Question

1. X is normally distributed with mean 30 and standard deviation 2.5

a) P(X > 33) = Answer

b)  P(X < 32) = Answer

c) P(32 < X < 33) = Answer

d) What value of X cuts off the top 10% of the distribution X=Answer

e) What value of X cuts off the bottom 16% of the distribution X=Answer

2. A recent article in the Daily Telegraph newspaper reported that only one in five new college graduates are able to obtain a job. The reason stated is that there is an overabundance of college graduates and a recession due to Coronavirus.

A survey of 250 college graduates revealed that 77 had jobs. Using a significance level of α=0.05, conduct a hypothesis test to examine if a larger proportion of college graduates have jobs than the newspapers claim.

(a)	State the null hypothesis and the alternate hypothesis. AnswerH0: π ≤ 0.20 ; H1: π > 0.20H0:π = 0.20 ; H1: π ≠ 0.20H0: π ≥ 0.20 ; H1: π < 0.20H0: π >0.20; H1: π = 0.20

(b)	State the decision rule for 0.05 significance level. (State your answer to 2 decimal places) Reject H0 if z is > Answer

(c)	Compute the value of the test statistic. (State your answer to 2 decimal places) Test Statistic = Answer

(d) Testing at the 0.05 level of significance whether the proportion of college graduates that have jobs is larger than the newspapers claim.

Answer Reject/Do not reject H₀. There is Answer sufficient/insufficient evidence to conclude that a larger proportion of college graduates have jobs than the newspapers claim.

3. Twelve sets of identical twins were enrolled in a study to measure the effect of home environment on certain social attitudes. One twin in each set was randomly assigned to a foster environment or a home environment. The twin assigned to the foster environment went to live with a low income family on state welfare for a period of 1 year. At the end of the year, an attitude survey was administered.

Their results follow.

ID	1	2	3	4	5	6	7	8	9	10	11	12
Foster	83	75	72	76	88	80	72	85	70	78	77	71
Home	65	67	75	77	69	65	73	78	70	72	73	79

Perform a dependent samples t-test (paired t-test) to examine if the hypothesis that living in the foster environment leads to higher scores on the attitudinal survey. Use a 5% significance level.

(a)	State the null hypothesis and the alternate hypothesis.
	AnswerH0:μd ≤ 0 ; H1: μd > 0H0:μd = 0 ; H1: μd ≠ 0H0:μd ≥ 0 ; H1: μd < 0H0:μd >0 ; H1: μd = 0
	(2 Marks)

(b)	Compute the value of the test statistic.    (state your answer accurate to 2 decimal places)
	Test Statistic t = Answer
	(2 Marks)

(c)	State the Degrees of Freedom?
	df = Answer
	(2 Marks)

(d)	Look up the critical value tc   (state to 3dp as given in your t-tables)
	tc = Answer
	(2 Marks)

Answer 1

1. X is normally distributed with mean 30 and standard deviation 2.5

a) P[X>33]= P[(X-30)/2.5>(33-30)/2.5]=P[Z>1.2] where Z~N(0,1)

=1-P[Z<1.2]=1-0.8849303=0.1150697 [answer] [using pnorm() function in R]

b) P[X<32]= P[(X-30)/2.5<(32-30)/2.5]=P[Z<0.8] where Z~N(0,1)

=0.7881446 [answer] [using pnorm() function in R]

c) P[32<X<33]= P[(32-30)/2.5< (X-30)/2.5 < (33-30)/2.5]

=P[0.8< Z< 1.2] =P[Z<1.2]-P[Z<0.8]= 0.8849303- 0.7881446=0.0967857 [answer] [using pnorm() function in R]

d) let k be the value that cuts off the top 10% of the distribution, which means

P[X>k]=0.10

or, P[X<k]=0.90

or, P[(X-30)/2.5< (k-30)/2.5] =0.90

or, P[Z< (k-30)/2.5]=0.90=P[Z< 1.281552] [using qnorm() function in R]

or, (k-30)/2.5=1.281552 or, k=30+2.5*1.281552=33.20388 [answer]

e) let r be the value that cuts off the bottom 16% of the distribution, then

P[X<r]=0.16

or, P[(X-30)/2.5<(r-30)/2.5]=0.16

or, P[Z< (r-30)/2.5]=0.16=P[Z<-0.9944579] [using qnorm() function in R]

or, (r-30)/2.5=-0.9944579

or, r=30-2.5*0.9944579=27.51386 [answer]