Question

In a fishing lodge brochure, the lodge advertises that 75% of its guests catch northern pike over 20 pounds. Suppose that last summer 58 out of a random sample of 79 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 standard normal, 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 Student's t, 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.

(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

Given : n=79 , X=58

The estimate of the sample proportion is ,

(a) The level of significance =

Hhypothesis : Vs  

(b) Here , np=79*0.75=59.25>5 and nq=79*(1-0.75)=19.75>5

Therefore , the sampling distribution will use is the standard normal , since np>5 and nq>5

The value of the sample test statistic is ,

(c) The p-value is ,

p-value=

; From Z-table

(d) Here , p-value=0.7490 >

Therefore , fail to reject the null hypothesis.

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

(e) Conclusion : 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%.