Question

1. A prison official wants to estimate the proportion of cases of recidivism. Examining the records of 238 convicts, the official determines that there are 92 cases of recidivism. Find the lower limit of the 90% confidence interval estimate for the population proportion of cases of recidivism.
(Round to 3 decimal places.)

2. A prison official wants to estimate the proportion of cases of recidivism. Examining the records of 157 convicts, the official determines that there are 47 cases of recidivism.
Find the margin of error for a 90% confidence interval estimate for the population proportion of cases of recidivism.   (Use 3 decimal places.)

3. A prison official wants to estimate the proportion of cases of recidivism. Examining the records of 221 convicts, the official determines that there are 47 cases of recidivism. A confidence interval will be obtained for the proportion of cases of recidivism. Part of this calculation includes the estimated standard error of the sample proportion.

Calculate the estimated standard error.   (Use 3 decimal places in calculations and in reporting your answer.)

4. The head of a computer science department is interested in estimating the proportion of students entering the department who will choose the new computer engineering option. A preliminary sample indicates that the proportion will be around 0.231.
What size sample should the department head take if she wants to be 95% confident that the estimate is within 0.02 of the true proportion?

5. A quality control engineer is interested in estimating the proportion of defective items coming off a production line. In a sample of 351 items, 48 are defective. Calculate a  95.0% confidence interval estimate for the proportion of defectives from this production line. (Use 3 decimal places in calculations and in reporting your answers.)

Lower Limit: ?

Upper Limit: ?

Answer 1

Answer:

1.

Given,

p^ = x/n = 92/238

= 0.387

Here ata 90% CI, z value = 1.645

CI = p^ +/- z*sqrt(p^(1-p^)/n)

substitute values

= 0.387 +/- 1.645*sqrt(0.387(1-0.387)/238)

= 0.387 +/- 0.0519

= (0.3351 , 0.4389)

lower limit = 0.3351

2.

p^ = 47/157

= 0.2994

At 90% CI, z value = 1.645

Margin of error E = z*sqrt(p^(1-p^)/n)

substitute values

= 1.645*sqrt(0.2994(1-0.2994)/157)

E = 0.060

3.

Standard error = sqrt(p^(1-p^)/n)

p^ = x/n = 47/221 = 0.27

SE = sqrt(0.27(1-0.27)/221)

= 0.0299

4.

Given,

p^ = 0.231

E = 0.02

Here for 95% CI, z value is 1.96

n = p^(1-p^)*(z/E)^2

substitute values

= 0.231(1-0.231)*(1.96/0.02)^2

= 1706.044956

= 1706

5.

p^ = x/n = 48/351 = 0.1368

Here for 95% CI, z value is 1.96

CI = p^ +/- z*sqrt(p^(1-p^)/n)

substitute values

= 0.1358 +/- 1.96*sqrt(0.1358(1-0.1358)/351)

= 0.1358 +/- 0.0358

= (0.1 , 0.1716)

Lower limit = 0.1

Upper limit = 0.1716