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). 91 86 80 107 97 111 84 87

The sample mean is x ≈ 92.9. 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.

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

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

What steps are needed to find these answers?

Thanks!

Answer 1

Solution :

= 85

= 92.9

= 12.5

n = 8

This is the right tailed test .

The null and alternative hypothesis is ,

H₀ :   = 85

H_a : >  1

significance level is α=0.05, and the critical value for a right-tailed test is z_c ​=1.64.

Test statistic = z

= ( - ) / / n

= (92.9 - 85) /12.5 / 8

= 1.79

P(z >1.79 ) = 1 - P(z < 1.79 ) = 0.0369

= 0.05

p = 0.0369 < 0.05, it is concluded that the null hypothesis is rejected.

Reject the null hypothesis .

There is enough evidence to claim that the population mean μ is greater than 85, at the 0.05 significance level.