Question

1) Construct 90%, 95%, and 99% confidence intervals for the population mean of total daily sales.

2) Run a hypothesis test on the population mean for total daily sales. Use a hypothesized value of $1600 and test at levels of significance of 0.01, 0.05, and 0.10. Use both the critical value and p-value approaches when testing at each level of significance.

TABLE 4:TOTAL DAILY SALES

$1,800	$975	$1,814	$1,421	$1,335
$1,273	$1,592	$1,305	$1,103	$1,105
$1,438	$1,219	$1,661	$1,692	$1,709
$1,584	$1,111	$1,613	$1,547	$1,715
$1,384	$1,595	$1,139	$2,017	$2,014
$1,497	$1,820	$1,770	$1,799	$1,386
$1,368	$1,506	$1,852	$1,642	$1,355
$1,199	$1,416	$1,716	$982	$1,851
$1,649	$1,864	$1,532	$1,318	$1,367
$1,939	$1,384	$1,867	$1,389	$1,152

Answer 1

The given data is as follow

TABLE 4:TOTAL DAILY SALES

$1,800	$975	$1,814	$1,421	$1,335
$1,273	$1,592	$1,305	$1,103	$1,105
$1,438	$1,219	$1,661	$1,692	$1,709
$1,584	$1,111	$1,613	$1,547	$1,715
$1,384	$1,595	$1,139	$2,017	$2,014
$1,497	$1,820	$1,770	$1,799	$1,386
$1,368	$1,506	$1,852	$1,642	$1,355
$1,199	$1,416	$1,716	$982	$1,851
$1,649	$1,864	$1,532	$1,318	$1,367
$1,939	$1,384	$1,867	$1,389	$1,152

n = 50     ( total observation )

Now we will calculate sample mean and variance of given data

Sample Mean :-

=

= ( 1800 + 975 +1814+ 1421+ 1335+ 1273 + ......... + 1939 +1384 +1867 +1389 +1152)/50

    = 75781 / 50 = 1515.62

= 1515.62

Sample Variance s² :-

s² =

     = { (1800-1515.62)²+(975-1515.62)²+(1814-1515.62)²+ ........+(1389-1515.62)²+(1152-1515.62)²} / (50-1)

     = 3667200 / 49 = 74840.81

s² = 74840.81

Hence sample standard deviation s will be

s = = 273.5705

Now we have

= 1515.62 ; s = 273.5705 ; n = 50

Now 100(1-)% confidence interval is given by

CI = { - * ,   + * }

where is t-distributed , with n-1 = 50-1 = 49 degree of freedom and level of significance .

We will find for = 0.1 , 0.05 and 0.01 i.e for 90%, 95%, and 99% confidence respectively .

We will R-Software to find t-critical value , any other softaware can be use or statistical book can be used for the same

• for = 0.1 i.e for 100(1-)%   = 90% confidence

> qt(1-0.1/2,df=49)
[1] 1.676551

Hence for 90% confidence we have = = 1.676551

• for = 0.05 i.e for 100(1-)%   = 95% confidence

> qt(1-0.05/2,df=49)
[1] 2.009575

Hence for 95% confidence we have = = 2.009575

• for = 0.01 i.e for 100(1-)%   = 99% confidence

> qt(1-0.01/2,df=49)
[1] 2.679952

Hence for 99% confidence we have = = 2.679952

Also = 1515.62 ; s = 273.5705 ; n = 50

So = = 38.68871

Thus

Now 100(1-)% confidence interval is given by

CI = { - * ,   + * }

CI = { 1515.62    - * 38.68871 ,   1515.62 + * 38.68871 }

• 90% confidence interval

CI = { 1515.62    - 1.676551 * 38.68871 ,   1515.62 + 1.676551 * 38.68871 }

CI = { 1450.756 , 1580.484 }

• 95% confidence interval

CI = { 1515.62    - 2.009575 * 38.68871 ,   1515.62 + 2.009575 * 38.68871 }

CI = { 1437.872 , 1593.368 }

• 99% confidence interval

CI = { 1515.62    - 2.679952 * 38.68871 ,   1515.62 + 2.679952 * 38.68871 }

CI = { 1411.936 , 1619.304 }

So we have

100(1-)% confidence intervals for the population mean of total daily sales as follow.

