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).

93

88

84

106

99

109

83

87

The sample mean is x ≈ 93.6. 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?

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

b.H₀: μ = 85; H₁: μ < 85; left-tailed    

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

d.H₀: μ = 85; H₁: μ ≠ 85; two-tailed

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

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

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

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

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

What is the 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.)

Answer 1

(a) What is the level of significance?

Level of significance=0.05

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

Ho:mu=85

Ha:mu>85

Right tail z test

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

since population standard deviation is known ,use z statistic

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

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

z=xbar-mu/sigma/sqrt(n)

z=(93.6-85)/(12.5/sqrt(8))

z=1.945958

z=1.95

test statistic,

z=1.95

Solution-d:

p avlue in excel for left tail

=NORMSDIST(1.95)

=0.974412

1-0.974412=0.025588

ANSWER:

p-value=0.0255