An institute reported that % of its members indicate that lack of ethical cultur

Question

An institute reported that % of its members indicate that lack of ethical culture within financial firms has contributed most to the lack of trust in the financial industry. Suppose that you select a sample of 100 institute members. Complete parts (a) through (d) below. 60 a. What is the probability that the sample percentage indicating that lack of ethical culture within financial firms has contributed the most to the lack of trust in the financial industry will be between 59 % and 62 %? .2403 (Type an integer or decimal rounded to four decimal places as needed.) b. The probability is % that the sample percentage will be contained within what symmetrical limits of the population percentage? 70 The probability is 70 % that the sample percentage will be contained above 68.2 % and below 71.8 %. (Type integers or decimals rounded to one decimal place as needed.) c. The probability is % that the sample percentage will be contained within what symmetrical limits of the population percentage? 97 The probability is 97 % that the sample percentage will be contained above % and below %. (Type integers or decimals rounded to one decimal place as needed.) d. Suppose you selected a sample of 400 institute members. How does this change your answers in (a) through (c) ? A. An increase in sample size increases the standard error, causing the value of the answer to part (a) to increase and the ranges in parts (b) and (c) to widen. B. An increase in sample size decreases the standard error, causing the value of the answer to part (a) to decrease and the ranges in parts (b) and (c) to narrow. C. An increase in sample size increases the standard error, causing the value of the answer to part (a) to decrease and the ranges in parts (b) and (c) to widen. D. An increase in sample size decreases the standard error, causing the value

Explanation / Answer

a) p = 0.64

Z = (p^ - 0.64)/(sqrt(0.64*0.36/100))

= (p^ - 0.64)/0.048

P(0.55 < p^< 0.71)

=P ( 1.88<Z<1.46 )=0.8978

b)

P( - z< Z<z) = 0.80

z = 1.282

hence

below = 0.64 - 1.282 * 0.048 =0.578464

upper = 0.64 + 1.282 * 0.048 =0.701536

c)

91 %

z = 1.696

lower = 0.64 - 1.696 * 0.048 = 0.558592

upper = 0.64 + 1.696* 0.048 = 0.721408

d)

C. An increase in sample size decreases the standard error, causing the value of the answer to part (a) to decrease and the ranges in parts (b) and (c) to narrow

is correct

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