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 51 fish caught, the mean length was x = 18.6 inches, with estimated standard deviation s = 2.6 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.

H₀: μ = 19 in; H₁:  μ < 19 in H₀: μ = 19 in; H₁:  μ ≠ 19 in       H₀: μ < 19 in; H₁:  μ = 19 in H₀: μ > 19 in; H₁:  μ = 19 in H₀: μ = 19 in; H₁:  μ > 19 in

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

The Student's t, since the sample size is large and σ is unknown.The standard normal, since the sample size is large and σ is unknown.     The standard normal, since the sample size is large and σ is known.The Student's t, 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.

P-value > 0.2500.100 < P-value < 0.250     0.050 < P-value < 0.1000.010 < P-value < 0.050P-value < 0.010

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

Answer 1

Part a)

α = 0.05

To Test :-

H0 :- µ = 19
H1 :- µ < 19

Part b)

distribution.

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

Test Statistic :-
t = ( X̅ - µ ) / (S / √(n) )
t = ( 18.6 - 19 ) / ( 2.6 / √(51) )
t = -1.099

Part c)

P - value = P ( t > 1.099 ) = 0.1386
0.100 < P-value < 0.250

part d)

Reject null hypothesis if P value < α = 0.05 level of significance
P - value = 0.1386 > 0.05 ,hence we fail to reject null hypothesis
Conclusion :- Fail to reject null hypothesis

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

Part e)

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