Question

1.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

89

80

103

98

109

84

91

The sample mean is x ≈ 93.4. 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).Compute the z value of the sample test statistic. (Round your answer to two decimal places.)


(b) Find (or estimate) the P-value. (Round your answer to four decimal places.)

2. Total blood volume (in ml) per body weight (in kg) is important in medical research. For healthy adults, the red blood cell volume mean is about μ = 28 ml/kg.† Red blood cell volume that is too low or too high can indicate a medical problem. Suppose that Roger has had seven blood tests, and the red blood cell volumes were as follows.

34

27

39

37

28

37

28

The sample mean is x ≈ 32.9 ml/kg. Let x be a random variable that represents Roger's red blood cell volume. Assume that x has a normal distribution and σ = 4.75. Do the data indicate that Roger's red blood cell volume is different (either way) from μ = 28 ml/kg? Use a 0.01 level of significance.

(a) What is the level of significance?

(b)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.)

Answer 1

We would be looking at first question here both the parts.

Q1) a) The sample test statistic here is computed as:

Therefore 1.90 is the required test statistic value here.

b) As we are testing here that Gentle Ben has an overall average glucose level higher than 85, therefore this is a one tailed test. The p-value here is computed from the standard normal tables here as:

p = P(Z > 1.90) = 0.0287

Therefore 0.0287 is the required p-value here.