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). 91 90 82 106 97 109 83 87 The sample mean is x ≈ 93.1. 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.

We want to know is Gentle Ben has an overall average glucose level higher than 85, so we need to define the null hypothesis H0 and alternative hypothesis H1 as:

H0: m ≤ 85

H1: m > 85

Where m is the overall average glucose level.

Taking into account that we may assume that x has a normal distribution and we know the standard deviation, we can calculated the statistic as:

[tex]z=\frac{x-m}{s/\sqrt{n}}[/tex]

Where x is the mean of the sample, m is 85, s is equal to σ and n is the size of the sample, so the statistic is equal to:

[tex]z=\frac{93.1-85}{12.5/\sqrt{8}}=1.83[/tex]

Then, we can calculated the p-value using the normal standard table as:

P(z>1.83) = 0.0336

Now, we need to compare the p value with [tex]\alpha[/tex], so we can conclude to reject H0, because [tex]\alpha =0.05[/tex] is bigger than the [tex]p-value=0.0336[/tex].

It means that with a level of significance of 0.05, these data indicate that Gentle Ben has an overall average glucose level higher than 85.