Question

OS A company selling anses for new e-commerce computer software advertises that fos using this software obtain on average during the first year, a yield of 115 on the hill investments A random sample of 10 of these franchises produced the yields shown below for the first year of operation 62 94 114 135 57 86 107 92 Assuming that population yields are normal distributed test the company's claim at the significance level Click the icon to view the upper rical values of the Student's distribution syn Rej RH si Determine the value for the appropriate for test Round to tredecimal places and Use a comma to separate aer as needed

Answer 1

Given 10 random samples are as follow ( n= 10 )

6.2 9.4 11.4   8.7 13.5   5.7 8.6 10.7 9.2 8.8

Now we will find sample mean and sample standard deviation

Formula

Sample mean
T
=
n
= ( 6.2 + 9.4 + 11.4+ 8.7+ 13.5 + 5.7 + 8.6 + 10.7 + 9.2+ 8.8 ) / 10

                                          = 92.2 / 10 = 9.22

Sample standard deviation s =

Στί - )2 η – 1

        s = = 2.300628               { after calculation }

{47.636 10-1

       

Hence

= 9.22 ;    s = 2.300628     

T

Given level of significance $\alpha$ = 0.01   ( or 1% )

Since company selling licence for new e-commerce software advertisement that firm obtaine on an average during the first year , a yeild of 11% on their initial investment .

Now we need to test weather a firm obtaine on an average , a yeild of 11% on their initial investment or not.

To Test :

H0 : $\mu$ 0 = 11         { firm obtaine on an average a yeild of 11% on their initial investment }

H1 : $\mu$ 0 $\neq$ 11         { firm may not obtaine on an average a yeild of 11% on their initial investment }

Test Statistics TS :

TS = $\frac{\bar{x}-\mu0}{s/\sqrt{n}}$

      = $\frac{9.22-11}{2.300628 /\sqrt{10}}$

TS = -2.44666

Thus calculated test statistics TS value is -2.44666

Rejection Criteria :-

Since we need to given licence to those firm which obtains on an average , a yeild of 11% on their initial investment ,

{ Note that it is mentioned that it will give licence to firm obtains a yeild of more than 11% / less than 11% , so this is two-tail test , if not we would be testing one-tail test }

So for this two tail test Rejection criteria would be

Reject H0 if | TS | > $t_{n-1,\alpha /2}$

i.e If   TS > $t_{n-1,\alpha /2}$     or   TS < - $t_{n-1,\alpha /2}$                

Where TS = $\frac{\bar{x}-\mu0}{s/\sqrt{n}}$

{ Note that in above given image not able to see all given options , but look like option (a) might be something like this i.e

option (a) is : Reject H0   if   $\frac{\bar{x}-\mu0}{s/\sqrt{n}}$ > $t_{n-1,\alpha /2}$     or   $\frac{\bar{x}-\mu0}{s/\sqrt{n}}$ < - $t_{n-1,\alpha /2}$ ,

So this would be correct option }

To find t-critical value :-

Since this is two-tail test , Then its t-critical value would be given by $t_{n-1,\alpha /2}$

Here $t_{n-1,\alpha /2}$ is t-distributed with n-1 = 10-1 = 9 degree of freedom at $\alpha$ = 0.01 .

Since above table of critical values of students t-distribution is not visible in given image , but it can be obtained from any software like R/Excel or from statistical table , finding critical value corresponding to 9 degree of freedom at $\alpha$ /2 = 0.01/2 = 0.005

Thus its critical value is $\boldsymbol{t_{n-1,\alpha /2}}$ = 3.250        { from table for DF = 9}

Conclusion :-

Since Test statistics value | TS | = | -2.44666 | = 2.44666 < 3.250

i.e    TS = $\frac{\bar{x}-\mu0}{s/\sqrt{n}}$ = -2.44666 > -3.250        ( - $t_{n-1,\alpha /2}$ = -3.250 )

also TS = $\frac{\bar{x}-\mu0}{s/\sqrt{n}}$ = -2.44666 < 3.250 ( $t_{n-1,\alpha /2}$ = -3.250 )

So we do not reject null hypothesis at 1% of level of significance .