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Paul is conducting a study on the dollar value of orders placed with various suppliers. A sample of 10 orders showed amounts in thousands of dollars of 43.6, 61.3, 54, 70.7, 46, 56.3, 58.8, 73.7, 47.4, 45.6. Paul wants to develop a 99% confidence interval on the average value dollar of orders.
A. State the value for the point estimate.
B. Find the appropriate standard error.
C. Develop the confidence interval.
Construct a 90 percent confidence interval for the true mean weight. What sample size would be necessary to estimate the true weight with an error of +/-} 0.03 grams with 90 percent confidence?
How many dissimilar computer passwords are possible?
1. students take an average of 45 minutes and standard deviation of 8 minutes to complete a certain statistics test. if
A check of dorm rooms on a large college campuses revealed that 38% had refrigerators, 52 percent had TVs and 21 percent had both a TV and a refrigerator. What's the probability that a randomly selected dorm room has.
you want to compare response rates percents of two regimens. a power of 90 and a two-sided test of size 0.05 are
question 1based on a survey of 1000 adults by greenfield online and resported in a may 2009 usa today snapshot adults
Jim Sears manufactures farm equipment. His work requires the use of steel bars which must have a mean length of at least 50 inches. The bars can be purchased from a supplier in Kansas
The effectiveness of treating respiratory infections with herbal remedies was studied. "Days of fever" was used to measure effects. Among 356 children treated with herbal remedies, teh mean number of days with was 0.36,
Suppose we have observations x1, x2, . . . , xn. Prove that the sum of deviations of the observations from their arithmetic mean is always equal to zero.
question 1you have been given sample data from two offices and told that the 95 confidence interval for the difference
Use regression analysis to develop an estimated regression equation that could be used to predict the value score given the price of the car.
The weight of the oranges, in pounds (1 pound = 16 ounces) is a random variable with standard deviation
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