Question

Random samples of of outh bass and smallmouth bass were taken from a lake, and their lengths in millimeters) were deter mined. We wish to know if the mean standard length differs between the two species in this lake. The results were as follows:

Largemouth Bass Smallmouth Bass
x 164.8 272.8
s 96.4 40.0
n 97 125

Answer 1

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:

`\=x_1 =164.8`

`s_1=96.4`

`n_1=125`

Smallmouth Bass:

`\=x_2 =272.8`

`s_2=40`

`n_2=97`

Assume

`\alpha =0.05`

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

`T=frac{( \=x_2 - \=x_1 )}{\sqrt{(s^(1))/(n_1) + (s_1^(2))/(n_2)}}`

`T =\frac{(272.8 - 164.8)}{\sqrt{(96.4^(2))/(125) + (40^(2))/(97)}}`

`T=(108)/(9.530925)`

`T=11.33`

Therefore

From table

Critical Value

`T_(\alpha,n_2-1)`

`T_(0.05,96)=1.661`

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.