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Type I and Type II error
Interpretation of Data using SPSS (--provide detailed response)
I'm in the number crunching phase of some research. I had an advanced stats class two years ago, hence, the haziness on the interpretation end. I compiled the data from the Metropolitan Chicago Information Center (MCIC) and developed two hypotheses relative to data about residents' perception of crime in their neighborhoods for ...continues
Determining the sampling distribution of a proportion.
According to a well-known newspaper journal, 58.3% of the 222,900 households in a particular county get the Sunday edition of the local newspaper. Suppose that random samples of 200 households are selected. In what proportion of the samples will between 55% and 60% of the households get the Sunday edition of the newspaper? ...continues
1. Only two doctors would make a diagnosis on patients coming into the same STD clinic for gonorrhea. The probability of the two doctors making a diagnosis of gonorrhea is (independent, dependent, synergistic, cannot tell). 2. Assume that birth weights are normally distributed with a mean of 3400g and a standard deviation of ...continues
Statistics/Phstat - Complete a decision tree and opportunity loss table
Purchase Price= 10 Selling price = 20 on Christmas day 2 otherwise Profit/loss 10 on Christmas day (8) otherwise a) Compute the payoffs for purchasing 100,200,500 or 1000 trees for each level of demand b) Set off the payoff table indicating the event and alternative course of action c) Set up a decisi ...continues
Confidence intervals for one population mean when S.D. is unknown.
A Variable has a mean of 100 and a standard deviation of 16. Four observations of this variable have a mean of 108 and a sample standard deviation of 12. Determine the observed value of the a. Standardized version of x (sample mean). b. Studentized version of x (sample mean).
Hypothesis Testing: Compute the statistical z test and deciding what to do with the null.
In a recent national survey, the mean weekly allowance for a nine-year-old child from his/her parents was reported to be $3.65. A random sample of 45 nine-year-olds in northwestern Ohio revealed the mean allowance to be $3.69 with a standard deviation of 0.24. At the 0.05 level of significance is there a difference in the mean a ...continues
One of the Rocky Mountain School District Principal's claims that each of his assistant principals makes 40 calls on parents per week. Several other Principals said that this estimate is too low and that their assistant principals make even more calls. To research the claim, a random sample of 28 assistants revealed that mean ...continues
finding expected value in 2x2 table
Myocardial Infarction Status Yes No OC User 683 2537 Non-OC user 1498 8747 The question is what the expected value for an OC user and no myocardial infarction? I found all kinds of things to calculate in 2x2 tables, but not this. Probably am overlooking something simple.
Given the following 2x2 table: Myocardial Infarction Status Yes No OC User 683 2537 Non-OC user 1498 8747 Calculate the expected value for an OC user and no myocardial infarction. The TA said "2537", which may be correct, but I think some kind of chi-square goodness of fit or a similar test is needed to se ...continues