# A. Out of 100 people sampled,

A. Out of 100 people sampled, 92 had kids. Based on this,construct a 95% confidence interval for the true populationproportion of people with kids.As in the reading, in your calculations:–Use z = 1.645 for a 90% confidence interval–Use z = 2 for a 95% confidence interval–Use z = 2.576 for a 99% confidence interval.Give your answers to three decimalsGive your answers as decimals, to three decimal places.

B. CNNBC recently reported that the mean annual cost of autoinsurance is 994 dollars. Assume the standard deviation is 282dollars. You take a simple random sample of 54 auto insurancepolicies.Find the probability that a single randomly selected value is lessthan 969 dollars.*P*(*X* < 969) =Find the probability that a sample of size n=54n=54 is randomlyselected with a mean less than 969 dollars.*P*(¯xx¯ < 969) =

C. Out of 400 people sampled, 332 preferred Candidate A. Basedon this, estimate what proportion of the voting population (pp)prefers Candidate A.Use a 95% confidence level, and give your answers as decimals, tothree places.

Answer:

(A)

n = 100

p = 0.92

% = 95

Standard Error, SE = √{p(1 – p)/n} = √(0.92(1- 0.92))/100 = 0.02712932

z- score = 1.959963985

Width of the confidence interval = z * SE = 1.95996398454005 * 0.0271293199325011 =0.05317249

Lower Limit of the confidence interval = P – width = 0.92 – 0.0531724899927667 = 0.86682751

Upper Limit of the confidence interval = P + width = 0.92 + 0.0531724899927667 = 0.97317249

**The 95% confidence interval is [0.867,0.973]**

(B)

(a) μ = 994, σ = 282, x = 969

z = (x – μ)/σ = (969 – 994)/282 = -0.0887

P(x < 969) = P(z < -0.0887) = **0.4647**

(b) μ = 994, σ = 282, n = 54, x-bar = 969

z = (x-bar – μ)/(σ/√n) = (969 – 994)/(282/√54) = -0.6515

P(x-bar < 969) = P(z < -0.6515) =**0.2574**

(C)

n = 400

p = 0.83

% = 95

Standard Error, SE = √{p(1 – p)/n} = √(0.83(1- 0.83))/400 = 0.01878164

z- score = 1.959963985

Width of the confidence interval = z * SE = 1.95996398454005 * 0.018781639970993 =0.03681134

Lower Limit of the confidence interval = P – width = 0.83 – 0.0368113379137441 = 0.79318866

Upper Limit of the confidence interval = P + width = 0.83 + 0.0368113379137441 = 0.86681134

**The 95% confidence interval is [0.793,0.869]**