**Problem Description:**A random sample of \( n \) measurements was selected from a population with an unknown mean \(\mu\) and known standard deviation. Calculate a 90% confidence interval for \(\mu\) for each of the situations given in parts a through e.---**a.**- Sample size (\( n \)): 70- Sample mean (\( \bar{x} \)): 26- Population variance (\( \sigma^2 \)): 13**Confidence Interval:**\( (25.29, 26.71) \)(Note: Round to two decimal places as needed.)---**b.**- Sample size (\( n \)): 150- Sample mean (\( \bar{x} \)): 118- Population variance (\( \sigma^2 \)): 24**Confidence Interval:**\( (\underline{\phantom{00}}, \underline{\phantom{00}}) \)(Note: Round to two decimal places as needed.)---**Instructions:**For each scenario, the confidence interval is provided partially or needs completion. Ensure that all calculations are rounded to two decimal places.

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Answered: A random sample of n measurements was…

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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In order to construct a confidence interval for the population variance, a random sample of n observations is drawn from a normal population. Use this information to find x²a/2, df and x²1-a/2, df under the following scenarios. (Round your answers to 3 decimal places. You may find it useful to reference the appropriate table: chi-square table or Ftable) X²al2,df a. A 95% confidence level with n = 9. b. A 95% confidence level with n = 40. c. A 99% confidence level with n = 9. d. A 99% confidence level with n = 40. X²1-a/2,df
7. What is the 95% confidence interval for the difference in the population means, µ1 - µ2, given the data from the two independent samples: n1 =56, x1= 58.2, s1= 4.3, n2=45, x2=60.2, s2=5.5 B. [-3.28 , -0.72] C. [-3.60, -0.40] D. [-3.65 , -0.35] A. [-3.96 , -0.04] 8. What is the 98% confidence interval for the difference in the population means, µ1 - µ2, given: n1= 22, x1= 41.4, s1= 5.9, n2= 19, x2= 43.2, s2= 6.5 D. [-3.48 , -0.12] B. [-3.07 , -0.53] A. [-3.12, -0.48] C. [-3.18, -0.42]
A researcher wishes to compare the mean of time spent watching TV in a day between men and women to test if the mean watching time for men equals the mean time for women. Assume that he performed the needed test and produce the following results: Gender N Mean Standard deviation Female 116 1.95 1.51 Male 59 2.37 1.85 95% confidence Interval for (µ,-µ,):(0.79,0.94) I. Which of the following is the correct conclusion about these results using a 5% significant level? OThe mean TV watching times of men and women are equal OThere is a statistically significant difference between the mean TV watching times of men and women. OThere is not a statistically significant difference between the mean TV watching tímes of men and women. OThere is not enough information to conclude II. Which statement is correct about p-value based on your conclusion about the test? Op-value0.05 Opvalue>0.005 OCannot be determine from the information given
A nutritionist selected a random sample of n = 50 adult males. The average daily intake of dairy products was found to be x (bar above) = 756 grams per day, with a standard deviation of s = 35 grams per day. Use this sample information to construct a 95% confidence interval for the mean daily intake of dairy products for men.
Suppose a randomly selected sample of n = 69 men has a mean foot length of x = 28.3 cm, and the standard deviation of the sample is 3 cm. Calculate an approximate 95% confidence interval for the mean foot length of men. (Round the answers to one decimal place.) to cm
If X=115, 25, and n = 31, construct a 99% confidence interval estimate of the population mean, μ. ☐ sμs ☐ (Round to two decimal places as needed.)
A sample of n = 7 scores is selected from a population with an unknown mean ( µ). The sample has a mean of M = 40 and a variance of s 2 = 63. Which of the following is the correct 95% confidence interval for µ?
2. Scored obtained by a large group of students taking a test are known to be normally distributed. A random sample of twenty-five students test scores yielded the following statistics: 25 と=1Xi 25 1508 x - 95628 Find a 95% confidence interval for the population mean.
Bob selects independent random samples from wo populations and obtains the values pi 0.700 and p2 confidence interval for p1 – P2 and gets: 0.500. He constructs the 95% 0.200 ± 1.96(0.048) = 0.200 ±0.094. %3| Note that 0.048 is called the estimated standard error of pi – P2 (the ESE of the estimate). Tom wants to estimate the mean of the success rates: Pi + p2 (a) Calculate Tom's point estimate. (b) Given that the estimated standard er- ror of (p1 + p2)/2 is 0.024, calculate the 95% confidence interval estimate of (Pi + P2)/2. Hint: The answer has our usual form: Pt. est. ±1.96 × ESE of the estimate.
The random sample shown below was selected from a normal distribution. 3, 7, 7, 5, 10, 4 D Complete parts a and b. a. Construct a 95% confidence interval for the population mean μ. (3.34,8.66) (Round to two decimal places as needed.) b. Assume that sample mean x and sample standard deviation s remain exactly the same as those you just calculated but that are based on a sample of n = 25 observations. Repeat part a. What is the effect of increasing the sample size on the width of the confidence intervals? The confidence interval is (4.96.7.04). (Round to two decimal places as needed.) What is the effect of the sample size on the width of the confidence interval? A. As the sample size increases, the width decreases. OB. As the sample size increases, the width increases. OC. As the sample size increases, the width stays the same.

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