1) a) studentised t distribution is used to find out the margin of error, because the standard deviation of the population is unknown. In such cases we go to t test for testing the significance
Coefficient of vairation CV is the ratio of Standard deviation to mean is not possible as the standard deviation of the population is unknown
2)As the standard deviation of population is known a normal or z-test can be applied in this case
Here again coefficient of variation is not possible to find as the mean of the sample is not given
3) Since the assumptions of a parametric tests are violated , and the sample size is too small to perform a normal test there are many options to perform a test
!) Non parametric test (such as wilcoxon signed rank test )when the assumptions of parametric tests are not met
ii) Transformation of the data especially in the case of skewed distributions a log transformation will do to make the distribution to be approximately normal
Ulse tl Example 1: Given that n = 25, σ is unknown, the CL-95%, and the...
1. Find the margin of error for the given values of c, σ, and n. c = 0.90, σ = 10.2, n = 75 2. Find the margin of error for the given values of c, σ, and n. c = 0.95, σ = 677, n = 40
For each example below you would start to construct a confidence interval by identifying which distribution should be used to find the margin of error. It is possible that you won't be able to use either distribution. (a) a confidence interval for a population proportion where n = 66 and there are 63 successes in your sample leading to a sample proportion of 0.955 A. z-distribution B. t-distribution C. neither distribution (b) a confidence interval for a population proportion where...
where p denote the population mean of the original random variable 5.7 Problems . Assume X is a normally distributed random variable with mean u and stan- dard deviation σ. A sample of size n-5 from this distribution is given as 1. Assume we are interested in the properties of the mean of the sam- pling distribution of the sample mean. Describe why this quantity is a 2. State an estimator for the parameter given in question 1. Use this...
Find the margin of error for the given values of c, σ, and n. c=0.95 , σ=3.4 n=36 E=____ (Round to three decimal places as needed.)
1) Random samples of size n were selected from populations with the means and variances given here. Find the mean and standard deviation of the sampling distribution of the sample mean in each case. (Round your answers to four decimal places.) (a) n = 16, μ = 14, σ2 = 9 μ=σ= (b) n = 100, μ = 9, σ2 = 4 μ=σ= (c) n = 10, μ = 118, σ2 = 1 μ=σ= 3) A random sample of size...
a.Construct the confidence interval for the population mea μ. c=0.98 x=16.3, σ=5.0, and n=95 A 98% confidence interval for μ is ( ___ , ___ ) (Round to one decimal place as needed. b. Find the critical value tc for the confidence level c=0.98 and sample size n=13. tc= _____ (Round to the nearest thousandth as needed.) c. Find the margin of error for the given values of c, s, and n.c=0.80, s=4, n=28 The margin of error is _____...
1. Which formula gives the standard error for all xbar values? _______ a) σ/√n b) σ c) npq d) pq/n e) 1.96σ/√n 2. Prof. Gersch knows that the score on a standardized test is perfectly normal with a mean of 200 and a standard deviation of 40. He then takes a random sample of 16 tests. Prof Gersch wants to check the probability that the average of these 16 tests is less than 210. Which of these is equivalent to the probability...
x =30 n=2323, σ=66, confidence =90% a. Use the one-mean z-interval procedure to find a confidence interval for the mean of the population from which the sample was drawn. The confidence interval is from to b. Obtain the margin of error by taking half the length of the confidence interval. What is the length of the confidence interval? c. Obtain the margin of error by using the formula E=z(a/2) x o/ square root of n Identify the critical value. What...
If n < 30 and σ İs unknown, then the 100(1-α)% confidence interval for a population mean is_ 7. assuming the population is normally distributed xtra 21 1(degrees of freedom = n-2) a. 1 (degrees of freedom-n-1) c. t(degrees of freedom-n-1) 1 d. x ± te |--| (degrees of freedom n-2) 1 (degrees of freedom = n-1) If n interval when the level of confidence is 99% 8. 16, then find the number ta2 needed in the construction of a...
Use the t-distribution to find a confidence interval for a mean μ given the relevant sample results. Give the best point estimate for μ, the margin of error, and the confidence interval. Assume the results come from a random sample from a population that is approximately normally distributed. A 95% confidence interval for u using the sample results x̄ = 10.7, s=4.5, and n = 30