A group of 26 canadians were randomly selected among a local group of fitness enthusiasts to participate in an innovative fitness program to lower their heart rate. After four months the group was evaluated to identify if their average heart rate was lower than the national average of 72 beats/minute. The mean heart rate for the group was 68. 5 bpm with a standard deviation of 6 bpm. Is there sufficient evidence to believe that the average heart rate for canadians who participate in this innovative fitness program is lower than the national average heart rate? what statistical procedure should be used to answer this research question?.

The term "standard deviation" (or "σ") refers to a measurement of the data's dispersion from the mean.
A low standard deviation implies that the data are grouped around the mean, whereas a large standard deviation shows that the data are more dispersed.

Explanation:

Since we are interested in comparing here the mean heart rate for Canadians who participated in this fitness program, to that of the national average,
We have to have to conduct a hypothesis test here for sample mean
Since we are not trying to find the true population mean based on the sample mean, we do not have to construct a confidence interval
As the sample size, n = 26 < 30,therefore we will use the t-test instead of the z-test.

Therefore, A one-sample t-test for means' is the correct choice.

To know more about sample-test check below link:

https://brainly.com/question/24466382

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