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vitfil [10]
3 years ago
6

Random samples of of outh bass and smallmouth bass were taken from a lake, and their lengths in millimeters) were deter mined. W

e 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
Mathematics
1 answer:
KatRina [158]3 years ago
8 0

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:

\=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{\frac{s^{1}}{n_1} + \frac{s_1^{2}}{n_2}}}

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

T=\frac{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.

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Step-by-step explanation:

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Answer: y=-8

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Suppose that a recent poll found that 49​% of adults believe that the overall state of moral values is poor. Complete parts​ (a)
Lunna [17]

Answer:

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Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they believe that the overall state of moral values is poor, or they do not believe this. The probability of an adult believing this is independent of other adults. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

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The standard deviation of the binomial distribution is:

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In this problem, we have that:

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So

E(X) = np = 250*0.49 = 122.5

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