A sample of 16 overweight people take a new reducing pill for 6 weeks and have a mean weight loss of 20 lbs . The standard deviation is 8 lbs . You may assume that the distribution is normal .a) Find a 95 % confidence interval for the mean weight loss .b) If the same information was found using a sample of 50 , how would that cffect the length of the confidence interval the standard standard deviation of S.16.) Find a 99% confidence interval for the mean price of regular gasoline.b) If we had used a 95% confidence level, how would that have effected thelength of the confidence interval?c) Based on your answer in part a) of this problem do you accept or rejectthe Retailers claim that the mean price is $2.18 Please explain youranswer.4) A sample of 16 overweight people take a new reducing pill for 6 weeksand have a mean weight loss of 20 lbs. The standard deviation is 8 Ibs.You may assume that the distribution is normal.a) Find a 95% confidence interval for the mean weight loss.b) If the same information was found using a sample of 50, how would thatcffect the length of the confidence interval5) The mean score for the Verbal SAT in the USA is M =480, the standarddeviation is6- 90 A sample of 100 students in a school district had a samplemeanX 512.At the 1 % significance level is the data suflicient to show that the SATscores are higher in this district

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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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A sample of 16 overweight people take a new reducing pill for 6 weeks and have a mean weight loss of 20 lbs . The standard deviation is 8 lbs . You may assume that the distribution is normal .

a) Find a 95 % confidence interval for the mean weight loss .

b) If the same information was found using a sample of 50 , how would that cffect the length of the confidence interval the standard

standard deviation of S.16.
) Find a 99% confidence interval for the mean price of regular gasoline.
b) If we had used a 95% confidence level, how would that have effected the
length of the confidence interval?
c) Based on your answer in part a) of this problem do you accept or reject
the Retailers claim that the mean price is $2.18 Please explain your
answer.
4) A sample of 16 overweight people take a new reducing pill for 6 weeks
and have a mean weight loss of 20 lbs. The standard deviation is 8 Ibs.
You may assume that the distribution is normal.
a) Find a 95% confidence interval for the mean weight loss.
b) If the same information was found using a sample of 50, how would that
cffect the length of the confidence interval
5) The mean score for the Verbal SAT in the USA is M =480, the standard
deviation is6- 90 A sample of 100 students in a school district had a sample
meanX 512.
At the 1 % significance level is the data suflicient to show that the SAT
scores are higher in this district

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Expert Solution

Step 1

The sample mean = 20.

Sample standard deviation, s is 8.

The sample size, n is 16.

From the EXCEL, using the formula, =T.INV.2T(0.05,15), the critical value is 2.1314.

The 95% confidence interval mean is calculated as follows:

$\begin{array}{l} C I = \bar{x} \pm t (\frac{s}{\sqrt{n}}) \\ = 20 \pm 2.1314 \times \frac{8}{\sqrt{16}} \\ = 20 \pm 4.2628 \\ = (15.7372, 24.2628) \end{array}$

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