Assume that the mean length of time required to complete the Columbus Marathon was 4.5 hours and that the standard deviation of the times was 0.70 hours.  Assume that the racing times were approximately normally distributed.  What is the probability that a randomly selected runner completed the race in less than 3.8 hours?

A college professor wants to estimate the difference in mean test scores of students who have taken his statistics and genetics classes in the past 10 years.  He selects a random sample of 20 student records from the statistics course and a random sample of 22 student records from the genetics course.  These two samples were independent random samples.  The study provided the results shown in the table below.  Calculate the pooled estimate of the variance.   Statistics Genetics Sample size 20 22 Sample mean 78 75 Sample standard dev. 10 12

The weights of 6 dogs (in pounds) are as follows:  30  32  28  42  40  44 Find the median.

Find the probability of an observation lying more than z = 2.25 standard deviations below the mean.

The average weight of a kennel of dogs is 40 lb and the standard deviation of the weights is 5 lb.  Find the probability that a randomly selected dog will weigh more than 48 lb.

A plant breeder develops a new wheat variety she hopes will yield more in the plains than the three most popular varieties under dry-land farming conditions.  She sets up a randomized block experiment with the three major types of soil in the region – sand, clay, and loam – as the blocks.  She selects a field with each soil type, divides it into four sections, and randomly selects one of the four wheat varieties for planting in each section.  The yields in bushels per acre in each section are shown in the following table. Variety Block A B C D Total Sand 20 21 21 18 80 Clay 25 24 21 20 90 Loam 30 28 22 20 100 Total 75 73 64 58 270 The correction factor for this Analysis of Variance is 6,075 and the total sums of squares is 141.  Calculate the sums of squares for treatments (i.e., varieties) and sums of squares for blocks (i.e., soil types) for this randomized block experiment.

The Analysis of Variance for a randomized block design produced the partially completed ANOVA table shown below. Source             df          SS            MS            F  Total                 24      131.3 Treatments        3        28.2 Blocks               5         69.0 Error                16        34.1                                  Find the mean squares for treatments, blocks, and error.

An Animal Scientist wants to estimate the average weaning weight of the Hereford breed of beef cattle.  Therefore, he selects a random sample of 100 Hereford calves and weighs them on the day they are weaned from their mothers.  The sample mean for these 100 calves is 450 lb and the sample standard deviation is 50 lb.  What is the 95% confidence interval for the mean of the entire population of all Hereford calves?

Frame score in beef cattle is based on height at the hips and is used as a measure of skeletal size.  Frame scores range from 1 to 10 with a higher number indicating a taller animal.  Independent random samples of frame scores were selected from the Angus and Simmental breeds of beef cattle with the following results: Angus Simmental 5 7 6 7 7 8 5 6 7 7 6   Using the appropriate table, find the critical F value needed to test the null hypothesis that the breed means for frame score are equal (use α = 0.05).

A randomized block design was used to compare the mean responses for three treatments.  Four blocks of three homogeneous experimental units were selected, and each treatment was randomly assigned to one experimental unit within each block.  The partially completed ANOVA table for this experiment is as follows: Source               df        SS             MS          F   Total                  11       84.489 Treatments         2       12.032       6.016 Blocks                3       71.749      23.916 Error                  6         0.708        0.118             Calculate the F value for treatments and for blocks.