Question

In: Statistics and Probability

Independent random samples of times served for 10 released prisoners for fraud offense showed a mean...

Independent random samples of times served for 10 released prisoners for fraud offense showed a mean of 10.12 months and a standard deviation of 4.90 months whereas times served for 10 released prisoners for firearms offense showed a mean of 18.78 months and a standard deviation of 4.64 months. a) At  = 0.05, do the data suggest that mean time served for fraud is less than that of firearm offenses? Assume that times are normally distributed in both groups and have same variances. (Use the critical value approach and the P-value approach) b) Find a 90% confidence interval for the difference between the mean times served by prisoners in the fraud and firearms offense categories. c) Use the Mann-Whitney test at  = 0.05 using the data below: Fraud: 3.6 5.3 10.7 8.5 11.8 17.9 5.9 7.0 13.9 16.6 Firearms: 25.5 10.4 18.4 19.6 20.9 23.8 17.9 21.9 13.3 16.1

Solutions

Expert Solution


Related Solutions

Independent random samples of released prisoners in the fraud and firearms offense categories yielded the given...
Independent random samples of released prisoners in the fraud and firearms offense categories yielded the given information on time served, in months. At the 1% significance level, do the data provide sufficient evidence to conclude that the mean time served for fraud is less than that for firearms offenses? Assume that populations standard deviations are same for both group. (i.e., we assume σ1 = σ2 .) What is the parameter of interest? What is the underlying distribution? Fraud: 15.2, 11.2,...
URGENT!! Independent random samples of the selling prices (in 10 000 TL ) of houses in...
URGENT!! Independent random samples of the selling prices (in 10 000 TL ) of houses in three popular districts of four big cities were taken. The selling prices are as shown below. Assume that the populations from which the sample drawn are normally distributed. (use 2 digits after decimal point) District A District B District C Ankara 45 52 54 İstanbul 46 51 58 İzmir 40 60 50 Bursa 37 49 46 1. Complete the following analysis of variance table....
In a controlled laboratory environment, independent random samples of 10 adults and 10 children were tested...
In a controlled laboratory environment, independent random samples of 10 adults and 10 children were tested by a psychologist to determine the room temperature that each person finds most comfortable. The study provided the following results:   Adults     Children Sample size    10 10 Sample mean (in degrees)   77.5              74.5 Sample variance 4.5    2.5 Which one of the following is the correct 99% confidence interval for the true difference in population mean temperatures that adults and children find most comfortable?
Given two independent random samples with the following results:
Given two independent random samples with the following results: n1=170x1=36    n2=123x2=65 Use this data to find the 90% confidence interval for the true difference between the population proportions.Step 1 of 4 : Find the values of the two sample proportions, pˆ1 and pˆ2. Round your answers to three decimal places.Step 2 of 4: Find the margin of error. Round your answer to six decimal places.Step 3 of 4: Construct the 90% confidence interval. Round your answers to three decimal places.
Table 2 shows two random independent samples of stock return over 10 years’ time. Assume the...
Table 2 shows two random independent samples of stock return over 10 years’ time. Assume the rate of return ?? for stock A and ?? for stock B are normally distributed with population standard deviation ?? = 0.01 and ?? = 0.02 respectively. To investigate whether the average rate of return for stock A different from the average rate of return for stock B, please answer the following questions. State the null hypothesis and the alternative hypothesis. Select an appropriate...
Two random samples are selected from two independent populations. A summary of the samples sizes, sample...
Two random samples are selected from two independent populations. A summary of the samples sizes, sample means, and sample standard deviations is given below: n1=37,n2=44,x¯1=58.9,x¯2=74.7,s1=5.5s2=10.1 n 1 =37, x ¯ 1 =58.9, s 1 =5.5 n 2 =44, x ¯ 2 =74.7, s 2 =10.1 Find a 95.5% confidence interval for the difference μ1−μ2 μ 1 − μ 2 of the means, assuming equal population variances. Confidence Interval
Two random samples are selected from two independent populations. A summary of the samples sizes, sample...
Two random samples are selected from two independent populations. A summary of the samples sizes, sample means, and sample standard deviations is given below: n1=39,n2=48,x¯1=52.5,x¯2=77.5,s1=5s2=11 Find a 97.5% confidence interval for the difference μ1−μ2 of the means, assuming equal population variances. Confidence Interval =
A random sample of 20 purchases showed the amounts in the table​ (in $). The mean...
A random sample of 20 purchases showed the amounts in the table​ (in $). The mean is ​$50.76 and the standard deviation is ​$19.48 ​a) Construct a 80​% confidence interval for the mean purchases of all​ customers, assuming that the assumptions and conditions for the confidence interval have been met. ​b) How large is the margin of​ error? ​c) How would the confidence interval change if you had assumed that the standard deviation was known to be $20? 60.91 64.3...
A random sample of 20 purchases showed the amounts in the table (in $). The mean...
A random sample of 20 purchases showed the amounts in the table (in $). The mean is $49.88 and the standard deviation is $20.86. B) What is the margin of error ?A) Construct a 80% confidence interval for the mean purchases of all customers, assuming that assumptions and conditions for confidence interval have been met. (What is the confidence interval) ? C) How would the confidence interval change if you had assumed that the standard deviation was known to be...
A random sample of 2020 purchases showed the amounts in the table​ (in $). The mean...
A random sample of 2020 purchases showed the amounts in the table​ (in $). The mean is ​$49.8149.81 and the standard deviation is ​$21.9521.95. ​a) Construct a 8080​% confidence interval for the mean purchases of all​ customers, assuming that the assumptions and conditions for the confidence interval have been met. ​b) How large is the margin of​ error? ​c) How would the confidence interval change if you had assumed that the standard deviation was known to be ​$2222​? A) The...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT