A) Mean sales per week exceeds 41.5 per salesperson . Ho: mu=41.5 HA: mu.>41.5 bar over x=51.02...

A) Mean sales per week exceeds 41.5 per salesperson

. Ho: mu=41.5
HA: mu.>41.5 bar over x=51.02
n=100 mu0=41.5
Square root of 100=10

	t=51.02-41.5 / 6.8 square root 100 51.02-41.5/6.8 / square root 100 CI =( 50.17, 52.83) Bar over X +/- t. s/sq.rt of n Excel command=T.DIST.RT(t-1) p-value close to 0 Bar over x,t=1.96,S=6.8,N=100 51.5 +/- (1.96) times (6.8 / sq.rt. 100 )=1.3328 (50.1672, 52.8328) 95 % C.I. FOUND THE INFORMATION ABOVE WHAT IS THE F-VALUE

Expert Solution

This is 1 -sample t-test

F = t^2 = 14^2 = 196

Please rate

orchestra answered 2 years ago

To test: Ho:: mu >= 26 Ha: mu < 26 We took...

To test: Ho:: mu >= 26 Ha: mu < 26 We took a sample of size 400. The sample mean was 25.98 and s = 0.80 . Find the p-value. a. 0.3085 b. 0.01 c. 0.1587 d. 0.025 e. 0.6915

What is wrong with these hypotheses? H0: x bar = 23 Ha: x bar > 23

Given Hypothesis: Ho: population mean μo = 200 Ha: population mean μo < 200 This is...

Given Hypothesis: Ho: population mean μo = 200 Ha: population mean μo < 200 This is Left-Tailed test. Population Standard Deviation σ = 50. Given Significance Level is 0.10 (10%). Critical z-value for 0.10 significance level and Left-Tailed test is (-1.28). Rejection Region will be to the left of z= - 1.28. -3…………..-2………….-1…………0……….1 Choose your sample mean, x̅ (any integer between 180 and 195) and your sample size, n, ( any integer between 16 and 36). Perform the steps below...

MEAN CI x-bar (Grand Mean) LCL x-bar (CI-A2R-bar) UCL x-bar (CI+A2R-bar) Sample Range (Max-Min) R-bar (Mean)...

MEAN CI x-bar (Grand Mean) LCL x-bar (CI-A2*R-bar) UCL x-bar (CI+A2*R-bar) Sample Range (Max-Min) R-bar (Mean) UCL R (R-bar*D4) 1 68.51 68.46 68.54 68.34 68.46 68.46 2 68.94 68.2 68.54 68.56 68.7 68.7 3 68.66 68.44 68.55 68.77 68.7 68.64 4 68.49 68.94 68.56 68.62 68.69 68.56 5 68.64 68.63 68.62 68.32 68.34 68.24 6 68.34 68.42 68.99 68.02 68.03 68.47 7 68.99 68.94 68.95 68.95 68.94 68.97 8 68.92 68.91 68.97 68.93 68.96 68.95 GRAND MEAN n= A2= D4=

A 95% confidence interval for mu using the sample results x bar = 14.0 s =...

A 95% confidence interval for mu using the sample results x bar = 14.0 s = 6.9 n = 30 Point Estimate ____ Margin of Error ____ The 95% Confidence interval is _____ to _____

A 99% Confidence interval for mu using the sample results x bar = 90.9 s =...

A 99% Confidence interval for mu using the sample results x bar = 90.9 s = 34.3 n =15 Point Estimate = ______ Margin of Error = _____ 99% CI is ____ to ____

Ho u = 2 vs Ha u not equal to 2 and o=20. sample mean =12...

Ho u = 2 vs Ha u not equal to 2 and o=20. sample mean =12 and use alpha = 0.05 a) if n= 10, what is p-value? b) n= 15, what is p-value? c) n=20, what is p-value? d) summarize your findings from above.

1)What is a sampling distribution for the sample mean, x-bar? Explain. Is x-bar a random variable?...

1)What is a sampling distribution for the sample mean, x-bar? Explain. Is x-bar a random variable? Why do you think so? 2) We learned standard error of the mean ( read sigma xbar in denotation) or standard deviation of sample mean. When the size(n) of the sample collected from a infinite population is greater than 1, why is the standard error of the mean smaller than the standard deviation of X, population ? Explain conceptually or/and algebraically. Think of formula for...

Complete the following four hypotheses, using α = 0.05 for each. a. Mean sales per week...

Complete the following four hypotheses, using α = 0.05 for each. a. Mean sales per week exceeds 41.5 per salesperson b. Proportion receiving online training is less than 55% c. Mean calls made among those with no training is less than 145 d. Mean time per call is greater than 15 minutes 1. Using the same data set from part A, perform the hypothesis test for each speculation in order to see if there is evidence to support the manager's...

Question: Test the mean of population X for equality to zero (mu=0) using the sample x...

Question: Test the mean of population X for equality to zero (mu=0) using the sample x and t-test at a significance level 0.05 set.seed(88) x <- rt(150,df=2)

Question