I have to conduct a Hypothesis Test on my sample Size.
sample size=58
(the first process involved 40 expenses and the second process was 18 expenses)
I need to conduct a hypothsis test on this evaluation.
•Confidence Interval: Before Improvement
•X=40
•N=58
•p=40/58=0.68
•Confidence Interval: After the improvement
•X=18
•N=58
•P=18/58=0.31
•Confidence Interval: (0.191, 0.429)
•95% confidence (1.96)that the interval contains the true population proportion. With a Margin of error:1
•
•Before the Improvement:
•
•Sample Standard Deviation, s 48.908185591827
•Variance (Sample Standard), s2 2392.0106178846
•Population Standard Deviation, σ 48.292963798441
•Variance (Population Standard), σ2 2332.2103524375
•Total Numbers, N 40
•Sum: 1484.89
•Mean (Average): 37.12225
•Standard Error of the Mean (SEx̄): 7.7330631348202
•
•After the Improvement:
•
•
•Sample Standard Deviation, s 17.276746092814
•Variance (Sample Standard), s2 298.48595555556
•Population Standard Deviation, σ 16.789979227776
•Variance (Population Standard), σ2 281.90340246914
•Total Numbers, N 18
•Sum: 277.82
•Mean (Average): 15.434444444444
•Standard Error of the Mean (SEx̄): 4.0721681063556
I have to conduct a Hypothesis Test on my sample Size. sample size=58 (the first process...
The annual earnings of 12 randomly selected computer software engineers have a sample standard deviation of $ 3626 Assume the sample is from a normally distributed population. Construct a confidence interval for the population variance sigma squared σ2 and the population standard deviation sigma σ. Use a 95% level of confidence. Interpret the results. What is the confidence interval for the population variance sigma squared σ2? What is the confidence interval for the population standard deviation sigma σ? Please show...
a) Suppose we know the population variance σ2-400, there is a randorm sample with i-135, and the sample size is n the mean. I. 100. Please construction a 95% confidence interval for b) Suppose we do not know the population variance, and only know the sample variable s2-400, there is a random sample with -135, and the sample size is n-25. Please construct a 95% confidence interval for the mean.
A sample of size n = 18 is drawn from an approximately normal population whose 3) standard deviation is σ = 7.5 sample mean is x = 46.1Construct a 98% confidence interval for μ.
Given the confidence interval for a mean of (74.5738,77.4262), from a sample of size 34 with a population standard deviation of σ=3.6, find the following: Margin of Error(ME)= Standard Error(SE)= Zc= what was the confidence level for this confidence interval?= (Show your work please)
5.6.1. A random sample of size 20 is drawn from a population having a normal distribution. The sample mean and the sample standard deviation from the data are given, respectively, as 2.2 and s-1.42. Construct a 90% confidence interval for the population variance σ2 and interpret.
I1. Follow the steps below to show that the pooled estimator $p is an unbi- ased estimator for the common standard deviation of two independent sam ples Let Yi, Yi2, ..., Yini denote the random sample of size n from the first population with population mean μ| and population variance σ, and let Y21, Y22, ..., Y2na denote an independent random sample of size n2 from the second population with population mean μ2 and population mean ơ3. Sup- pose that...
Here is an example with steps you can follow: sample size n=9, sample mean=80, sample standard deviation s=25 (population standard deviation is not known) Estimate confidence interval for population mean with confidence level 90%. Confidence Interval = Sample Mean ± Margin of Error Margin of Error = (t-value)×s/√n t-value should be taken from Appendix Table IV. For n=9 df=n-1=9-1=8 For Confidence Level 90% a = 1 - 0.90 = 0.10, a/2 = 0.10/2 = 0.05 So, we are looking for...
A sample of size n = 42 is drawn from a population whose standard deviation is σ = 7.5. Find the margin of error for a 99% confidence interval for μ.
I want you to conduct a hypothesis test for a difference of means for cholesterol levels between male and female students. There are 148 females and 164 males in our sample. You can treat this as a large sample problem and use z-values for confidence intervals and hypothesis tests (however, Excel uses a t-value in anything it calculates). The output from Microsoft Excel is given below to help. Your job will be to find the right numbers in the output...
A simple random sample with n = 58 provided a sample mean of 26.5 and a sample standard deviation of 4.4. (Round your answers to one decimal place.) (a) Develop a 90% confidence interval for the population mean. 25.5 to 27.5 (b) Develop a 95% confidence interval for the population mean. to 27.7 25.3 (C) Develop a 99% confidence interval for the population mean. 24.9 x to 28.1 x