Question

In a fishing lodge brochure, the lodge advertises that 75% of its guests catch northern pike over 20 pounds. Suppose that last summer 65 out of a random sample of 88 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 α = 0.05. (a) What is the level of significance? State the null and alternate hypotheses. H0: p = 0.75; H1: p ≠ 0.75 H0: p < 0.75; H1: p = 0.75 H0: p = 0.75; H1: p > 0.75 H0: p ≠ 0.75; H1: p = 0.75 H0: p = 0.75; H1: p < 0.75 (b) What sampling distribution will you use? The Student's t, since np > 5 and nq > 5. The Student's t, since np < 5 and nq < 5. The standard normal, since np < 5 and nq < 5. The standard normal, since np > 5 and nq > 5. 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.) Sketch the sampling distribution and show the area corresponding to the P-value. WebAssign Plot WebAssign Plot WebAssign Plot WebAssign Plot (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α? At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant. At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant. At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. (e) Interpret your conclusion in the context of the application. There is sufficient evidence at the 0.05 level to conclude that the true proportion of guests who catch pike over 20 pounds differs from 75%. There is insufficient evidence at the 0.05 level to conclude that the true proportion of guests who catch pike over 20 pounds differs from 75%.

Answer 1

a)

Ho :   p =    0.75
H1 :   p ╪   0.75

b)

The standard normal, since np > 5 and nq > 5

Number of Items of Interest,   x =   65                  
Sample Size,   n =    88                  
                          
Sample Proportion ,    p̂ = x/n =    0.7386                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0462                  
Z Test Statistic = ( p̂-p)/SE = (   0.7386   -   0.75   ) /   0.0462   =   -0.25

c)

p-Value   =   0.8055 [excel formula =2*NORMSDIST(z)]  

d)

Decision:   p value>α ,do not reject null hypothesis

At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

e)

There is insufficient evidence at the 0.05 level to conclude that the true proportion of guests who catch pike over 20 pounds differs from 75%.