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A study is conducted to find out whether the wait times at two local banks are different. The sample statistics are listed be
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Answer #1

in above question , we need to test

To Test :-

H0 : u1 = u2               [ wait time at two local bank do not differ significantly , i.e they are same ]

H1 : u1 \neq u2              [ wait time at two local bank differ significantly , i.e they are not same ]

Given level of significance \alpha = 0.05 , and we assume equal variance i.e \sigma 1 = \sigma 2

Test Statistics TS :-

TS = \frac{\bar{x1}-\bar{x2}}{SE}

where

S.E = \sqrt{sd^{2}/n1 +sd^{2}/n2 }

and sd^2 = \frac{(n1-1)*s1^{2}+(n2-1)*s2^{2}}{n1+n2-2}

From Above given data we have

   sd^2 = \frac{(n1-1)*s1^{2}+(n2-1)*s2^{2}}{n1+n2-2}

           = \frac{(15-1)*1.1^{2}+(16-1)*1.0^{2}}{15+16-2}

   sd^2 =1.101379           { after calculation }

Thus

S.E = \sqrt{sd^{2}/n1 +sd^{2}/n2 }

       = \sqrt{1.101379 /15 +1.101379 /16 }

S.E = 0.3771756

hence we have

TS = \frac{\bar{x1}-\bar{x2}}{SE}

     = \frac{5.3-5.6}{0.3771756}

TS = -0.7953855

So , calculated test statistics value is TS = -0.7953855

Rejection criteria :-

We reject null hypothesis if absolute value of calculated test statistics is greater than t-critical value t_{\alpha /2}

i.e Reject H0 if | TS | > t_{\alpha /2}

where t_{\alpha /2} is t-distributed with n1+n2-2 = 15+16-2 = 29 degree of freedom at \alpha = 0.05

It can be computed from from statistical book or more accuratly from any software like R,Excel

From R

> qt(1-0.05/2,df=29)
[1] 2.04523

Thus t-critical value t_{\alpha /2} = 2.04523

Now | TS |= | -0.7953855 | = -0.7953855   < 2.04523

i.e   | TS | < t_{\alpha /2}

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

Conclusion :-

Since we do not have enough evidence against null hypothesis H0 , so we conclude that wait time at two local bank may not be different

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