Question

Because of the outbreak of the coronavirus, students of UM are allowed to opt for Pass/Fail...

Because of the outbreak of the coronavirus, students of UM are allowed to opt for Pass/Fail grade for courses they enroll in the second semester of 2019/2020. An instructor of ISOM2002 believes that there will be more than 35% of the students in the course choosing this option. A random sample of 40 students of ISOM2002 reveals that 16 of them prefer to take the Pass/Fail option. What would be the two hypotheses if we want to test the instructor’s belief?

Select one:

a. H0: π ≥ 0.40 and H1: π < 0.40

b. H0: π ≥ 0.35 and H1: π < 0.35

c. H0: π ≤ 0.35 and H1: π > 0.35

d. H0: π ≤ 0.40 and H1: π > 0.40

A business school placement director wants to estimate the mean annual salaries 5 years after students graduate. A random sample of such graduates found a sample mean of $427,400. A confidence interval for the population mean annual salaries 5 years after students graduate is then estimated to be from $419,020 to $435,780. If the placement director considers that this confidence interval is too wide and would like to have an interval of half width only, what should he do?

Select one:

a. Keeping all others constant, reduce the confidence level by half.

b. Keeping all others constant, decrease the sample size by half.

c. None of the other choices.

d. Keeping all others constant, choose another sample with half sample mean.

Homework Answers

Answer #1

Solution-1:

from the statement believes that there will be more than 35% of the students in the course choosing this option.

Ho:p=0.35

Ha:p>0.35

right tail z test for proportion

c. H0: π ≤ 0.35 and H1: π > 0.35

Solution-2

Sample mean+margin of error=435,780.

margin of error=435,780.-sample mean

=435780.-427400

= 8380

we have

xbar-z*sigma/sqrt(n),xbar+z*sigma/sqrt(n)

if we reduce sample size ,confidence interval becomes wider

(2) is ruked out

if we decrease the confidence level z crit reduces but the interval do not reduce to half

if its with half sample mean to217890

CI is

217890-8380,217890+8380

209510,226270
Half of 419,020 to $435,780 is
209510,217890

which is not same as

209510,226270

c. None of the other choices.

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