MA 123 Quiz 3 Sample
Spring, 2012

1. (10 pts)  A researcher claims that adult hogs fed a certain diet will have an average weight of 200 pounds, with a population standard deviation of 3 pounds, but a consumer group thinks the claim is wrong. To check the claim, a sample of 9 hogs is put on the diet and is found to have an average weight of 198 pounds. Does the consumer group have sufficient evidence to reject the researcher’s claim? Use a 5% significance level, and estimate the P-value of the test.
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   

2. (10 pts) At the College of Weber City, an SRS of 25 students was asked how much time they spent per week in the library.  The sample average was 16 hours, and the population standard deviation is 1 hour.  Provide a 98% confidence interval for the mean time spent in the library.

Answers

1. See p. 371.  α = 0.05, μ₀ = 200 lbs., σ = 3 lb, x̅ = 198 lbs and n = 9.  Thus H₀ is μ = 200 lbs and H_a is μ ≠ 200 lbs. Then z = ( x̅ -  μ₀) / (σ /√n ) = -2, so P-value = 2 · P(Z ≥ |-2|) = 2*P(Z ≤ -2) = 2 · 0.0228 = 0.0456 < α = 0.05.  Thus we reject the null hypothesis, i.e., that the average weight of hogs on a certain diet is 200 lbs.
   

2. See p. 349.  C = 98%, n = 25, x̅ = 16 hrs, σ = 1 hr.  Look z* up in Table D based on the value of C, obtaining z* = 2.326. Then calculate m = z* σ / √n  = 2.326 ·1 / √25 = 0.4652.  The 98% confidence interval we obtain is x̅ ± m = 16 ± 0.4652.  Another way to write the confidence interval is (15.5348, 16.4652).