A researcher claims that the stomachs of blue crabs from Location A contain more

Question

A researcher claims that the stomachs of blue crabs from Location A contain more fish than the stomachs of blue crabs from Location B. The stomach contents of a sample of 1616 blue crabs from Location A contain a mean of 193193 milligrams of fish and a standard deviation of 3535 milligrams. The stomach contents of a sample of 1111 blue crabs from Location B contain a mean of 181181 milligrams of fish and a standard deviation of 4141 milligrams. At alphaequals=0.010.01, can you support the researcher's claim? Assume the population variances are equal. Complete parts (a) through (d) below (b) Find the standardized test statistic for mu 11minusmu 22. tequals=nothingm (Round to three decimal places as needed.) (c) Calculate the P-value. Pequals=nothingm (Round to four decimal places as needed.)

Explanation / Answer

B)

Formulating the null and alternative hypotheses,              
              
Ho:   u1 - u2   <=   0  
Ha:   u1 - u2   >   0  
At level of significance =    0.01          
As we can see, this is a    right   tailed test.      
Calculating the means of each group,              
              
X1 =    193          
X2 =    181          
              
Calculating the standard deviations of each group,              
              
s1 =    35          
s2 =    41          
              
Thus, the pooled standard deviation is given by              
              
S = sqrt[((n1 - 1)s1^2 + (n2 - 1)(s2^2))/(n1 + n2 - 2)]               
              
As n1 =    16   , n2 =    11  
              
Then              
              
S =    37.5153302          
              
Thus, the standard error of the difference is              
              
Sd = S sqrt (1/n1 + 1/n2) =    14.69380667          
              
As ud = the hypothesized difference between means =    0   , then      
              
t = [X1 - X2 - ud]/Sd =    0.816670606   [ANSWER, TEST STATISTIC]      
              
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c)

Also,
          
df = n1 + n2 - 2 =    25          
              
Getting the p value using technology,              
              
p =    0.210915513   [ANSWER]

*************************************************      
As P > 0.01, we   FAIL TO REJECT THE NULL HYPOTHESIS.          

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Hi! If you use another method/formula in calculating the degrees of freedom in this t-test, please resubmit this question together with the formula/method you use in determining the degrees of freedom. That way we can continue helping you! Thanks!

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