A sample consist of 12 observations, for all of which a given feature X is measured. If the sample variance of X is 100, what is the sum of squared deviations from the sample mean? O 1200 O 1350 O 1100 O 250 O 1000
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A: Given: The sample has following data. n= 16 hours Mean (M) = 56 Sample standard deviation (s) = 12
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A: Answer in step 2 Note please post second question seperately. Thankyou
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A: As given data A bar = 9.2 Standard deviation = 1.8 N = 784.
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A: μ=66000σ=21000
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A: Given that, Sample observations (n ) = 60 population proportion (p) = 0.85
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A: Given: Sum of squared errors (SSE) = 50 Number of observations (n) = 12 Number of treatments (k) = 6
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Q: The variance of sample means is:
A: To find : Variance of sample given.
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A: We have given that H0 = 50HA ≠50 x¯= 55, s = 8, n= 20 and α =0.10 We have to check which…
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A: Number of observations = 16 μ = 8δ = 2x¯ = 7
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Q: onsider the following data: −11,−5,−5,−11,13,−11,−5−11,−5,−5,−11,13,−11,−5 Copy Data Step 2 of 3:…
A:
Q: Given the following data .Find the OLS estimators using the deviation method. Y 6 10 9 14 7 5 X1…
A: Y X1 X2 6 1 3 10 3 -1 9 2 4 14 -2 6 7 3 2 5 5 4 Using excel, we get the regression…
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A: We can find out the OLS regressor using the formula's of b0 and b1
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A: Given that, Population proportion (P) = 0.60 1-P = 1-0.60 = 0.4
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A: A. Below is the calculation for range. Range = Maximum value - Minimum valueRange = 6 - 1Range = 5
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Q: A health expert evaluates the sleeping patterns of adults. Each week she randomly selects 40 adults…
A: * SOLUTION :-
6
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- A total of 280 observations of Bob Ramos, an assembly-line worker, were made over a 40-hour work week. The sample also showed that Bob was busy working (assembling the parts) during o 230 bservations. (Round all intermediate calculations to at least two decimal places before proceeding with further calculations.) The number of observations that need to be taken of Bob to be 99% confident with 4% acceptablee error=19 The mean of z-scores in a data set a is zero only when the x-values are balanced around the mean of x. b is greater than zero for data sets that have large outlying values. c is always equal to one. d is always equal to zeroSCCoast, an Internet provider in the Southeast, developed the following frequency distribution on the age of Internet users. Find the mean and the standard deviation. (Round squared deviations to nearest whole number and final answer to 2 decimal places.) Age (years) Frequency 10 up to 20 3 20 up to 30 7 30 up to 40 18 40 up to 50 20 50 up to 60 12
- Given the following data .Find the OLS estimators using the deviation method. Y 6 10 9 14 7 5 X1 1 3 2 -2 3 5 X2 3 -1 4 6 2 4ASAP Exercise 7-48 (Algo) (LO7-3) According to a government study among adults in the 25- to 34-year age group, the mean amount spent per year on reading and entertainment is $2,125. Assume that the distribution of the amounts spent follows the normal distribution with a standard deviation of $439. (Round your z-score computation to 2 decimal places and final answers to 2 decimal places.) a. What percent of the adults spend more than $2,400 per year on reading and entertainment? b. What percent spend between $2,400 and $3,400 per year on reading and entertainment? c. What percent spend less than $1,000 per year on reading and entertainment?Consider the following data: −11,−5,−5,−11,13,−11,−5−11,−5,−5,−11,13,−11,−5 Copy Data Step 2 of 3: Calculate the value of the sample standard deviation. Round your answer to one decimal place.
- Calculate the sample size to study the average age of a Central State College student, plus or minus 0.4 years. We’d like to be 85% confident about our result. From a previous study, we know that the standard deviation for the population is 2.8.A random sample of residents from various income brackets was asked their opinion on tax reform for the country of Guyana. The data so gathered were analyzed by a statistician and the results he obtained using MINITAB are shown below: Expected counts are printed below observed counts Low Middle High Total For 26 (30.67) 75 (51.11) 14 (33.22) 115 Against 60 (69.33) 95 (115.56) 105 (**) 260 Neutral 34 (20) 30 (33.33) 11 (21.67) 75 Total * 200 130 450 Chi-sq = 0.71 + 11.17 + 11.12 + *** + 3.66 + 11.89 + 9.8 + 0.33 + 5.25 = 55.2 DF =??, p-value =??? Carefully define the null and alternative hypotheses of the x2 test underlying the generation of the above table. Find the missing values ‘*’, ‘**’, ‘***’, ‘??’ and ‘???’. What is the conclusion of this test? Give reasons for your answer.A sample of 20 observations from a normal distribution with standard deviation 4.51 gives amean of 32.8. Does this data support that the mean of the population is greater than 30 at0.05 significant level?
- Let x be a random variable representing dividend yield of Australian bank stocks. We mayassume that x has a normal distribution with σ = 2.4%. A random sample of 15 Australianbank stocks has a sample mean of x = 6.83%. For the entire Australian stock market, themean dividend yield is μ = 6.9%. Do these data indicate that the dividend yield of allAustralian bank stocks is higher than 6.9%? Use α = 0.05. What is the value of the teststatistic?What do you mean by Sample Variance, Sample Standard Deviation,and Standard Error?In your own words, describe what the difference is between an error term and a residual. How does sample size affect the variance of each?