1. Find the variance of the t - distribution with 35 degrees offreedom.2. Find the value of ta for a 95% confidence interval when thesample size is 25.23. Find an interval of values for Z using the standard normaldistribution, corresponding to an area of 91.3%

1) The variance of t-distribution can be obtained using this formula:Variance = df−2dfwhere:df =…

Answered: 1. Find the variance of the t -…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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3. Consider a standard normal distribution. a. If z is a standard normal variable, then P(-0.73
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Test the claim below about the mean of the differences for a population of paired data at the level of significance a. Assume the samples are random and dependent, and the populations are normally distributed. Claim: 0 В. Но На > 0 Ha: Ha s0 O C. Ho: Ha = 0 Ha: Hd #0 D. Ho: Ha <0 Ha: Hd 20 O E. Ho: Hd + 0 Ha: Hd F. Ho: Hd 20 Ha: Hd = 0 <0 The test statistic is t= (Round to two decimal places as needed.)
A population of values has a normal distribution with u = 36 and o = 31.3. You intend to draw a random sample of size n = 12. Please answer the following questions, and show your answers to 1 decimal place. Find the value separating the bottom 11% values from the top 89% values. Find the sample mean separating the bottom 11% sample means from the top 89% sample means.
use the t- distribution and the sample results to complete the test of the hypothesis. use a 5% significance level. Assume the results come from a random sample , and if the sample size is small, assume the underlying distribution is relatively normal . H0: miu=4 vs Ha: miu not equal to 4 using the sample results x bar =4.8 s= 2.3 with n=15. WHAT IS THE P-VALUE?
1.Find the z* values based on a standard normal distribution for the following. a.An 82% confidence interval. b.A 93% confidence interval. c.A 78% confidence interval.
This assignment is worth 1 points. The extra point will be added to your overall course grade. For example: if you receive an 88% in the course you can receive up to 1 point giving you a new score of 89%. The following rubric will be used. 0.25 point for drawing the normal distribution curve with the mean value labeled on the curve and the appropriate area shaded. 0.25 point for determining the value of the standard deviation of the sample mean. 0.5 point for finding the correct probability. All work must be shown in order to receive any credit. Please upload your completed assignment here. The length of time taken on the SAT for a group of students is normally distributed with a mean of 2.5 hours and a standard deviation of 0.25 hours. A sample size of n = 60 is drawn randomly from the population. Find the probability that the sample mean is between two hours and three hours.
Can a normal approximation be used for a sampling distribution of sample means from a population with u = 55 and o = 10, when n = 4? Answer E Tables E Keypad Keyboard Shortcuts O Yes, because the mean is greater than 30. O Yes, because the sample size is less than 30. O No, because the sample size is less than 30. O No, because the standard deviation is too small.
Test the claim below about the mean of the differences for a population of paired data at the level of significance x. Assume the samples are random and dependent, and the populations are normally distributed. Claim: μ ≥0; x = 0.10. Sample statistics: d= -2.4, s = 1.3, n = 19 Identify the null and alternative hypotheses. Choose the correct answer below. OA. Ho: Hd 0 The test statistic is t = (Round to two decimal places as needed.) The P-value is (Round to three decimal places as needed.) Since the P-value is the level of significance, C OB. Ho: Hd>0 Ha: Hd ≤0 O D. Ho: Hd #0 Ha: Hd = 0 OF. Ho: Hd 20 Ha: Hd <0 the null hypothesis. There statistically significant evidence to reject the claim.
10) A county is considering raising the speed limit on a road because they claim that the mean speed of vehicles is greater than 40 mi/hr. A random sample of 15 vehicles has a mean speed of 43 mil/hr and a standard deviation of 4.4 mi/hr. Assume that speeds follow a normal distribution. What statistical test should be used to test the claim? (pick one of the options) 1. t-test 2. 2.73 3. 15 4. 10.9 5. 14 6. z-test 7. 2.52 8. 2.64 Calculate the value of standardized test statistic. (pick one of the options) 1. t-test 2. 2.73 3. 15 4. 10.9 5. 14 6. z-test 7. 2.52 8. 2.64 What is the degree of the freedom for the distribution of the test statistics. (pick one of the options) 1. t-test 2. 2.73 3. 15 4. 10.9 5. 14 6. z-test 7. 2.52 8. 2.64
You have data drawn from a normal distribution with a known variance of 16. You set up the following NHST: • Ho: data follows a N(2, 4²) • HÃ: data follows a N(µ, 4²) where µ ‡ 2. Test statistic: standardized sample mean z. Significance level set to a = .05. You then collected n = 16 data points with sample mean 1.5. (a) Find the rejection region. Draw a graph indicating the null distribution and the rejection region. (b) Find the z-value and add it to your picture in part (a). (c) Find the p-value for this data and decide whether or not to reject Ho in favor of HA.
Data on the weights (lb) of the contents of cans of diet soda versus the contents of cans of the regular version of the soda is summarized to the right. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.05 significance level for both parts. a. Test the claim that the contents of cans of diet soda have weights with a mean that is less than the mean for the regular soda. What are the null and alternative hypotheses? OA. Ho: H₁ H₂ H₁: H₁ H₂ O C. Ho: H#2 H₁ H₁

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1. Find the variance of the t - distribution with 35 degrees of freedom. 2. Find the value of ta for a 95% confidence interval when the sample size is 25. 2 3. Find an interval of values for Z using the standard normal distribution, corresponding to an area of 91.3%