Question

1. Estimation of Mean Monthly Rentals

a. Assume a sample of 10 monthly rentals from a large housing complex of 300,000 units is: 980, 1230, 1320, 1440, 1180, 1050, 1520, 1630, 880, 920. Suppose the landlord association claims that the tue average rental of the low income 31/2 room apartment is $1100. The housing authority believes that the average rental is higher. State and test the null hypothesis of the landlord association's claim versus the alternative hypothesis of the housing authority claim at the a = 0.05 significance level. Does the data support the landlord's association claim? What would you report to the housing authority?
b. Suppose that a sample of 10,000 apartments from a much larger complex of apartments, with identical characteristics as the complex described in question 1, yields the following results: X bar = 1110, Standard deviation = 250. Using this data, what are your conclusions? Use a = 0.05 (or your favorite type I error). What would you report to the housing authority?

2. Applications to Auditing

a. Suppose it is claimed that more efficient record keeping has resulted in a lower error rate in accounting entries than the 3% which has been true in the past. The null hypothesis is that the error rate is .03; the alternate hypothesis is that the error rate is greater than 0.03. In a sample of 15 entries, 2 errors were found. Does the data support the claim in the null hypothesis at the a = 0.05 significance level?
b. In a large study of 1000 entries, 40 errors were found. Does the data support the null hypothesis at the a = 0.05 significance level?

Answer 1

a)

Sample of 10 monthly rentals

Hence n = 10   and =3.162278

Assume a sample of 10 monthly rentals from a large housing complex of 300,000 units is: 980, 1230, 1320, 1440, 1180, 1050, 1520, 1630, 880, 920.

Landlord association claims that the tue average rental of the low income 31/2 room apartment is $1100

The housing authority believes that the average rental is higher.

Thus ,hypothesis to test are

H0 : = 1100       ( there is no differnce between Landlord association claims )

H1 : > 1100      ( average rental is higher that Landlord association claims )

Given a = 0.05 significance level.

we will perform one smple t-test

Test Statistics T.S -

T.S =

Calculation -

Sample Mean =

= (980+ 1230+1320+1440+1180+ 1050+ 1520+1630+ 880+ 920)/10

    = 12150 / 10

= 1215

Sample Variance s2 =

s2 = [ (980-1215)2+ (1230-1215)2+(1320-1215)2+.........+(1630-1215)2+(880-1215)2+(920-1215)2 ] / (10-1 )

s2 = 610050 / 9

s2 = 67783.33

Thus Sample standard deviation s = (67783.33)1/2   = 260.3523

Thus s =260.3523 , = 1215

and   =3.162278, = 1100

Thus test statistics will be

T.S = = = 1.396807

T.S = 1.396807

Critical Region -

We reject null hypothesis is calclated test statistics value T.S is greater than t-table value

To calculate t-value with n-1 = 9 degree of freedom

t-value = = 1.833113

We can calculated respective t-value from statistical book or from any software

{

> qt(1-0.05,9)        # calculating t-value from R
[1] 1.833113

}

Thus t-value = = 1.833113

Now   T.S = 1.396807 < 1.833113

                    or T.S < t-value

Hence we do not reject null hypothesis at 5 % of level of significance .

So we conclude that Landlord association claims may be true

b. Suppose that a sample of 10,000 apartments from a much larger complex of apartments, with identical characteristics as the complex described in question 1, yields the following results: X bar = 1110, Standard deviation = 250. Using this data, what are your conclusions

Thus given , s =250 , = 1110 and n = 10,000

to test

H0 : = 1100       ( there is no differnce between Landlord association claims )

H1 : > 1100      ( average rental is higher that Landlord association claims )

Test Statistics T.S -

T.S =

       =

T.S   = 4

Critical value would be -

t-value with n-1 = 10000-1 degree of freedom

{

> qt(1-0.05,10000-1)             # calculating t-value from R
[1] 1.645006

}

Now   T.S = 4 > 1.645006

                    or T.S > t-value

So we reject null hypothesis at 5% of level of significance .

Thus , Using this data, what are your conclusions is that we reject null hypothesis and average rental may be higher that Landlord association claims

We would report to the housing authority that average rental is highe landlord's calim.

a. Suppose it is claimed that more efficient record keeping has resulted in a lower error rate in accounting entries than the 3% which has been true in the past. The null hypothesis is that the error rate is .03; the alternate hypothesis is that the error rate is greater than 0.03. In a sample of 15 entries, 2 errors were found. Does the data support the claim in the null hypothesis at the a = 0.05 significance level?

Now ,there are sample of n=15 entries , x= 2 errors were found.

i.e n = 15 , x = 2

thus = x / n = 2 /15 = 0.1333333      ( proportion estimated from sample )

Thus ,hypothesis to test are

H0 : po = 0.03    ( error rate is .03 )

H1 : po > 0.03      ( error rate is greater than 0.03)

Test Statistics T.S

T.S =

where SE =

calculation - SE = = = 0.007703702

                       SE = 0.00194

    Thus            =   ( 0.00194)1/2 = 0.04404543

Hence = 0.04404543

Now

T.S = = = 2.346062

T.S = 2.346062

Critical Region -

We reject null hypothesis is calclated test statistics value T.S is greater than Z-value

a = 0.05 significance level    Z-value = 1.64

Since calculated value of test statistics is greater than 1.64

That is T.S = 2.346062 > Z-value = 1.64

Hence , It is concluded that the null hypothesis H0 is rejected at 5% of level of significnce .

So , the data do not support the claim in the null hypothesis at the a = 0.05 significance level

b. In a large study of 1000 entries, 40 errors were found. Does the data support the null hypothesis at the a = 0.05 significance level?

Now ,there are sample of n=15 entries , x= 2 errors were found.

i.e n = 1000 , x = 40

thus = x / n = 40 /1000 = 0.04    ( proportion estimated from sample )

         = 0.04

To test

H0 : po = 0.03    ( error rate is .03 )

H1 : po > 0.03      ( error rate is greater than 0.03)

Test Statistics T.S

T.S =

where SE =

calculation - SE = = = 0.0291/1000

                    

    Thus            =   ( 0.0291/1000)1/2 = 0.005394442

Hence = 0.005394442

Now

T.S = = = 1.85376

T.S = 1.85376

Since calculated value of test statistics is greater than 1.64

That is T.S = 1.85376 > Z-value = 1.64

Hence , It is concluded that the null hypothesis H0 is rejected at 5% of level of significnce .

So , the data do not support the claim in the null hypothesis at the a = 0.05 significance level