Sorry i dont mean to be annoying, statistics help (this will be one of the last i swear!)?

Question

1.A sample of 500 respondents was selected in a large metropolitan area to study consumer behavior. Among the questions asked was “Do you enjoy shopping for clothing?” Of 240 males 136 answered yes. Of 260 females 224 answered yes. Construct a contingency table to evaluate the probabilities. What is the probability that a respondent chose at random:

a. enjoys shopping for clothing?

b. is a female and enjoys shopping for clothing

c. is a female or enjoys shopping for clothing?

d. is male or female?

2. The increase of decrease in the price of a stock between the beginning and the end of a trading day is assumed to be an equally likely random event. What is the probability that a stock will show and increase in its closing price on fOf 43 bank customers depositing a check, 18 received some cash back.

a. Construct a 90% confidence interval for the proportion of all depositors who ask for cash back.

b. Check the normality assumption.

ive consecutive days?

Answer 1

1(a).

P(Enjoys shopping for clothing)

= No. of men and women enjoy shopping for clothing/Total respondents

= (136+224)/500

= 360/500

= 18/25

= 0.72

1(b).

P(is a female and enjoys shopping for clothing)

= No.of female who is enjoying shopping for clothing/Total respondents

= 224/500

= 56/125

= 0.448

1(c).

P(is a female or enjoys shopping for clothing)

= P(female) + P(enjoys shopping for clothing) - P(female and enjoys shopping for clothing)

= 260/500 + (136+224)/500 - 56/125

= 99/125

= 0.52 + 0.72 - 0.448

= 0.792

1(d).

P(Male or female)

= P(Male) + P(female)

= 240/500 + 260/500

= 1

2(a)&(b).

Probability that the number of bank customers who would ask for cash back

= Sample proportion

= ^p

= 18/43

Since both n*^p = 43*(18/43) = 18 and n(1-^p) = 43*[1-(18/43)] = 25 are both at least greater than 5, then n is considered to be large and so it is appropriate to use the z distribution in this case due to the distribution of all depositors who would ask for cash back would be very approximately normally distributed.

Thus, a 90% confidence interval for the proportion of all depositors who ask for cash back

= [^p ± Z∞/2 * √^p(1-^p)/n]

= [(18/43)±z(0.1/2) * √(18/43)*[1-(18/43)]/43]

= [(18/43)±z0.05 * √(18/43)*(25/43)/43]

= [(18/43)±1.645* √(450/79507)]

= [0.418604651±0.123756915]

= [0.294847736, 0.542361566]

≈ [29.48%, 54.24%]

The above interval indicates that we are 90% confident that the true proportion of all depositors who would ask for cash back should be any value between 29.48% to 54.24%.

The following question is from your pervious post:

3. Consider a population of 1,024 mutual funds that primarily invested in large companies. You determined that μ, the mean one-year total percentage return achieved by the funds, is 8.20 and that σ, the standard deviation, is 2.75. IN addition, suppose you determined that the range in the one year total returns is from -2.0 to 17.1 and that the quartiles are, respectively, 5.5 (Q1) and 10.5 (Q3). According to the empirical rule, what percentages of these funds is expected to be

a. within +/-1 standard deviation?

b. within +/- 2 standard deviations of the mean?

c. According to Cebyshev’sule what percentage of these funds are expected to be within +/- 1, +/- 2 or +/- 3 standard deviations?

d. According to Chebyshevs rule, at least 93.75% of these funds is expected to have one-year total returns between what two amounts?

δ

My solution:

3(a). According to the empirical rule, about approximately 68% percentage of these funds is expected to be lie within +/-1 standard deviation of the mean.

3(b). According to the empirical rule, about approximately 95% percentage of these funds is expected to be lie within +/- 2 standard deviations of the mean.

3(c).

According to Cebyshev’s rule,

At least (1 − 1/(√2)^2] × 100% = 50% of these funds are within 1 standard deviations from the mean.

At least (1 − 1/(2)^2] × 100% = 75% of these funds are within 2 standard deviations from the mean.

At least (1 − 1/(3)^2] × 100% = 89% of these funds are within 3 standard deviations from the mean.

3(d). We are given that

(1 − 1/k^2)*100% = 93.75%

i.e. 1 − 1/k^2 = 0.9375

    Ã¢ÂˆÂ’ 1/k^2 = 0.9375 - 1

− 1/k^2 = 0.9375 - 1

   − 1/k^2 = -0.0625

   1/k^2 = 0.0625

       k^2 = 1/0.0625

      k^2 = 16

         k = ±√16

         k = ±4

At least 93.75% of these funds is expected to have one-year total returns between the two amounts

= [μ±4σ]

= [8.20±4*2.75]

= [8.20±11]

= [-2.8, 19.2]

Hope this helps you.