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A university will use a support course model to improve student grades. The average grade before the support course model was used was 70%. The university gives 10 classes, which total 300 students, to test out the support course model before using the model on the entire schedule of statistics classes. The average grade of the 300 students was 72% with a standard deviation of 12%. We use a 5% significance level to determine if the support course model is successful at improving overall grades. Give the null and alternative hypothesis. Null: The population average is 70, Alternative: The population average is less than 70 Null: The population average is 70, Alternative: The population average is greater than 70 Null: The population average is 72%, Alternative: The population average is less than 72% The population average is 72%, Alternative: The population average is greater than 72% Give the rejection region. z greater than 1.64 z less than -1.64 t greater than 1.64 t less than -1.64 Give the test statistic. t = 2.89 z = 2.89 z = 0.002 t = 0.002 Give the p-value. 2.89 0.05 0.001 1.64 Give the conclusion. Reject the null since the test statistic is in the rejection region and the p value is smaller than the significance Reject the null since the test statistic is in the rejection region and the p-value is larger than the significance Fail to reject the null since the test statistic is not in the rejection region and the p-value is larger than the significance Fail to reject the null since the test statistic is not in the rejection region and the p-value is smaller than the significance

A university will use a support course model to improve student grades. The average grade before the support course model was used was 70%. The university gives 10 classes, which total 300 students, to test out the support course model before using the model on the entire schedule of statistics classes. The average grade of the 300 students was 72% with a standard deviation of 12%. We use a 5% significance level to determine if the support course model is successful at improving overall grades. Give the null and alternative hypothesis. Null: The population average is 70, Alternative: The population average is less than 70 Null: The population average is 70, Alternative: The population average is greater than 70 Null: The population average is 72%, Alternative: The population average is less than 72% The population average is 72%, Alternative: The population average is greater than 72% Give the rejection region. z greater than 1.64 z less than -1.64 t greater than 1.64 t less than -1.64 Give the test statistic. t = 2.89 z = 2.89 z = 0.002 t = 0.002 Give the p-value. 2.89 0.05 0.001 1.64 Give the conclusion. Reject the null since the test statistic is in the rejection region and the p value is smaller than the significance Reject the null since the test statistic is in the rejection region and the p-value is larger than the significance Fail to reject the null since the test statistic is not in the rejection region and the p-value is larger than the significance Fail to reject the null since the test statistic is not in the rejection region and the p-value is smaller than the significance

MATLAB: An Introduction with Applications
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 university will use a support course model to improve student grades. The average grade before the support course model was used was 70%. The university gives 10 classes, which total 300 students, to test out the support course model before using the model on the entire schedule of statistics classes. The average grade of the 300 students was 72% with a standard deviation of 12%. We use a 5% significance level to determine if the support course model is successful at improving overall grades. Give the null and alternative hypothesis. Null: The population average is 70, Alternative: The population average is less than 70 Null: The population average is 70, Alternative: The population average is greater than 70 Null: The population average is 72%, Alternative: The population average is less than 72% The population average is 72%, Alternative: The population average is greater than 72% Give the rejection region. z greater than 1.64 z less than -1.64 t greater than 1.64 t less than -1.64 Give the test statistic. t = 2.89 z = 2.89 z = 0.002 t = 0.002 Give the p-value. 2.89 0.05 0.001 1.64 Give the conclusion. Reject the null since the test statistic is in the rejection region and the p value is smaller than the significance Reject the null since the test statistic is in the rejection region and the p-value is larger than the significance Fail to reject the null since the test statistic is not in the rejection region and the p-value is larger than the significance Fail to reject the null since the test statistic is not in the rejection region and the p-value is smaller than the significance
Transcribed Image Text:A university will use a support course model to improve student grades. The average grade before the support course model was used was 70%. The university gives 10 classes, which total 300 students, to test out the support course model before using the model on the entire schedule of statistics classes. The average grade of the 300 students was 72% with a standard deviation of 12%. We use a 5% significance level to determine if the support course model is successful at improving overall grades. Give the null and alternative hypothesis. Null: The population average is 70, Alternative: The population average is less than 70 Null: The population average is 70, Alternative: The population average is greater than 70 Null: The population average is 72%, Alternative: The population average is less than 72% The population average is 72%, Alternative: The population average is greater than 72% Give the rejection region. z greater than 1.64 z less than -1.64 t greater than 1.64 t less than -1.64 Give the test statistic. t = 2.89 z = 2.89 z = 0.002 t = 0.002 Give the p-value. 2.89 0.05 0.001 1.64 Give the conclusion. Reject the null since the test statistic is in the rejection region and the p value is smaller than the significance Reject the null since the test statistic is in the rejection region and the p-value is larger than the significance Fail to reject the null since the test statistic is not in the rejection region and the p-value is larger than the significance Fail to reject the null since the test statistic is not in the rejection region and the p-value is smaller than the significance
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