The following data represent the length of time, in days, to recovery for patients randomly treated with one of two medications to clear up severe bladder infections. Find a 90% confidence interval for the difference \( \mu_2 - \mu_1 \) between the mean recovery times for the two medications, assuming normal populations with equal variances.**Medication 1**- \( n_1 = 13 \)- \( \bar{x}_1 = 11 \)- \( s_1^2 = 1.8 \)**Medication 2**- \( n_2 = 19 \)- \( \bar{x}_2 = 16 \)- \( s_2^2 = 1.1 \)Links:- [Click here to view page 1 of the table of critical values of the t-distribution.](#)- [Click here to view page 2 of the table of critical values of the t-distribution.](#)---The confidence interval is \([ \, ] < \mu_2 - \mu_1 < [ \, ] \).*(Round to two decimal places as needed.)* A graduate student wanted to estimate the average time spent studying among graduate students at her school. She randomly sampled graduate students from her school and obtained a 99% confidence interval of (17, 25) hours/week. Which of the following would be true if the level of confidence was lowered to 95%?---Choose the correct answer below.- A. The width of the confidence interval would be smaller.- B. The width of the confidence interval would be larger.- C. The width of the confidence interval would remain the same.- D. More information is needed to determine what would happen to the width of the confidence interval.

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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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2. The average pre-school cost for tuition fees last year was P74,200. The following year, 20 institutions had a mean of P13,100 and a standard deviation of P2,250. Is there sufficient evidence at a = 0.10 level of significance to conclude that the mean cost has increased? Solution: Step 1: State the hypotheses. Ho: H;: Step 2: The level of significance and critical region. a = df = and tertical = Step 3: Compute for the value of one sample t test. teomputed =, Step 4: Decision rule. Step 5: Conclusion.
A medical researcher wants to investigate the amount of time it takes for patients' headache pain to be relieved after taking a new prescription painkiller. She plans to use statistical methods to estimate the mean of the population of relief times. She believes that the population is normally distributed with a standard deviation of 15 minutes. How large a sample should she take to estimate the mean time to within 3 minutes with 93% confidence? ... Sample Size = Note:
4.27 please. I already had 4.26 answered. Thank you!
You are conducting a study to see if the proportion of women over 40 who regularly have mammograms is significantly different from 0.6.With H1 : p ≠ 0.6 you obtain a test statistic of z=-2.971.Use a normal distribution calculator and the test statistic to find the P-value accurate to 4 decimal places. It may be left-tailed, right-tailed, or 2-tailed. P-value=
A sample of size 5 was collected from an unknown population 15 15 16 16 17 1. Use the table of expected z-scores (rounded to the nearest tenth) below to construct the normality plot with expected z-scores on the x-axis and the observations on the y-axis: 6 -1.3 -0.6 -0.2 0.2 0.6 1.3 5 -1.2 -0.5 0 0.5 1.2 -1 19 18 normal O not normal Clear All Draw: 17- 16- 14 T 15- رها 1 2 7 -1.4 -0.8 -0.4 0 0.4 0.8 1.4 8 -1.4 -0.9 -0.5 -0.2 0.2 0.5 0.9 1.4 9 -1.5 -0.9 -0.6 -0.3 0 0.3 0.6 0.9 1.5 2. Based on whether the pattern above is linear or not, in your opinion, was the sample drawn from a normal population or not?
The mean pulse rate for adult males is approximately 61.4 beats per minute. In a sample of 27 overweight males, the mean pulse rate was 62.1 beats per minute with a standard deviation of 4.19 beats per minute. Test at a = 0.10 whether these findings suggest that the mean pulse rate for all such overweight males is significantly greater than 61.4. You obtain a sample of size 27 with a mean of 62.1 and a standard deviation of 4.19. You believe that the population is normally distributed. (a). The null and alternative hypotheses are O Ho: ữ = 62.1 Ha:ữ > 62.1 Ho:µ = 61.4 Ha: µ > 61.4 Ο H0: μ = 61.4 На:р 62.1 Part 2 of 4 (b). The test statistic is (use 4 decimal places)
Need step by step directions to determine the P-value. Thanks
A school district decided to measure the BMI of all 3rd graders and identify the highest 1% and lowest 1% of those students. If the distribution is converted to a z distribution, what would be the upper critical z value and lower critical z value that mark the two significant areas at the tails?
Hoaglin, Mosteller, and Tukey (1983) presented data on blood levels of beta-endorphin as a function of stress. They took beta-endorphin levels for 19 patients 12 hours before surgery and again 10 minutes before surgery. The data are presented below, in fmol/ml Construct a 95% confidence limits on the true mean difference between endorphin levels at the two times described. Participant 12 hours before 10 minutes before 1 10 6.5 2 6.5 14.0 3 8.0 13.5 4 12 18 5 5.0 14.5 6 11.5 9.0 7 5.0 18.0 8…
Hello! I need answer for letter F only using Anderson-Darling test to support the normal distribution of the data.
Suppose you had obtained a test statistic of 2.45 (this is incorrect). What would the P-value be? Draw a conclusion based on a = 0.01. Select one: O a. 0.01 < P-value < 0.02, we fail to reject Ho and do not have sufficient evidence to conclude that the mean difference in LDL from switching from an omnivorous to a plant-based diet for 6 weeks is not the same for men and women. O b. 0.005 < P-value < 0.01, we fail to reject Ho and do not have sufficient evidence to conclude that the mean difference in LDL from switching from an omnivorous to a plant-based diet for 6 weeks is not the same for men and women. O c. 0.01 < P-value < 0.02, we reject Ho and have sufficient evidence to conclude that the mean difference in LDL from switching from an omnivorous to a plant-based diet for 6 weeks is not the same for men and women. O d. 0.005 < P-value < 0.02, we reject Ho and have sufficient evidence to conclude that the mean difference in LDL from switching from an omnivorous to a plant-based diet…
The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.7 days and a standard deviation of 2.1 days.What is the 80th percentile for recovery times? (Round your answer to two decimal places.)

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