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Show Arizona municipal bond Problem - Normal Distribution Recently the national average yield on municipal bonds has been μ = 4.19%. A random sample of 16 Arizona municipal bonds gave an average yield of 5.11% with a sample standard deviation s = 1.15%. Does this indicate that the populations mean yield for all Arizona municipal bonds is greater than the national average? Use α = 0.05. Assume x is normally distributed.
a) Position the null and the alternate hypothesis.
b) Recognize the sampling distribution to be used: the standard normal distribution or the Student's t distribution. Find critical value(s).
c) Calculate the z or t value of the sample test statistic.
d) Find out the P value or an interval containing the P value for the sample test statistic.
e) Based on the answers to a through d, decide whether or not to reject the null hypothesis at the given significance level. Explain your conclusion in the context of the problem.
The conservative degrees of freedom for the t distribution are and The standard error, conservative degrees of freedom
For all questions, clearly explain/show your answers. For the Analysis section, justify your answers with the appropriate graphs, statistics, equations, etc.
Assume the population standard deviations are not the same. At the .05 significance level, is there a difference in the mean waiting time?
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Construct a 99% confident interval for the standard deviation of the amount of vitamin B6 per pill.
An experiment was carried out to assess the effects of four tomato varieties and four planting densities on yield. There were 64 plots, and each of the 16 factor combinations was randomly assigned to 4 plots
What advantage does correlation have over a cross-tab?
Determine the most economical monthly shipping schedule. Provide a table showing the routes and total costs.
Multiple choice question using Forecasting techniques. Which among of the following models are best suited to forecast a time series with a stable seasonality component.
A game involving a pair of dice pays out $4 with probability 16/36, costs you $2 with probability 14/36, and costs you $6 with probability 6/36.
The mean television viewing time for Americans is 15 hours per week (Money, November 2003). Assume a sample of 60 Americans is taken to further investigate viewing habits. Suppose the population standard deviation for weekly viewing time is Q = 4..
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