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# Hypothesis Testing & Confidence Interval

See attached file.

Problem 4, Chapter 10

This problem asks you to test a hypothesis concerning mean heights of Navy recruits. It specifically asks you to compute a t-statistic and a p-value. The easiest way to calculate the p-value is using the Excel =TDIST(...) function which returns the probability that a sample from the population could return a mean at least as extreme as the level you are testing. o However, the Excel =TDIST(...) function cannot calculate over a negative probability. If the t-statistic you calculate is negative, you still want to determine the likelihood that the observed mean height (calculated from the sample called "new recruits") could have been drawn from a population whose mean was 68. That is, you want to determine the probability that this sample drawn from the population could have created a t-statistic such that the observed value of the sample mean could have been drawn from the population. To accomplish this, just calculate the complement. That is, =1-TDIST(...). o The Excel help system provides details on the =TDIST(...) function. It states: &#61607; Since x < 0 is not allowed, to use TDIST when x < 0, 1. note that TDIST(-x,df,1) = 1 - TDIST(x,df,1) = P(X > -x) and 2. TDIST(-x,df,2) = TDIST(x df,2) = P(|X| > x).

Problem 40, Chapter 10

o Create a table of appropriate descriptive statistics of the grades
o Create a frequency table of grades with 8 bins (&#8804;65, 66-70, 71-75, ... 91-95, >95)
&#61607; Using different bins will result in different answers and may affect your grade. This is one downside of using a Chi-Square test for normality.
o Add a column to this table of the expected values if the distribution were normal
o Create a histogram including both the expected values and the observed values (something like figure 10.24 on page 537)
o Calculate the Chi-Square statistic
o Calculate the p-value &#61607; The degrees of freedom to use in a chi square test are based on:

1. The number of bins used in the histogram or frequency table for the data being analyzed. Call this number k. 2. The number of parameters being estimated for the distribution. Call this number c. 3. The degrees of freedom for the Chi Square is given by k-c-1 &#61607; In the example in class, there were 10 bins created for the frequency table and resulting histogram. Hence k=10. The analysis for the sample estimates of a normal distribution has two parameters - the mean and the standard deviation. Hence c =2. This gives us df = 10-2-1 = 7 degrees of freedom.

Problem C3, Chapter 10

o You can read this problem as stating, "With 99% confidence the mean proportion (percentage) of residents of Lewiston whose annual salary is greater than \$80,000 is between 10% and 18%."
o You can deduce some information regarding the appropriateness of sample sizes by viewing other statistics such as confidence interval limits. Remember the "rules of thumb" for sample sizes we discussed.
o To discuss issues of normality, consider how the Chi-Square test might be applied.

#### Solution Summary

The solution provides step by step method for the calculation of descriptive statistics, confidence interval and testing of hypothesis. Formula for the calculation and Interpretations of the results are also included. Interactive excel sheet is included. The user can edit the inputs and obtain the complete results for a new set of data.

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