Answer:
Hence , the means of Largemouth Bass and Smallmouth Bass are significantly different.
Step-by-step expl;anation:
From the question we are told that:
Largemouth Bass:
[tex]\=x_1 =164.8[/tex]
[tex]s_1=96.4[/tex]
[tex]n_1=125[/tex]
Smallmouth Bass:
[tex]\=x_2 =272.8[/tex]
[tex]s_2=40[/tex]
[tex]n_2=97[/tex]
Assume
[tex]\alpha =0.05[/tex]
Generally The hypothesis is given as
H_0: The Largemouth Bass and Smallmouth Bass are equal
H_1: The Largemouth Bass and Smallmouth Bass are not equal
Generally the equation for Test statistics is mathematically given by
[tex]T=frac{( \=x_2 - \=x_1 )}{\sqrt{\frac{s^{1}}{n_1} + \frac{s_1^{2}}{n_2}}}[/tex]
[tex]T =\frac{(272.8 - 164.8)}{\sqrt{\frac{96.4^{2}}{125} + \frac{40^{2}}{97}}}[/tex]
[tex]T=\frac{108}{9.530925}[/tex]
[tex]T=11.33[/tex]
Therefore
From table
Critical Value
[tex]T_{\alpha,n_2-1}[/tex]
[tex]T_{0.05,96}=1.661[/tex]
Conclude
Since 11.33 is greater that 1.661 we eject the null hypothesis that the means are the same.
Hence , the means of Largemouth Bass and Smallmouth Bass are significantly different.
