Working Hypothesis Report - Assignment Example

P. 313#28) Rutter Nursery Company packages its pine bark mulch in 50-pound bags. From a long history, the production department reports that the distribution of the bag weights follows the normal distribution and the standard deviation of this process is 3 pounds per bag. At the end of each day, Jeff Rutter, the production manager, weighs 10 bags and computes the mean weight of the sample. Below are the weights of 10 bags from today’s production.45.6 47.7 47.6 46.3 46.2 47.4 49.2 55.8 47.5 48.5a. Can Mr.

Rutter conclude that the mean weight of the bags is less than 50 pounds?Use the .01 significance level.b. In a brief report, tell why Mr. Rutter can use the z distribution as the test statistic.c. Compute the p-value.P. 314#35) In recent years the interest rate on home mortgages has declined to less than 6.0 percent. However, according to a study by the Federal Reserve Board the rate charged on credit card debt is more than 14 percent. Listed below is the interest rate charged on aSample of 10 credit cards.14.6 16.7 17.4 17.0 17.8 15.4 13.1 15.8 14.3 14.

5Is it reasonable to conclude the mean rate charged is greater than 14 percent? Use the.01 significance levelP. 313#28)a) Based on the dataset, we conclude that the Mean of the sample is 48.18. Accepting the given population Mean of 50 and the standard deviation of 3, this sample is within .01 or is included within the 99% range of probability.A z-test provides the following information:Ho: = 50Ha: < 50z(48.18) = (48.18-50)/[3/sqrt(10)] = -1.91845…p-value = 0.0551Since p-value > 0.01, fail to reject Ho.

Based on this, we fail to reject 50 as the mean, so we cannot say with confidence that the mean weight of the population is less than 50 pounds.b) Using a z-statistic is applicable due to the range of data that fall evenly above and below a given mean on a normal distribution. The z-statistic tests to see if the sample follows the expectation of the population. In this case, the sample complies with the population parameters, as the mean falls within 1.918 standard deviations of the population mean.

A z-test is appropriate, because the population variance is known. If we did not have this information, then a t-test would be more appropriate.Given the fact that the sample contained an outlier of 55.8, one would expect that the sample is skewed based on that data point. This is true, because if we were to remove the outlier, the new sample mean would be 47.33 and the corresponding z-stat would be -2.67, giving a p-value of 0.0076. This result would lead us to accept Ha, since p-value < 0.01. Based on this, one would suggest taking an additional sample.c) p-value = 0.0551P. 314#35)Ho: = 14Ha: > 14z(15.66) = (15.66-14)/[1.

5436/sqrt(10)] = 3.400764…p-value = 0.0007 Since p-value of 0.0007 < 0.01, we reject Ho, and accept Ha, while concluding that the mean interest rate is greater than 14%.Do not turn in this pageHere is some additional information to help you study for your exam.Ho: This is the accepted norm, taking into account the assumed population mean and standard deviation. Ha: This is the “alternative hypothesis” or test. When you are given a sample data set, you are testing this sample to see if you can disprove Ho.

Mean: This is the average of the data set you are given. Add the data together and divide by the number of items in the set.Variance: This is the average of the sum of squares of the variation from the mean. In English, this means that you take the difference between each data item and subtract the mean. You then square each number. Total them together and divide by the number of data points minus the number of variables. You will see this as ∑(x-u)2 / (n-1)Sample calculation of variance: (u = 5)X(x-u)(x-u)22-393-244-11500611∑-515If this is the entire population of data, the variance would be 15 / 5 = 3.

If this is only a sample population, we must subtract one from the denominator, so the variance would be 15 / (5-1) = 15 / 4 = 3.75. Standard Deviation: This is the square root of the variance. In the case above, this would be 1.732 for the entire population or 1.936 for a sample population. Variance and Standard Deviation will always be greater for a sample population than for the entire population, in order to allow for the fact that not all information is known to the sampler.Calculating z-stat: The z-stat calculation is simple, once you know the mean and standard deviation of the population and the sample mean.

Just plug the numbers in as shown:z = (data point – mean) / (std deviation of sample)Data point is the average for your sample data, while mean is the population mean and std deviation is the std deviation for the population. You will normally be given these numbers for the population, so you are comparing your sample.Once you know the z-score, you can find the p-score either by using the function on your calculator or referring to a table in your textbook.Good luck!