Simplify the expression below 3x² - 5x² + 4x²
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Using the t-distribution, as we have the standard deviation for the sample, it is found that there is a significant difference between the wait times for the two populations.
<h3>What are the hypothesis tested?</h3>At the null hypothesis, we test if there is no difference, that is:
At the alternative hypothesis, it is tested if there is difference, that is:
For each sample, we have that:
For the distribution of differences, we have that:
It is given by:
In which is the value tested at the null hypothesis.
Hence:
Considering a one-tailed test, as stated in the exercise, with 100 - 1 = 99 df, using a t-distribution calculator, the p-value is of 0.014.
Since the p-value is less than the significance level of 0.05, it is found that there is a significant difference between the wait times for the two populations.
More can be learned about the t-distribution at brainly.com/question/16313918