a.) Using technology (calculator or StatCrunch), find the sample mean and sample standard deviation. (Round two decimal places).b.) H0: __________________ H1: __________________Type of Test: ________________Test Statistic: _________________ (2 decimal places)Critical Value(s): ________________ (3 decimal places)P-Value: __________________ (4 decimal places)Label the critical value and test statistic on the curve and shade the rejection region(s).Reject or Fail to Reject: ______________________Conclusion (express in a sentence in the context of the problem) Many consumers pay careful attendant to stated nutritional content on packaged foods when makingpurchases, so it is important that the information on packages be accurate. A random sample of 14 frozendinners of a certain type was selected and the calorie content of each one was determined. Below are theresulting observations. Assume the data is normally distributed. The stated mean calorie content is 240.Perform a hypothesis test to determine if the actual mean calorie content is different from the stated packagevalue of 240. Use a 0.01 level of significance.255 244 239 242 265 245 243259 248 225226 254 233 235

Name: a.) Using technology (calculator or StatCrunch), find the sample mean
Uploaded: 2024-03-20T07:06:16.000Z
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Description: a.) Using technology (calculator or StatCrunch), find the sample mean and sample standard deviation. (Round two decimal places).b.) H0: __________________ H1: __________________Type of Test: ________________Test Statistic: _________________ (2 decima

a) For the given data mean and standard deviation is calculated using Ti 83+ is as follows: Choose Stat>Edit>Edit. Enter the given values in List L1. Select Stat>Calc>1 Var stats. Select ‘2nd’ and then press 1. Output: Sample Mean is 243.79 Sample standard deviation is 11.83b) Hypothesis: H0:μ=240H1:μ≠240 Two tailed test. Since population standard deviation is not known. it is appropriate to use One sample t test. Procedure for 1 sample t test using TI-83+ calculator: Select STAT>TESTS>T test Select Input as DATA Enter µ0 as 240 Select List as L1 and select µ≠ µ0 Select calculate. Output screen: Thus, Test statistic is 1.19 p-value is 0.2527 Critical values: Computation of critical-value: The two tailed critical-value for the t-test at 13 degrees of freedom and 0.01 level of significance is 3.012 can be obtained using the excel formula, “=T.INV.2T(0.01,13)” Critical values are ±3.012.Graph: Decision rule: If p-value ≤α, then reject the null hypothesis. Otherwise, do not reject the null hypothesis. Conclusion: Here, p-value (0.2527) is greater than the level of significance (0.01). Therefore, fail to reject the null hypothesis. The actual mean calorie content is same from the stated package value of 240.

Answered: Using technology (calculator or…

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.) Using technology (calculator or StatCrunch), find the sample mean and sample standard deviation. (Round two decimal places).

b.) H0: __________________
H1: __________________
Type of Test: ________________
Test Statistic: _________________ (2 decimal places)
Critical Value(s): ________________ (3 decimal places)

P-Value: __________________ (4 decimal places)
Label the critical value and test statistic on the curve and shade the rejection region(s).
Reject or Fail to Reject: ______________________
Conclusion (express in a sentence in the context of the problem)

Many consumers pay careful attendant to stated nutritional content on packaged foods when making
purchases, so it is important that the information on packages be accurate. A random sample of 14 frozen
dinners of a certain type was selected and the calorie content of each one was determined. Below are the
resulting observations. Assume the data is normally distributed. The stated mean calorie content is 240.
Perform a hypothesis test to determine if the actual mean calorie content is different from the stated package
value of 240. Use a 0.01 level of significance.
255 244 239 242 265 245 243
259 248 225
226 254 233 235

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 + ...

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