1. 90% confidence interval : { 1450.756 , 1580.484 }
2. 95% confidence interval : { 1437.872 , 1593.368 }
3. 99% confidence interval : { 1411.936 , 1619.304 }

2) Run a hypothesis test on the population mean for total daily sales. Use a hypothesized value of $1600 and test at levels of significance of 0.01, 0.05, and 0.10. Use both the critical value and p-value approaches when testing at each level of significance.

To test :-

H0 :                      { given hypothesized value }

H1 :                      { the mean is significantly different from 1600 }

Test statistics TS :-

TS =

   =               { using abova calculated values of and }

TS = -2.180998

Thus calculated Test statistics value is TS = -2.180998

To calculate P-Value

It is calcuated as follow

P-Value = Pr ( t < - |TS| ) + Pr ( t > |TS| )

P-Value = Pr ( t < -2.180998 ) + Pr ( t > 2.180998 )

P-Value = Pr ( t < -2.180998 ) + [ 1 - Pr ( t < 2.180998 ) ]

where t is t-distributed with n-1 = 49 degree of freedom

Now we need to calculate Pr ( t < -2.180998 ) and Pr ( t < 2.180998 )

We will use R-Software to calculate required probabilites using "pt()" command as follow

> pt(-2.180998,49)+(1-pt(2.180998,49))             # Pr ( t < -2.180998 ) + [ 1 - Pr ( t < 2.180998 ) ]
[1] 0.03401257

Thus Pr ( t < -2.180998 ) + [ 1 - Pr ( t < 2.180998 ) ] = 0.03401257

Hence P-Value = 0.03401257

Thus

Test Statistics Value TS = -2.180998 and   P-Value = 0.03401257

Criteria:-

1. Using Critical value approach

We reject null hypothesis if absolute of calculated Test statistics value is greater than t-critical value .

i.e | TS | >

Now we have already calculated t-critical values for levels of significance of 0.01, 0.05, and 0.10 which were as follow

for = 0. 1     :- = 1.676551

for = 0. 05 :- = 2.009575

for = 0. 01 :- = 2.679952

• a) at levels of significance of 0.1

|TS| = |-2.180998| = 2.180998 > 1.676551 ( )

Hence | TS | >

i.e test statistics value is greater than t-critical value , so we reject null hypothesis at 0.1 level of significance

• b) at levels of significance of 0.5

|TS| = |-2.180998| = 2.180998 > 2.009575( )

Hence | TS | >

i.e test statistics value is greater than t-critical value , so we reject null hypothesis at 0.05 level of significance

• c) at levels of significance of 0.01

|TS| = |-2.180998| = 2.180998 < 2.679952( )

Hence | TS | <

i.e test statistics value is less than t-critical value , so we fail to reject null hypothesis at 0.01 level of significance

Thus using critical value approach we reject null hypothesis H0 , at levels of significance 0.05, and 0.10 but we fail to reject H0 at 0.01 level of significance

2. Using p-value value approach

Using P-value , we reject null hypothesis is P-Value is less than given level of significance .

Now P-Value = 0.03401257

For given level of significance of 0.01, 0.05, and 0.10

P-Value = 0.03401257 > 0.01         { Reject H0 , since P-Value greater than given significance = 0.01 }

P-Value = 0.03401257 < 0.05         { Fail to Reject H0 , since P-Value less than given significance = 0.01 }

P-Value = 0.03401257 < 0.10         { Fail to Reject H0 , since P-Value less than given significance = 0.01 }

  

Conclusion :-

Using both the critical value and p-value approaches when testing at each level of significance , we conclude that hypothesized value of $1600 i.e null hypothesis get rejected at significance level = 0.05 and = 0.10 but we fail to reject ( or do not reject ) the null hypothesis at significance level = 0.01