### Confidence Interval AnalysisThe 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. We need to find a 95% confidence interval for the difference \(\mu_2 - \mu_1\) between the mean recovery times for the two medications, assuming normal populations with equal variances.#### Data Summary:| Medication | \( n \) (Sample Size) | \( \overline{x} \) (Sample Mean) | \( s^2 \) (Sample Variance) ||------------|-----------------------|---------------------------------|----------------------------|| Medication 1 | \( n_1 = 13 \) | \( \overline{x}_1 = 14 \) | \( s_1^2 = 1.7 \) || Medication 2 | \( n_2 = 17 \) | \( \overline{x}_2 = 18 \) | \( s_2^2 = 1.3 \) |#### Confidence Interval Formula:The formula for the confidence interval for the difference between two means (\(\mu_2 - \mu_1\)) with equal variances is given by:\[ \left( \overline{x}_2 - \overline{x}_1 \right) \pm t_{\alpha/2, \, df} \cdot \sqrt{ \left( \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} \right) \left( \frac{1}{n_1} + \frac{1}{n_2} \right) } \]#### Calculations:- Calculate the pooled variance: \[ s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} \] - Insert the sample values: \[ s_p^2 = \frac{(13 - 1) \cdot 1.7 + (17 - 1) \cdot 1.3}{13 + 17 - 2} = \frac{12 \cdot 1.7 + 16 \cd

Answered: Image /qna-images/answer/840543e5-a860-47d5-bf51-d62753c52284.jpg

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 95% confidence interval for the difference p,-H, between in the mean recovery times for the two medications, assuming normal populations with equal variances. Medication 1 s =17 s3-13 n = 13 X1 = 14 Medication 2 n2 = 17 X2 = 18 The confidence interval is <µ,-< (Round to two decimal places as needed.) Enter your answer in the edit fields and then click Check Answer All parts showing Clear All Fal Cherk

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 95% confidence interval for the difference p,-H, between in the mean recovery times for the two medications, assuming normal populations with equal variances. Medication 1 s =17 s3-13 n = 13 X1 = 14 Medication 2 n2 = 17 X2 = 18 The confidence interval is <µ,-< (Round to two decimal places as needed.) Enter your answer in the edit fields and then click Check Answer All parts showing Clear All Fal Cherk

