A study at the University of Colorado at Boulder shows that...

A study at the University of Colorado at Boulder shows that running increases the percent resting metabolic rate ( $\mathrm{RMR})$ in older women. The average RMR of 30 elderly women runners was $34.0 \%$ higher than the average RMR of 30 sedentary elderly women, and the standard deviations were reported to be 10.5 and $10.2 \%$, respectively. Was there a significant increase in RMR of the women runners over the sedentary women? Assume the populations to be approximately normally distributed with equal variances. Use a $P$-value in your conclusions.

Key Concepts

Hypothesis Testing

Hypothesis testing is a statistical method used to decide whether there is enough evidence in a sample of data to infer that a certain condition holds for the entire population. In this context, it involves comparing the mean percent RMR of runners versus sedentary women to determine if running has a significant effect on RMR.

Two-Sample t-Test for Independent Means

The two-sample t-test for independent means is used to compare the means of two populations and decide if they are statistically different from each other. It assumes that the data from both groups are normally distributed and have equal variances, which is essential for the validity of the test.

Null and Alternative Hypotheses

The null hypothesis (H0) typically states that there is no difference between the population means, meaning that running does not affect RMR. The alternative hypothesis (Ha) posits that there is a difference, often suggesting that the runners have a higher RMR than the sedentary group. Establishing these hypotheses is a critical first step in conducting any hypothesis test.

P-Value

The P-value is a measure that helps determine the significance of the results obtained from a hypothesis test. It represents the probability of obtaining test results at least as extreme as the ones observed during the study, assuming that the null hypothesis is true. A small P-value indicates that the observed difference is unlikely to have occurred by random chance, leading to the rejection of the null hypothesis.

Assumptions of Normality and Equal Variance

When performing a t-test for two independent samples, it is important to assume that the samples are drawn from populations following a normal distribution and that these populations have equal variances. These assumptions ensure the accuracy and validity of the test's results and conclusions.

Transcript

00:01 For this problem.

00:01 We are told that a study at the university of colorado, excuse me colorado at boulder shows that running increases the percent resting metabolic rate, or rmr in older women, we have that the average rmr of 30 elderly women runners was 34% higher than the average rmr of 30 sedentary elderly women.

00:23 So we can see we can say that x1 bar equals x two bar plus 34.

00:30 And we have that the standard deviations were reported to be 10.5 and 10.2% respectively.

00:36 So sigma one equals 10.5 in sigma two equals 10.2.

00:42 Were then asked.

00:43 Was there a significant increase in rmr of the women runners over the sedentary women so particularly were being asked, is it possible? or is it true that μ one is significantly greater than you two.

00:58 We then have that we assume the populations to be approximately normally distributed with equal variances.

01:05 And we are to use a p value in our conclusions.

01:09 So here the null hypothesis is that μ one is equal to μ two, and the alternative hypothesis is going to be that μ one is greater than you two.

01:21 So actually we can change the first or the null hypothesis to less than or equal to.

01:26 So, what we have then that we want to calculate the value when we're comparing two different samples, which would be equal to x 1 -12 divided by or excuse me? x one minus x two.

01:43 X one bar minus x two bar.

01:45 Oh, actually also slightly change how i'm writing the null hypothesis here.

01:52 So if they know hypothesis is that the distance or the difference between these is zero.

02:00 We can write that μ 1 -92 equals d.

02:05 Which in turn equals zero.

02:06 And then the alternative hypothesis is that μ one is greater than you two...

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