Question

Pyramid Lake is on the Paiute Indian Reservation in Nevada. The lake is famous for cutthroat trout. Suppose a friend tells you that the average length of trout caught in Pyramid Lake is μ = 19 inches. However, a survey reported that of a random sample of 46 fish caught, the mean length was x = 18.6 inches, with estimated standard deviation s = 3.1 inches. Do these data indicate that the average length of a trout caught in Pyramid Lake is less than μ = 19 inches? Use α = 0.05.

(a) What is the level of significance?


State the null and alternate hypotheses.

A.)H₀: μ = 19 in; H₁:  μ > 19 in

B.)H₀: μ = 19 in; H₁:  μ ≠ 19 in       

C.)H₀: μ = 19 in; H₁:  μ < 19 in

D.)H₀: μ > 19 in; H₁:  μ = 19 in

E.)H₀: μ < 19 in; H₁:  μ = 19 in

(b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.

A.)The Student's t, since the sample size is large and σ is unknown.

B.)The Student's t, since the sample size is large and σ is known.     

C.)The standard normal, since the sample size is large and σ is unknown.

D.)The standard normal, since the sample size is large and σ is known.

What is the value of the sample test statistic? (Round your answer to three decimal places.)


(c) Estimate the P-value.

A.)P-value > 0.250

B.)0.100 < P-value < 0.250     

C.)0.050 < P-value < 0.100

D.)0.010 < P-value < 0.050

E.)P-value < 0.010

(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?

A.)At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.

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

C.)At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.

D.)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.

A.)There is sufficient evidence at the 0.05 level to conclude that the average fish length is less than 19 inches.T

B.)here is insufficient evidence at the 0.05 level to conclude that the average fish length is less than 19 inches.  

Answer 1

(a)

The level of significance is α = 0.05

Hypotheses are:

C.)H0: μ = 19 in; H1:  μ < 19 in

(b)

A.)The Student's t, since the sample size is large and σ is unknown.

Here we have X =18.6, S=3.1, n =46

Test Statistics: t= X-u = = (18.6-19)/(3.1/sqrt(46)) (18. 5 = -0.875

(c)

Level of significance: a =0.05 Test is one tailed (left tailed). Degree of freedom: df =n-1=45

Using table, the bounds for p-value are:

B.)0.100 < P-value < 0.250     

(d)

Decision: Since p-value is not less than level of significance so we fail to reject the null hypothesis

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

(e)

B.)here is insufficient evidence at the 0.05 level to conclude that the average fish length is less than 19 inches.