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Construct a 90% confidence interval for μ1−μ2 with the sample statistics for mean calorie content of two bakeries' specialty pies and confidence interval construction formula below. Assume the populations are approximately normal with equal variances. Bakery A: x1 = 1890 calories, s1 = 134 and n1 = 8 Bakery B: x2 = 1712, s2 = 186 calories and n1 = 9 ------------------------------------------------------------------------- The confidence interval construction formula is in the attachment. Looking for __ <μ1 − μ2< __ (Round to the nearest integer as needed.)
b) Following regression output is based on a sample of 30 farms where Y = output of rice per acre in tonnes and X = quantity of manure applied per acre in kgs. = 384.105 + 3.67X¡ se = (151.54) (1.00) RSS = 6776 Construct a 95% confidence interval for mean output when 8kg of manure is applied given that the sample average of manure applied per acre is 5kgs.
Independent random samples from two normally distributed populations give the following results:nx =10x̅ =480sx=30 ny=12ȳ =520sy =25 If we assume that the unknown population variances are equal, find the 90% confidence interval for the difference of population means (40 points).
in a test of the effectiveness of garlic for lowering cholesterol. 45 subjects were treated with garlic in a processed tablet form Cholesterol levels were measured before and after the treatment The changes K(before-after) in their levels of LDL cholesterol (in mgid:) have a moan of 46 and a standard deviation of 15.5 Construct a 90% confidence interval estimate of the mean net change in LDL cholesterol after the geric treatment. What does the confidence interval suggest about the effectiveness of garic in reducing LDL cholestero Click here new.a.fdstribution tabee Click here to view page 1. of the standard.normal distribution table Cack here to view p900 2 of the standard.normal distobubun table. What is the confidence interval estimate of the population mean? mgmid (Round to two decimal places as needed) -cor
The distribution of study times for first year students follows a Normal distribution with standard deviation σσ = 66 minutes. Suppose a sample of 468 students found that the mean study time for these students was x¯x¯= 195 minutes per week. Use this to find a 88.58% confidence interval to estimate the mean study time for all students. We are 88.58% confident that the mean study time for first year students is between What is the margin of error for your confidence interval? Margin of error =
Construct a 99% confidence interval for H1 H2 with the sample statistics for mean cholesterol content of a hamburger from two fast food chains and confidence interval construction formula below. Assume the populations are approximately normal with unequal variances. Stats x₁ = 135 mg, s₁ = 3.82 mg, n₁ = 18 Confidence interval when (x-2)- variances are not equal 2 A
A researcher was interested in comparing the resting pulse of people who exercise regularly and people wo do not exercise regularly. Independent simple random samples of 16 people ages 30-40 who do no exercise regularly and 12 people ages 30-40 who exercise regularly were selected, and the resting pulse rate of each person was measured. The summary statistics are as follows. Do not exercise Do exercise X1= 73.9 X2= 69.1 S1= 10.9 S2= 10.9 N1= 16 N2=12 1. Verify the conditions needed to proceed with hypothesis testing or Confidence intervals. 2. At the 5% significance level, do the data provide sufficient evidence to conclude that the mean resting pulse rate of people who do not exercise regularly is greater than the mean resting pulse rate of people who exercise regularly? State your null, alternate hypothesis, calculate the test statistic, determine your decision and interpret your decision. 3. Find the 95% confidence interval around the difference between the sample…
Do men take less time than women to get out of bed in the morning? The 48 men observed averaged 7.7 minutes to get out of bed after the alarm rang. Their standard deviation was 1.8. The 55 women observed averaged 8 minutes and their standard deviation was 1.9 minutes. What can be concluded at the αα = 0.01 level of significance? For this study, we should use Select an answer t-test for the difference between two dependent population means z-test for the difference between two population proportions t-test for a population mean t-test for the difference between two independent population means z-test for a population proportion The null and alternative hypotheses would be: H0:H0: Select an answer μ1 p1 Select an answer > ≠ = < Select an answer p2 μ2 (please enter a decimal) H1:H1: Select an answer μ1 p1 Select an answer > < = ≠ Select an answer p2 μ2 (Please enter a decimal) The test statistic ? t z = (please show your answer to 3 decimal places.) The…
In a study of morphological variation in natural populations of fruit fly, it was reported that the mean wing length of 16 females, collected in a certain area, was 4.653 mm with s = 0.012 mm; and the mean wing length of 11 females, collected in a second area, was 4.274 mm with s = 0.02. Let the distribution of the wing length be normal, find a 98% confidence interval on the ratio of the two populations’ standard deviations.
3. Overproduction of uric acid in the body can be an indication of cell breakdown. Over a period of clam 2 months, an adult male patient has taken eight blood tests for uric acid. The mean concentration was x = 5.35 mg/dl. The distribution of uric acid in healthy adult males can be assumed to be sono normal, with σ = 1.86mg/dl. Find a 95% confidence interval for the population mean concentration of uric acid in this patient's blood. What is the margin of error?
Volunteers who had developed a cold within the previous 24 hours were randomized to take either zinc or placebo lozenges every 2 to 3 hours until their cold symptoms were gone. Twenty-five participants took zinc lozenges, and 23 participants took placebo lozenges. For the placebo group, the mean overall duration of symptoms was x, 7.7 days, and the standard deviation was 1.4 days. The mean overall duration of symptoms for the zinc lozenge group was x, 4.2 days, and the standard deviation of overall duration of symptoms was 1.4 days. n USE SALT (a) Calculate x, - x, difference in sample means. days 2x - 'x Compute the unpooled s.e.(x, – x,) standard error of the difference in means. (Round your answer to four decimal places.) s.e.(x1 - X2) = days (b) Compute a 95% confidence interval for the difference in mean days of overall symptoms for the placebo and zinc lozenge treatments. Use the unpooled standard error and use the smaller of n1 1 and n2 1 as a conservative estimate of degrees of…
A scientist wants to establish if the Mississippi’s river height has increased using data from 1970 and 2020. Historically, it’s known that the variance of the height within any given year is 4.5ft. The sample from 1970 has 80 measurements with ?1μ1 = 30ft and the sample from 2020 has 260 measurements with ?2μ2 = 31.5ft. Find a 95% confidence interval for the average increase of height.