2. Margin of error using critical value of standard normal:
z*standard deviation/√n = 1.96*12/√25 = 4.704
So the answer is 4.7
3. Using t critical value, t score is t 0.025,24 = 2.064
So margin of error= 2.064*12/√25 = 4.9536
So rounding to the nearest tenth, the answer will be 5.0
Watch later Share Info Sample of 25 Fish Sample mean-X = 41 cm Sample SD 12...
Question 202.5 pts If we consider the simple random sampling process as an experiment, the sample mean is _____. Group of answer choices always zero known in advance a random variable exactly equal to the population mean Flag this Question Question 212.5 pts The basis for using a normal probability distribution to approximate the sampling distribution of x ¯ and p ¯ is called _____. Group of answer choices The Law of Repeated Sampling The Central Limit Theorem Expected Value...
25> Consider a variable known to be Normally distributed with unknown mean μ and known standard deviation σ-10. (a) what would be the margin of error of a 95% confidence interval for the population mean based on a random sample size of 25? The multiplier for a z confidence interval with a 95% confidence level is the critical value z. 1.960. (Enter your answer rounded to three decimal places.) margin of error 25 (b) What would be the margin of...
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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...
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