# Respond to each classmate 100 words or more. Classmate 1 The major event or scandal during the global recession of 2008 and 2009 was complaints of unethical behaviors by Wall Street executives, finan

Respond to each classmate 100 words or more.

Classmate 1

Solution:

The following is the 95% confidence intervals for the proportion of students that were involved in some type of cheating.

P=.4111

Margin of error = z.025√ (p (1-p)/n = 1.96 √. 4111(1-.4111/90 =.1017

95% confidence interval = .4111 ± .1017 or .3094 to .5128

Male students

p= .4375

Margin of error = z.025√ (p (1-p)/n = 1.96 √. 4375(1-.4375/48 =.1403

95% confidence interval = .4375 ± .1403 or .2972 to .5778

Female students

p= .3810

Margin of error = z.025√ (p (1-p)/n = 1.96 √. 3810(1-.3810/42 =.1469

95% confidence interval = .3810 ± .1469 or .2341 to .5279

The analysis showed that Bayview University has a high rate of cheating among its business students, and adequate precautions needs to be taken to control the wide spread of heating.

Classmate 2

Cheating entails deception and trickery and goes against morals and affects business practices. Example includes militants, who do so to pass a certain type of message. (Marian 2020), “Most business ethics scholars consider that the very idea of cheating is indefensible on moral grounds” (p.1). This violates rules and brings about punishment in business and other area of life. We can see that from the readings in the class text, 36% of student has done any type of cheating (29% are female and 42% are male). 18% has copied form internet and 20% copied on exam.

Proportion of all students involved in any type of cheating (pa) = 0.3556

Critical Z at 95% confidence interval= 1.96

For Males,sample proportion, pcap = 0.4375sample size, n = 48Standard error, SE = sqrt (pcap * (1 – pcap)/n)SE = sqrt (0.4375 * (1 – 0.4375)/48) = 0.0716

Given CI level is 95%, hence α = 1 – 0.95 = 0.05α/2 = 0.05/2 = 0.025, ZC = Z(α/2) = 1.96

CI = (pcap – z*SE, pcap + z*SE)CI = (0.4375 – 1.96 * 0.0716, 0.4375 + 1.96 * 0.0716)CI = (0.2972, 0.5778)

For Females,sample proportion, pcap = 0.3809sample size, n = 42Standard error, SE = sqrt (pcap * (1 – pcap)/n)SE = sqrt (0.3809 * (1 – 0.3809)/42) = 0.0749

Given CI level is 95%, hence α = 1 – 0.95 = 0.05α/2 = 0.05/2 = 0.025, ZC = Z(α/2) = 1.96

CI = (pcap – z*SE, pcap + z*SE)CI = (0.3809 – 1.96 * 0.0749, 0.3809 + 1.96 * 0.0749)CI = (0.2341, 0.5277)