A researcher takes sample temperatures in Fahrenheit of 16 days from Miami and 14 days from Atlanta. Use the sample data shown in the table.Test the claim that the mean temperature in Miami greater than the mean temperature in Atlanta. Use a significance level of α=0.10α=0.10.Assume the populations are approximately normally distributed with unequal variances.Note that list 1 is longer than list 2, so these are 2 independent samples, not matched pairs. MiamiAtlanta68.373.683.169.379.154.97281.172.878.683.35482.736.180.744.38758.483.150.877.460.586.161.276.346.874.554.483.3 78.5 The Null Hypotheses is: H0: μ1 - μ2 = 0What is the alterative hypothesis? Select the correct symbols for each space. (Note this may view better in full screen mode.)HA: μ1 - μ2 Based on these hypotheses, find the following. Round answers to 4 decimal places.Test Statistic = p-value = The p-value is The correct decision is to .The correct summary would be: the claim that the mean temperature in Miami greater than the mean temperature in Atlanta.

For Miami : sample size (n1) = 16sample mean (x̄1) = Σx1 /n1 = 1268.2/16 = 79.26sample standard deviation (s1) =√ (x1-x̄1)2 /(n1-1) = √ (432.7776)/(16-1) = 5.3714For Atlanta : sample size (n2) = 14sample mean (x̄2) = Σx2 /n2 = 824/14 = 58.86sample standard deviation (s2) = √ (x2-x̄2)2 /(n2-1) = √ 2218.5344/(14-1) = 13.0636 1) Null and alternative hypothesis H0: μ1 = μ2 HA: μ1 > μ2 2) pooled sample variance ( sp2 ) = ( (n1-1)*s12 + (n2-1)*s22) /(n1+n2-2) = ((16-1)*5.37142 + (14-1)*13.06362)/(16+14-2) = 94.6903 Standard error (S.E.) = √sp2 * (1/n1 + 1/n2) = √ 94.6903*(1/16 +1/14) = 3.5611 Test statistic (t) = (x̄1-x̄2)/Standard error (S.E.) = (79.26-58.86)/3.5611 = 5.7286 3) degree of freedom (df) = n1+n2-2 = 16+14+2 = 28 significance level (α) = 0.10 4) For a one-tailed 't' test,P-value = 0.00001 ( from excel) 5) since P-value < significance level (α), we reject the null hypothesis. The result is significant at the 0.1 significance level. 6) Conclusion: We can not conclude that the mean temperature in Miami is greater than the mean temperature in Atlanta.

Answered: A researcher takes sample temperatures…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

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A researcher takes sample temperatures in Fahrenheit of 16 days from Miami and 14 days from Atlanta. Use the sample data shown in the table.

Test the claim that the mean temperature in Miami greater than the mean temperature in Atlanta. Use a significance level of α=0.10α=0.10.
Assume the populations are approximately normally distributed with unequal variances.
Note that list 1 is longer than list 2, so these are 2 independent samples, not matched pairs.

Miami	Atlanta
68.3	73.6
83.1	69.3
79.1	54.9
72	81.1
72.8	78.6
83.3	54
82.7	36.1
80.7	44.3
87	58.4
83.1	50.8
77.4	60.5
86.1	61.2
76.3	46.8
74.5	54.4
83.3
78.5

The Null Hypotheses is: H₀: μ1 - μ2 = 0

What is the alterative hypothesis? Select the correct symbols for each space. (Note this may view better in full screen mode.)

H_A: μ1 - μ2

Based on these hypotheses, find the following. Round answers to 4 decimal places.

Test Statistic =

p-value =

The p-value is

The correct decision is to .

The correct summary would be: the claim that the mean temperature in Miami greater than the mean temperature in Atlanta.

Definition Definition Measure of central tendency that is the value that occurs most frequently in a data set. A data set may have more than one mode if multiple categories repeat an equal number of times. For example, in a data set with five item—3, 5, 5, 29, 473—the mode is 5 because it occurs twice and no other value occurs more than once. On a histogram or bar chart, the element with the highest bar represents the mode. Therefore, the mode is sometimes considered the most popular option. The mode is useful for nominal or categorical data (e.g., the most common color car that users purchase), but it is problematic for continuous data because it is more likely not to have any value that is more frequent than the other.

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