In a fishing lodge brochure, the lodge advertises that 75% of its guests catch n

Question

In a fishing lodge brochure, the lodge advertises that 75% of its guests catch northern pike over 20 pounds. Suppose that last summer 57 out of a random sample of 82 guests did, in fact, catch northern pike weighing over 20 pounds. Does this indicate that the population proportion of guests who catch pike over 20 pounds is different from 75% (either higher or lower)? Use = 0.05.What is the value of the sample test statistic? (Round your answer to two decimal places.) (c) Find the P-value of the test statistic. (Round your answer to four decimal places.)

Explanation / Answer

Formulating the null and alternatuve hypotheses,          
          
Ho:   p   =   0.75
Ha:   p   =/=   0.75
As we see, the hypothesized po =   0.75      
Getting the point estimate of p, p^,          
          
p^ = x / n =    0.695121951      
          
Getting the standard error of p^, sp,          
          
sp = sqrt[po (1 - po)/n] =    0.047818253      
          
Getting the z statistic,          
          
z = (p^ - po)/sp =    -1.147638084   [ANSWER, TEST STATISTIC]

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c)  
          
As this is a    2   tailed test, then, getting the p value, by table/technology,  
          
p =    0.251118001   [ANSWER, P VALUE]

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As P > 0.05, we fail to reject Ho. There is no significant evidence that the population proportion of guests who catch pike over 20 pounds is different from 75%.

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