report three things for this distribution of sample means: its shape, its expected value, and its standard error.
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Imagine a distribution of sample means for samples of n = 25 selected from a population with a
Let , X→N(μ=25 , σ=5) , n=25
Our aim is to report three things for this distribution of sample means: its shape, its expected value, and its standard error.
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- Let X de a normally distributed variable with a population standard deviation = 43. A research believe that the standard deviation should increase when a new variable is manipulated. To test their claim, the research collects a sample of size n = 19 and finds a sample statistic of 8 = 53. Use the x²-table to find the largest value that the corresponding P-value can be. Type your answer...A population of N = 5 scores has a mean of u = 20 and a standard deviation of o = 4. Which of the following is the correct Interpretation of the standard deviation? O The average distance of scores from the mean of p- 20 is 5 points. O The average distance of scores from the mean of u = 20 is 1 point. The average distance of scores from the mean of u = 20 is 4 points. O The average squared distance of scores from the mean of u = 20 is 4 points.See pictures for question D and E and (A, B and Cand was submitted on a different question. According to a report done by S & J Power, the mean lifetime of the light bulbs it manufactures is 44 months. A researcher for a consumer advocate group tests this by selecting 16 bulbs at random. For the bulbs in the sample, the mean lifetime is 47 months. It is known that the population standard deviation of the lifetimes is 5 months. Assume that the population is normally distributed. Can we conclude, at the 0.10 level of significance, that the population mean lifetime, μ, of light bulbs made by this manufacturer differs from 44 months? Perform a two-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places, and round your responses as specified below.
- The mean age of graduate students at a University is at most 31 years with a standard deviation of 2 years. A random sample of 15 graduate students is taken. The sample mean is 32 years and the sample standard deviation is 3 years. Are the data significant at the 1% level? a. State the null and alternative hypothesesb. Performzscore and find p-valuec. Perform Cohen's d equationd. State answerUse the formula to find the standard error of the distribution of differences in sample means, I1 – X2. Samples of size 110 from Population 1 with mean 82 and standard deviation 14 and samples of size 80 from Population 2 with mean 76 and standard deviation 17 Round your answer for the standard error to two decimal places. standard error = iSuppose you take a random sample of n = 80 Douglas College students and ask each student their age. You find that the mean age of students in this sample is X = 22.0375 years and the sample standard deviation is s = 3.2014 years. Which of the following are sufficient conditions and are also true (to the best of our knowledge) for this study of Douglas College students’ ages? Select all that apply. A. X is normal variable B. n is greater or equal to 30 C. nq is greater or equal to 5 D. np is greater or equal to 5
- A variable of two populations has a mean of 30 and a standard deviation of 18 for one of the populations and a mean of 30 and a standard deviation of 32 for the other population. Complete parts (a) through (c). a. For independent samples of size 9 and 16, respectively, find the mean and standard deviation of X₁ -X₂. (Assume that the sampling is done with replacement or that the population is large enough.) The mean of X₁ -X₂ is. (Type an integer or a decimal. Do not round.) The standard deviation of X₁-X₂ is (Round to four decimal places as needed.) b. Must the variable under consideration be normally distributed on each of the two populations for you to answer part (a)? Choose the correct answer below. OA. No, the variable does not need to be normally distributed for the formulas for the mean and standard deviation of x₁-x₂ to hold as long as the sample sizes are large enough, as long as the sampling is done with replacement. B. No, the formulas for the mean and standard deviation of…In a population of a certain species of newt, the mean length of the newts is μ= 63 inches and the standard deviation is σ= 3.4 inches. You collect a random sample of n=16 newts. The bell curve below represents the distribution of these sample means. The scale on the horizontal axis is the standard error of the sampling distribution. Complete the indicated boxes, correct to two decimal places. mean of distribution= standard error= Note: The left box is 2 standard errors below the mean. The middle box is the mean. The right box is 2 standard errors above the mean.The weights of cans of Ocean brand tuna are supposed to have a net weight of 6 ounces. The manufacturer tells you that the net weight is actually a Normal random variable with a mean of 5.95 ounces and a standard deviation of 0.2 ounces. Suppose that you draw a random sample of 42 cans. Parti) Suppose the number of cans drawn is doubled. How will the standard deviation sample mean weight change? A. It will decrease by a factor of √2. B. It will increase by a factor of √2. C. It will increase by a factor of 2. D. It will decrease by a factor of 2. E. It will remain unchanged. Part ii) Suppose the number of cans drawn is doubled. How will the mean of the sample mean weight change? ▸ A. It will increase by a factor of √2. B. It will decrease by a factor of 2. C. It will increase by a factor of 2. D. It will decrease by a factor of √2. E. It will remain unchanged. Part iii) Consider the statement: The distribution of the mean weight of the sampled cans of Ocean brand tuna is Normal." A. It…
- Use the data in the Excel workbook, on the worksheet labeled "Sample," for this question. The first column is just an observation index labeled i to name each observation. The observations are a simple random sample with replacement from a much larger population. The second column contains specific values y of a random variable Y found in some sampled population. The population standard deviation is not known to you, but it is known that Y has a Normal distribution in the sampled population. Compute the lower bound of an 85% confidence interval for the population mean of the random variable Y. i y 1 37 2 40 3 31 4 27 5 23 6 33 7 29 8 26 9 39 10 28 11 34 12 39 13 31 14 34 15 39 16 33 17 31 18 42B Choosing a topic Sk.. For data that is not normally distributed we can't use z-scores. However, there is an equation that works on any distribution. It's called Chebyshev's formula. The formula 1 where p is the minimum percentage of scores that fall within k k2 is p = 1 - standard deviations on both sides of the mean. Use this formula to answer the following questions. a) If you have scores and you don't know if they are normally distributed, find the minimum percentage of scores that fall within 2.2 standard deviations on both sides of the mean? b) If you have scores that are normally distributed, find the percentage of scores that fall within 2.2 standard deviations on both sides of the mean? c) If you have scores and you don't know if they are normally distributed, how many standard deviations on both sides of the mean do we need to go to have 64 percent of the scores? Note: To answer part c you will need to solve the equation for k.Determine if the following statements are true or false. a. Jane took a sample of n = 100 observations from a population with mean and standard deviation. Mike also took a sample of n = 200 observations from the same population. Then, Mike’s sample mean will be closer to the true population mean than Jane’s sample. b. Jane took a sample of n = 100 observations from a population with mean and standard deviation. Mike also took a sample of n = 200 observations from a population with mean and standard deviation 2. Then, the sample mean from Mike’s sample is more likely to be close to the true mean than the sample mean from Jane’s sample. That is, the probability that X is between 90% and 110% of mean is higher for Mike’s sample than for Jane’s sample.
