# Calculating Statistical Analysis

Calculating Statistical Analysis – 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…

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## Calculating Statistical Analysis

### Paper details

For this exercise, use the attached dataset to complete the following:

**(Dataset available on request)**

- 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.

### 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.

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##### 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

- 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.

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