{"id":185554,"date":"2022-04-28T13:23:34","date_gmt":"2022-04-28T13:23:34","guid":{"rendered":"https:\/\/thestudycorp.com\/?p=185554"},"modified":"2022-04-28T13:24:21","modified_gmt":"2022-04-28T13:24:21","slug":"calculating-statistical-analysis-2","status":"publish","type":"post","link":"https:\/\/studyinghq.com\/essay-examples\/nursing\/calculating-statistical-analysis-2\/","title":{"rendered":"Calculating Statistical Analysis – Calculate the mean, median, mode and standard deviation for temperatures by sorting for Latitude 3."},"content":{"rendered":"\n

This article provides a sample solution for Calculating Statistical Analysis.<\/a><\/p>\n\n\n\n

Permalink: https:\/\/studyinghq.com\/essay-examples\/calculating-statistical-analysis<\/a><\/p>\n\n\n\n

As you continue, thestudycorp.com<\/a> has the top and most qualified writers to help with any of your assignments. All you need to do is place an order <\/a>with us. <\/p>\n\n\n\n

For this exercise, use the attached dataset to complete the following:<\/p>

Calculate the mean, median, mode and standard deviation for temperatures by sorting for Latitude 3.<\/p>

Conduct a hypothesis testing for temperatures in Latitude 1. A two-tailed hypothesis test for a single sample<\/a> t-test. Report reject or retain the null, and interpret what the results indicate.<\/p>

…<\/p><\/blockquote>\n\n\n\n

Solution<\/strong><\/h2>\n\n\n\n

Calculating Statistical Analysis<\/h2>\n\n\n\n

Question 1<\/p>\n\n\n\n

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.<\/p>\n\n\n\n

 Question 2<\/p>\n\n\n\n

H0<\/sub>: The temperature value in latitude 1 is insignificant.<\/p>\n\n\n\n

H1<\/sub>: The temperature value in latitude 1 is significant.<\/p>\n\n\n\n

Temperature 1 had the following inference on its test of hypothesis<\/p>\n\n\n\n

Mean<\/td>9.275362319<\/td><\/tr>
Variance<\/td>168.4083546<\/td><\/tr>
Observations<\/td>69<\/td><\/tr>
df<\/td>68<\/td><\/tr>
t Stat<\/td>-3.331449471<\/td><\/tr>
P(T<=t) one-tail<\/td>0.00069979<\/td><\/tr>
t Critical one-tail<\/td>1.667572281<\/td><\/tr>
P(T<=t) two-tail<\/td>0.00139958<\/td><\/tr>
t Critical two-tail<\/td>1.995468931<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Since the analysis were done one a one sample two tailed test, the t calculated value is<\/p>\n\n\n\n

-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.<\/p>\n\n\n\n

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.<\/p>\n\n\n\n

A conclusion is therefore made that the temperature value in latitude 1 are all significant in the dataset.<\/p>\n\n\n\n

Question 3<\/p>\n\n\n\n

H0<\/sub>: The mean temperature in all locations is not equal to 14.48<\/p>\n\n\n\n

H1<\/sub>: The mean temperature in all locations is equal to 14.48<\/p>\n\n\n\n

Mean<\/td>14.4787234<\/td><\/tr>
Variance<\/td>208.7468543<\/td><\/tr>
Observations<\/td>94<\/td><\/tr>
Hypothesized Mean<\/td>14.48<\/td><\/tr>
df<\/td>93<\/td><\/tr>
t Stat<\/td>-0.000856658<\/td><\/tr>
P(T<=t) one-tail<\/td>0.49965916<\/td><\/tr>
t Critical one-tail<\/td>1.661403674<\/td><\/tr>
P(T<=t) two-tail<\/td>0.999318321<\/td><\/tr>
t Critical two-tail<\/td>1.985801814<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

As you continue, thestudycorp.com<\/a> has the top and most qualified writers to help with any of your assignments. All you need to do is place an order <\/a>with us. (Calculating Statistical Analysis)<\/p>\n\n\n\n

\"Calculating<\/a>
Calculating Statistical Analysis<\/figcaption><\/figure>\n\n\n\n

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.<\/p>\n\n\n\n

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.<\/p>\n\n\n\n

A conclusion is therefore made that the temperature value in all the locations is not equal to 14.48.<\/p>\n\n\n\n

Question<\/strong><\/h2>\n\n\n\n

For this exercise, use the attached dataset to complete the following:<\/p>\n\n\n\n