This article provides a sample solution for Calculating Statistical Analysis.
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For this exercise, use the attached dataset to complete the following:
Calculate the mean, median, mode and standard deviation for temperatures by sorting for Latitude 3.
Conduct a hypothesis testing for temperatures in Latitude 1. A two-tailed hypothesis test for a single sample t-test. Report reject or retain the null, and interpret what the results indicate.
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Solution
Calculating Statistical Analysis
Question 1
From the dataset, temperature 3 had the following descriptive statistics after its analysis; the mean was 28.6, Mode at 25, Median was 28.5 and standard deviation was 2.674987.
Question 2
H0: The temperature value in latitude 1 is insignificant.
H1: The temperature value in latitude 1 is significant.
Temperature 1 had the following inference on its test of hypothesis
Mean | 9.275362319 |
Variance | 168.4083546 |
Observations | 69 |
df | 68 |
t Stat | -3.331449471 |
P(T<=t) one-tail | 0.00069979 |
t Critical one-tail | 1.667572281 |
P(T<=t) two-tail | 0.00139958 |
t Critical two-tail | 1.995468931 |
Since the analysis were done one a one sample two tailed test, the t calculated value is
-3.331449471 with the T critical value being 1.995468931. This leads to rejection of the null hypothesis since the absolute t calculated is greater than t critical.
The p value is 0.00139958 which is less than the alpha value of 0.05, this leads to the decision to reject the null hypothesis.
A conclusion is therefore made that the temperature value in latitude 1 are all significant in the dataset.
Question 3
H0: The mean temperature in all locations is not equal to 14.48
H1: The mean temperature in all locations is equal to 14.48
Mean | 14.4787234 |
Variance | 208.7468543 |
Observations | 94 |
Hypothesized Mean | 14.48 |
df | 93 |
t Stat | -0.000856658 |
P(T<=t) one-tail | 0.49965916 |
t Critical one-tail | 1.661403674 |
P(T<=t) two-tail | 0.999318321 |
t Critical two-tail | 1.985801814 |
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The analysis were done one a two tailed test, the t calculated value is -0.000856658 with the T critical value being 1.985801814. This leads to failure to reject the null hypothesis since the absolute t calculated is less than t critical.
The p value is 0.999318321 which is greater than the alpha value of 0.05, this leads to the decision to fail to reject the null hypothesis.
A conclusion is therefore made that the temperature value in all the locations is not equal to 14.48.
Question
For this exercise, use the attached dataset to complete the following:
- Calculate the mean, median, mode and standard deviation for temperatures by sorting for Latitude 3.
- Conduct a hypothesis testing for temperatures in Latitude 1. A two-tailed hypothesis test for a single sample t-test. Report reject or retain the null, and interpret what the results indicate.
- The average temperature for all locations is 14.48 degrees.
- Test to see if the temperature is different from 14.48 degrees.