The distribution of heights of a certain breed of terrier has a mean of 71 centimeters and a standard deviation of 9 centimeters, whereas the distribution of heights of a certain breed of poodle has a mean of 29 centimeters with a standard deviation of 5  centimeters. Assuming that the sample means can be measured to any degree of​ accuracy, find the probability that the sample mean for a random sample of heights of 64 terriers exceeds the sample mean for a random sample of heights of 100 poodles by at most 44.2 centimeters. What is the probability? Round to four decimals. Use standard normal distribution table.

According to Kaiser Family Foundation survey in 2011 and 2010, the average premium for employer-sponsored health insurance for family coverage was $15,073 in 2011 and $13,770 in 2010 (USA TODAY, September 29, 2011). Suppose that these averages were based on random samples of 250 and 200 employees who had such employer-sponsored health insurance plans for 2011 and 2010, respectively. Further assume that the population standard deviations for 2011 and 2010 were $2160 and $1990, respectively.

A parenting magazine reports that the average amount of wireless data used by teenagers each month is 5 Gb. For her science fair project, Ella sets out to prove the magazine wrong. She claims that the mean among teenagers in her area is less than reported. Ella collects information from a simple random sample of 25 teenagers at her high school, and calculates a mean of 4.7 Gb per month with a standard deviation of 0.9 Gb per month. Assume that the population distribution is approximately normal. Test Ella's claim at the 0.01 level of significance. Step 3 of 3: Draw a conclusion and interpret the decision. E Tables E Keypad Answer Keyboard Shortcuts We reject the null hypothesis and conclude that there is sufficient evidence at a 0.01 level of significance that the average amount of wireless data used by teenagers each month is less than 5 Gb. We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.01 level of significance that the average amount of…

If we meet these conditions, the sampling distribution of the mean will have a normal shape and ... The mean of the sampling distribution will be u, (i.e., the same as the population mean) and the standard deviation of the sampling distribution will be (that's the population standard deviation divided by the square root of the sample size). In the AP Stats Guy video, he talks about the number of text messages his students send during class. Suppose the average number of text messages his students send during class is u = 30 text messages. If we take samples of say, n = 36 students at a time, we would expect the mean of the sampling distribution we create to be the same as the population mean, 30 text messages. If we can further say that standard deviation of the number of text messages is 12 text messages, by how much would we expect the sample means to vary? (hint, use the formula above) text messages

The salaries of professional baseball players are heavily skewed right with a mean of $3.2 million and a standard deviation of $2 million. The salaries of professional football players are also heavily skewed right with a mean of $1.9 million and a standard deviation of $1.5 million. A random sample of 40 baseball players’ salaries and 35 football players’ salaries is selected. The mean salary is determined for both samples. Let  represent the difference in the mean salaries for baseball and football players. Which of the following represents the shape of the sampling distribution for ? skewed right since the populations are both right skewed skewed right since the differences in salaries cannot be negative approximately Normal since both sample sizes are greater than 30 approximately Normal since the sum of the sample sizes is greater than 30

If a sample size is 100 and the population standard deviation is 30 then the standard deviation of the sampling distribution for means is 3.  Is this true or false.

A parenting magazine reports that the average amount of wireless data used by teenagers each month is 10 Gb. For her science fair project, Ella sets out to prove the magazine wrong. She claims that the mean among teenagers in her area is less than reported. Ella collects information from a simple random sample of 16 teenagers at her high school, and calculates a mean of 9.1 Gb per month with a standard deviation of 1.3 Gb per month. Assume that the population distribution is approximately normal. Test Ella's claim at the 0.05 level of significance. Step 3 of 3: Draw a conclusion and interpret the decision.

According to the Bureau of Labor Statistics, the mean salary for registered nurses in Kentucky was $57,688. The distribution of salaries is assumed to be normally distributed with a standard deviation of $5,879. Someone would like to determine if registered nurses in Ohio have a greater average pay. To investigate this claim, a sample of 192 registered nurses is selected from the Ohio Board of Nursing, and each is asked their annual salary. The mean salary for this sample of 192 nurses is found to be $58,016.005.  1. Completely describe the sampling distribution of the sample mean salary when samples of size 192 are selected. 2. What conjecture has been made? 3. Using the distribution described in part a, what is the probability of observing a sample mean of 58,016.005 or more? 4. Based on the probability found, what conclusion can be reached?