Question

The Chartered Financial Analyst (CFA) Institute reported that 63% of its U.S. members indicate that lack of ethical culture within financial firms has contributed the most to the lack of trust in the financial industry. (Data extracted from Global Market Sentiment Survey 2014, cfa.is/1dzvdF8.) Suppose that you select a sample of 100 CFA members.
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 62% and 67


b. The probability is 90% that the sample percentage will be contained within what symmetrical limits of the population percentage?


c. The probability is 95% that the sample percentage will be contained within what symmetrical limits of the population percentage?


d. Suppose you selected a sample of 400 CFA members. How does this change your answers in (a) through (c)7

Answer 1

Answer:

a. The probability that the percentage will be between 62% and 67% is 0.378348  

b. For the central 90% the lack of trust in the financial industry is between 55.282% and  70.92%

c. For the probability within the central 95%, the lack of trust in the financial industry is between 56.77% and  56.77%%

d. Increasing the sample size increases  the sensitivity and narrows the ranges and values

Step-by-step explanation:

a. The percentage of the CFA members that indicated that lack of ethical culture within financial firms has contributed the most to the lack of trust in the financial industry, p = 63%

The test statistic is given as follows;

$z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p \cdot q}{n}}}$

p = 63% = 0.63

q = 1 - p = 0.37

a. When $\hat p$ = 62%, we get;

$z=\dfrac{0.62-0.63}{\sqrt{\dfrac{0.63 \times 0.37}{100}}} \approx -0.2071$

From the z-table the p-value p(z <-0.2071) = 0.42074, from the calculator, we get the more accurate value as p = 0.4179563

When $\hat p$ = 67%, we get;

(0.67 - 0.63)/((0.63*0.37)/100)^(0.5)

$z=\dfrac{0.67-0.63}{\sqrt{\dfrac{0.63 \times 0.37}{100}}} \approx 0.82849302133$

From the z-table the p-value p(z < 0.8) = 0.79673, from the calculator, we get the more accurate value as p(z < 0.828493) = 0.7963043

Therefore, using the calculator values, the probability that the percentage will be between 62% and 67% = 0.7963043 - 0.4179563 = 0.378348

The probability that the percentage will be between 62% and 67% = 0.378348  

b. For the central 90% symmetrical limit, we get;

$P\left(Z \le \dfrac{x-0.63}{\sqrt{\dfrac{0.63 \times 0.37}{100}}}\right) = 0.05$

Therefore;

$\dfrac{x-0.63}{\sqrt{\dfrac{0.63 \times 0.37}{100}}} = \phi^{-1}(0.05)$

z = -1.64

Therefore;

$\dfrac{x-0.63}{\sqrt{\dfrac{0.63 \times 0.37}{100}}} \approx -1.64$

x = -1.64×√(0.63*0.37/100) + 0.63 ≈ 0.55282, which is 55.282%

Similarly, we get;

$P\left(Z \geq \dfrac{x-0.63}{\sqrt{\dfrac{0.63 \times 0.37}{100}}}\right) = 0.95$

Therefore;

$\dfrac{x-0.63}{\sqrt{\dfrac{0.63 \times 0.37}{100}}} = \phi^{-1}(0.95)$

z = 1.64

Therefore;

$\dfrac{x-0.63}{\sqrt{\dfrac{0.63 \times 0.37}{100}}} \approx 1.64$

x = 1.64×√(0.63*0.37/100) + 0.63 ≈ 0.7092, which is 70.92%

Therefore, the central 90% is between 70.92% and 55.282%

Therefore, the lack of trust in the financial industry is between 55.282% and  70.92%

c. Given that the probability is 95%, we get

$\dfrac{x-0.63}{\sqrt{\dfrac{0.63 \times 0.37}{100}}} = \phi^{-1}(0.025)$

Therefore;

$\dfrac{x-0.63}{\sqrt{\dfrac{0.63 \times 0.37}{100}}} \approx -0.67$

x = -0.67×√(0.63*0.37/100) + 0.63 ≈ 0.5977 = 56.77%

Similarly, we have;

$\dfrac{x-0.63}{\sqrt{\dfrac{0.63 \times 0.37}{100}}} = \phi^{-1}(0.975)$

Therefore;

$\dfrac{x-0.63}{\sqrt{\dfrac{0.63 \times 0.37}{100}}} = 1.96$

x = 1.96×√(0.63*0.37/100) + 0.63 ≈ 0.72463 =%

Therefore, at a probability of 95%, the lack of trust in the financial industry is between 56.77% and  56.77%%

d. Increasing the sample size increases  the sensitivity and narrows the ranges and values such that we get;

For a) z = 0.62-0.63×√(0.63*0.37/400) ≈ 0.605

P ≈ 0.72907

0.67-0.63×√(0.63*0.37/400) ≈ 0.655

P ≈ 0.74527 - 0.72907 = 0.0162 = 1.62%

For b)

The p-value = 0.74527 -

-1.64×√(0.63*0.37/400) + 0.63 ≈ 0.5904 ≈ 59.04%

1.64×√(0.63*0.37/400) + 0.63 ≈ 0.7 = 70%

For c)

x = -0.67×√(0.63*0.37/400) + 0.63 ≈ 0.61383= 61.83%

x = 1.96×√(0.63*0.37/400) + 0.63 ≈ 67.73=%