.0249) = Pr(0.60 < p ; 0.55; 0.0249) - Pr(0.50 < p ; 0.55; 0.0249) = (Z2) - (Z1)

Z2 = (0.60 - 0.55)/ 0.0249 = 2.00

Z1 = (0.50 - 0.55)/ 0.0249 = -2.00

(Z2) - (Z1) = 0.9772 - 0.0228 = 0.9544

Here is the cumulative standard normal distribution. Here sample size is more that have reduced standard error of the proportion so probability is more than the earlier.

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