Question

1. A prison official wants to estimate the proportion of cases of recidivism. Examining the records of 217 convicts, the official determines that there are 80 cases of recidivism. Find the margin of error for a 95% confidence interval estimate for the population proportion of cases of recidivism.   (Use 3 decimal places.)

2. A prison official wants to estimate the proportion of cases of recidivism. Examining the records of 204 convicts, the official determines that there are 32 cases of recidivism. Find a point estimate of the population proportion of cases of recidivism.  (Round your answer to three decimal places.)

3. A prison official wants to estimate the proportion of cases of recidivism. Examining the records of 298 convicts, the official determines that there are 93 cases of recidivism. Find the standard error for the population proportion of cases of recidivism.  (Round to 3 decimal places.)

4. A prison official wants to estimate the proportion of cases of recidivism. Examining the records of 241 convicts, the official determines that there are 63 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.)

5. A quality control engineer is interested in estimating the proportion of defective items coming off a production line. In a sample of 336 items, 48 are defective. Calculate a upper confidence limit for a 95.0% confidence interval for the proportion of defectives from this production line.  (Use 3 decimal places in calculations and in reporting your answer.)

Answer 1

Answer:

1.

Given,

p^ = x/n = 80/217 = 0.369

Here at 95% CI, z value is 1.96

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

substitute values

= 1.96*sqrt(0.369(1-0.369)/217)

E = 0.064

2.

Point estimate p^ = x/n

= 32/204

= 0.157

3.

p^ = 93/298 = 0.312

SE = sqrt(p^(1-p^)/n)

substitute values

= sqrt(0.312(1-0.312)/298)

= 0.027

4.

p^ = 63/241 = 0.2614

Here at 90% CI, z value is 1.645

Lower limit = p^ - z*sqrt(p^(1-p^)/n)

substitute values

= 0.2614 - 1.645*sqrt(0.2614(1-0.2614)/241)

= 0.2614 - 0.0466

= 0.215

5.

p^ = 48/336 = 0.1429

Here at 95% CI, z value is 1.96

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

substitute values

= 0.1429 + 1.96*sqrt(0.1429(1-0.1429)/336)

= 0.1803