Question

Gentle Ben is a Morgan horse at a Colorado dude ranch. Over the past 8 weeks, a veterinarian took the following glucose readings from this horse (in mg/100 ml).

92 90 83 105 98 111 86 87

The sample mean is x ≈ 94.0. Let x be a random variable representing glucose readings taken from Gentle Ben. We may assume that x has a normal distribution, and we know from past experience that σ = 12.5. The mean glucose level for horses should be μ = 85 mg/100 ml.† Do these data indicate that Gentle Ben has an overall average glucose level higher than 85? Use α = 0.05.

(a) What is the level of significance?

State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test?

H₀: μ = 85; H₁: μ < 85; left-tailed

H₀: μ = 85; H₁: μ > 85; right-tailed    

H₀: μ = 85; H₁: μ ≠ 85; two-tailed

H₀: μ > 85; H₁: μ = 85; right-tailed

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

The Student's t, since we assume that x has a normal distribution with known σ.

The Student's t, since n is large with unknown σ.   

The standard normal, since we assume that x has a normal distribution with known σ.

The standard normal, since we assume that x has a normal distribution with unknown σ.

Compute the z value of the sample test statistic. (Round your answer to two decimal places.)

(c) Find (or estimate) the P-value. (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?

Correct Answer: 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) State your conclusion in the context of the application.

Correct Answer: There is sufficient evidence at the 0.05 level to conclude that Gentle Ben's glucose is higher than 85 mg/100 ml.

There is insufficient evidence at the 0.05 level to conclude that Gentle Ben's glucose is higher than 85 mg/100 ml.

Answer 1

Solution:

Given:

Sample Size = n = 8

The sample mean is  

x has a normal distribution, and σ = 12.5.

We have to test if these data indicate that Gentle Ben has an overall average glucose level higher than 85.

α = 0.05

Part a)

What is the level of significance?

the level of significance =  α = 0.05

State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test?

Since we have to test mean is higher than 85, this is right tailed test.

Thus:

H₀: μ = 85; H₁: μ > 85; right-tailed    

Part b) What sampling distribution will you use?

The standard normal, since we assume that x has a normal distribution with known σ.

Compute the z value of the sample test statistic.

Part c) Find (or estimate) the P-value.

P-value = P( Z > z )

P-value = P( Z > 2.04)

P-value = 1 - P( Z < 2.04)

Look in z table for z = 2.0 and 0.04 and find corresponding area.

P( Z< 2.04) = 0.9793

Thus

P-value = 1 - P( Z < 2.04)

P-value = 1 - 0.9793

P-value = 0.0207

Sketch the sampling distribution and show the area corresponding to the P-value.

Part 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?

Since P-value = 0.0207 < 0.05 level of significance , we reject null hypothesis H0.

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

Part e) State your conclusion in the context of the application.

There is sufficient evidence at the 0.05 level to conclude that Gentle Ben's glucose is higher than 85 mg/100 ml.