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- In drilling a vertical wellbore, the spacing between major fractures are recorded It has been found that this spacing is having a mean of 1800 ft and a variance of 3,240,000 ft2. In the first 2000 ft drilling, there was no present of major fracture in the well. Determine the probability that there is a major fracture at the depth of 2000-3500 ft.X is the number of red marbles that Suzan has in her hand after she selects 5 marbles from a bag containing 5 red marbles and 2 green ones. expected value variance standard deviationTimes for a procedure are normally distributed. There are 2 methods. Method A has a mean of 25 minutes and a standard deviation of 8 minutes. Method B has a mean of 29 minutes and a standard deviation of 4 minutes. Which procedure is preferred if it must be completed within 33 minutes?
- The data from an independent-measures research study produce a sample mean difference of 4 points, and each sample has a variance of 10. If there are n = 20 scores in each sample, then what is the estimated standard error of the difference between means?A researcher found a significant correlation of .12 between gender (coded as 0 = ‘male’ and 1 = ‘female’) and number of friends at university. If the assigned values of gender were changed to 1 = ‘female’ and 2 = ‘male’, how would this affect the correlation? The strength of the relationship would become weaker The amount of variance explained by the relationship would increase It would become nonsignificantThe sign would change to become negative A researcher is trying to predict anxiety (measured on a 7-point scale) from exposure to stress (operationalised as the number of stressful life events participants can recall in the last week). She has found a correlation coefficient of r = .54 between anxiety and stress, with a standard deviation of 4.45 for anxiety and a variance of 17.21 for stress. What other information is needed to calculate b? The sums of squares of stressThe sums of products of anxiety and stress The covariance of anxiety and stressNo other information is…One sample has n = 10 scores and a variance of s2 = 20, and a second sample has n = 15 scores and a variance of s2 = 30. What can you conclude about the pooled variance for these two samples?
- Records from previous years for a casualty insurance company show that its clients average a combined total of 1.9 auto accidents per day, with a variance of 0.31. The actuaries of the company claim that the current variance, oʻ, of the number of accidents per day is not equal to 0.31. A random sample of 17 recent days had a mean of 2 accidents per day with a variance of 0.62. If we assume that the number of accidents per day is approximately normally distributed, is there sufficient evidence to conclude, at the 0.05 level of significance, that the actuaries are correct? Perform a two-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places and round your answers as specified below. (If necessary, consult a list of formulas.) (a) State the null hypothesis H, and the alternative hypothesis H,. p H, :0 H, :0 (b) Determine the type of test statistic to use. (Choose one) ▼ D=0 OSO O20 (c) Find the value of the test statistic. (Round…a sample of n = 10 score has ss= 108 what is the variance for this sampleA certain IQ test is known to have a population mean of 100 and standard deviation of 15 in the general population. You want to test whether psychology majors have a different average IQ than the population as a whole. Assume the variance of IQ is the same for Psych majors as it is in the general population. Suppose that Psychology majors actually have an average IQ of 108. If you do a 2-tailed test at α= .05 with a sample of 56 Psychology majors, you will be able to reject the null hypothesis if the mean IQ of your sample is below [L] or above [H]. Find L and H values. Options listed below. [L] answer choices: 96.08, 98.00, 103.92, 104.08, 110.00, 111.92. [H] Answer choices: 96.08, 98.00, 103.92, 104.08, 110.00, 111.92.
- The service times for customers coming through a checkout counter in a retail store are independentrandom variables with mean 1.5 minutes and the variance 1.0. Approximate the probability that 100customers can be served in less than (a) 2 hours (b) 3 hours of total service time.Suppose Y = 6x-8, where x is a random variable with mean 2 and a variance of 4/3. Determine the variance of the random variable yA study was conducted to compare the length of time between men and women to assemble a certain product. Sample standard deviation in a random sample of times for 11 men was 8.6 minutes, and sample standard deviation in a random sample of times for 13 women was 6.9 minutes. Past experience indicate that population of times is normal for both men and women. Is it sufficient sample evidence to conclude that the variance of the lengths of time for men is larger than the variance of the lengths of time for women? Test appropriate hypothesis at a = 0.05 level of significance using critical value approach. Use S²/S2 as test statistic. Construct a 90% confidence interval for the ratio of two population variances. a. State null and research hypotheses H: H₁ b. The observed value of test statistic is c. State the rejection region d. Insert the word sufficient or insufficient into the following statement. There is of time for men is larger than the variance of the lengths of time for women.…
