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Confidence Interval

Conditional probability

4) In a survey conducted to determine cost of vacations, 164 individuals were randomly sampled. Each person was asked to assess the total cost of his or her most recent vacation. The average cost was $1,386. Assuming that the standard deviation was $400, find a 94% confidence interval for the average cost of all vacation trips.

Working with confidence levels.

I randomly selected 200 likely voters and created a 95 percent confidence interval for the true proportion of voters that will vote for my candidate. My confidence interval has endpoints of 0.39 and 0.45. Given the confidence interval that I created, is my candidate going to win? After constructing the confidence interval I call

Least square regression

1.The following data represent the number of calories per serving and the number of grams of sugar per serving for a random sample of high -fiber cerals. Here are the data to test this claim. Calories, x 200 210 170 190 200 180 210 210 210 190 190 200 Sugar, y 18 23 17

Suresh is comparing magazines and newspapers.

Context: It would normally follow on from work on setting up and testing hypotheses, and statistical analysis. Question: Suresh is comparing magazines and newspapers. He chooses a passage from one newspaper and one magazine. They each contain 100 words and he counts the lengths of all the words. now this is what you have

Confidence Interval

Using the following information, construct a 90% confidence interval for (:1 - :2). Sample 1: n = 8 mean = 151 standard deviation = 31 Sample 2: n=21 mean = 130 standard deviation = 72 please show results of all steps in the process.

Confidence Interval

The null hypothesis is V = 72.4 and alternative hypothesis is V does not equal 72.4. The sample mean is 68.325 and sample standard deviation is 14.172 from a sample of size 45. Find the 95% confidence interval. Show the results of all steps in the process.

Finding the best estimate from an average using a confidence limit.

Last year Groton, Georgia had 512 burglaries. The police chief wants to know the average economic loss associated with burglaries in Groton and wants to know it this afternoon. Since there is not time to analyze all 512 burglaries, the department's research analyst selects 10 burglaries at random, which show the following loss