Independent random samples, each containing 800 observations, were selected from two binomial populations. The samples from populations 1 and 2 produced 320 and 400 successes, respectively. Construct a 90% confidence interval for the true difference in population proportions of successes. In the interest of time you can assume that the sample sizes are large enough that it is appropriate to use a large-sample confidence interval.
Assume that the probabilities of two genetic defects (we wil…
Assume that the probabilities of two genetic defects (we will call them defect A and defect B) in horses are 0.05 and 0.08, respectively. If these two genetic defects represent independent events, what is the probability that a horse will have both of these genetic defects?
The average weight of a kennel of dogs is 40 lb and the stan…
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 between 44 and 48 lb.
Scores on a statistics exam indicate that the 25th percentil…
Scores on a statistics exam indicate that the 25th percentile was a 70 (out of 100 points possible), the 50th percentile was a 79, and the 75th percentile was an 88. Use this information to construct a box plot for the exam scores. The highest score on the exam was a perfect score of 100 out of 100. Is this score a suspect or highly suspect outlier?
A survey was conducted to determine how people rate the qual…
A survey was conducted to determine how people rate the quality of programming available on TV. Twenty-one respondents were asked to rate the overall quality from 0 (no quality at all) to 100 (extremely good quality). The stem-and-leaf display based on these data (using the first digit as the stem and the second digit as the leaf) is shown below: Stem Leaves 3 2 5 4 0 3 4 7 8 9 5 1 1 2 3 4 5 6 1 2 5 6 7 7 7 8 Based on the stem-and-leaf display, the upper quartile for these TV ratings is __________.
Find the probability of an observation lying more than z = 1…
Find the probability of an observation lying more than z = 1.56 standard deviations below the mean.
The mean length of time required to complete a 5K race was 2…
The mean length of time required to complete a 5K race was 20 minutes. The standard deviation of the times was 4 minutes. The racing times were approximately normally distributed. What proportion of the runners would be expected to require between 18 and 25 minutes to complete the race?
The weights (in pounds) of a sample of 6 dogs are as follows…
The weights (in pounds) of a sample of 6 dogs are as follows: Dog Weight 1 30 2 32 3 26 4 42 5 40 6 46 Find the standard deviation of the weights of this sample of 6 dogs.
We are interested in comparing the mean supermarket prices o…
We are interested in comparing the mean supermarket prices of two leading colas in the Columbus area. Assume that we have independent random samples of prices of six-packs at 8 supermarkets. The data are shown in the following table: Supermarket Brand 1 Brand 2 1 $2.25 $2.30 2 2.47 2.45 3 2.38 2.44 4 2.27 2.29 5 2.15 2.25 6 2.25 2.25 7 2.36 2.42 8 2.37 2.40 _______________________________________ Mean $2.3125 $2.3500 Standard deviation $0.1007 $0.0859 _______________________________________ Find the point estimate for the difference in mean price of Brand 1 and Brand 2, assuming that these are independent random samples.
The weaning weights of 3 herds of beef cattle are compared a…
The weaning weights of 3 herds of beef cattle are compared across the 4 seasons of the year. Thus, this is a 3 x 4 factorial experiment with 3 herds and 4 seasons. Three calves from each herd are sampled during each season (i.e., there are 3 replications). The resulting data were analyzed using Analysis of Variance and the partially completed ANOVA table is as follows: Source df F Total 35 Herd 2 17.2 Season 3 3.0 Herd x Season 6 1.2 Error 24 Was there a significant difference among the 4 seasons for mean weaning weight? Use a significance level of α = 0.01.