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A population of proportions has a distribution with p=0.46p=0.46 and σ=0.498σ=0.498. You intend to draw a random
According to the Central Limit Theorem:
What is the
ˆp=p^=
What is the standard deviation of the distribution of sample means?
(Report answer accurate to 3 decimal places.)
σˆp=
population proportion,p = 0.46
sample size n = 31
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- If bone density is normally distributed with 0 for the mean and 1 for the standard deviation, what is the probaility that a population has a bone density between the z scores of -0.84 and 1.2?A sample is selected from a population withu = 50. After a treatment is administered to the 16 individuals in the sample, the mean is found to be M = 55 and the standard deviation is s = 8, What is the test statistic for the sample mean?Have American males (AMs) gotten heavier over the last 20 years? A random sample 77 AMs in 2019 had a mean weight of x = 189.030 pounds. A random sample 93 AMs in 1999 had a mean weight of y = 184.795 pounds. It is recognized that the true standard deviation of 2019 AMs weights is σx = 14.04 pounds while it is recognized that the true standard deviation of 1999 AMs weights is σy = 10.03 pounds. The true (unknown) mean of 2019 AMs weights is μx pounds, while the true (unknown) mean of 1999 AMs weights is μy pounds. Weights are known to be a normally distributed. In summary: Type Sample Size Sample Mean Standard Deviation 2019 Data (X) 77 189.030 14.04 1999 Data (Y) 93 184.795 10.03 i) If we used this data to test H0:μx-μy=0 against the alternative Ha:μx-μy>0 then what would the p value have been? j)If we used this data to test H0:μx-μy=0 against the alternative Ha:μx-μy≠0 then what would the p value have been?
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- A random sample of 51 adult coyotes in a region of northern Minnesota showed the average age to be x = 2.01 years, with sample standard deviation s = 0.76 years. However, it is thought that the overall population mean age of coyotes is u = 1.75. Do the sample data indicate that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years? Use a = 0.01. (a) What is the level of significance? State the null and alternate hypotheses. O Ho: H 1.75 yr; Hi: и 1.75 yr; Hi: и %3D 1.75 yr II Но: и 3 1.75 уr; Hi: и > 1.75 yr Ho: H = 1.75 yr; H1: µ # 1.75 yr (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. O The standard normal, since the sample size is large and o is known. The Student's t, since the sample size is large and o is known. The standard normal, since the sample size is large and o is unknown. O The Student's t, since the sample size is large and o is unknown. What is the value of the…Have American males (AMs) gotten heavier over the last 20 years? A random sample 77 AMs in 2019 had a mean weight of x = 189.030 pounds. A random sample 93 AMs in 1999 had a mean weight of y = 184.795 pounds. It is recognized that the true standard deviation of 2019 AMs weights is σx = 14.04 pounds while it is recognized that the true standard deviation of 1999 AMs weights is σy = 10.03 pounds. The true (unknown) mean of 2019 AMs weights is μx pounds, while the true (unknown) mean of 1999 AMs weights is μy pounds. Weights are known to be a normally distributed. In summary: Type Sample Size Sample Mean Standard Deviation 2019 Data (X) 77 189.030 14.04 1999 Data (Y) 93 184.795 10.03 d) Calculate the standard deviation of X - Y? . e) If we wish to create an 98% confidence interval for μx-μy then what is the z critical value used? f)Create an 98% confidence interval for μx-μy .The lengths of pregnancies in a small rural village are normally distributed with a mean of 261 days and a standard deviation of 17 days. What percentage of pregnancies last fewer than 314 days? P(x < 314 days) = P(z<
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