g A test was conducted to determine if there is a difference between Canadian and American average heart rates. A group of 36 Canadians and 34 Americans were randomly selected and their heart rates measured in beats per minute (bpm). The mean heart rate for Canadians was 68.5 and 70 for Americans. The pooled standard deviation of 2.0 beats/minute was calculated. Estimate a plausible range of differences between the mean heart rate of Canadians and Americans. What statistical procedure should be used to answer this research question

Answer:

The test statistic  | t |  =4.45

Step-by-step explanation:

Step(i):-

Given that the first sample size n₁ = 36

Given that the second sample size n₂ = 34

Mean of the first sample x₁⁻ = 68.5

Mean of the second sample x₂⁻ = 70

Step(ii):-

Test statistic

       [tex]t = \frac{x^{-} _{1}-x^{-} _{2} }{\sqrt{S^{2} (\frac{1}{n_{1} }+\frac{1}{n_{2} } } )}[/tex]

     [tex]t = \frac{68.5-70}{\sqrt{2(\frac{1}{36}+\frac{1}{34} ) } }[/tex]

    [tex]t = \frac{-1.5}{0.337} = -4.45[/tex]

The test statistic  | t | = |-4.45| =4.45

The test statistic is given by t and t is calculated with the help of the test statistic formula. The value of t is 4.45.

The test statistic summarises the observation data into a unique number using central tendency, the variation, sample, and size in your statistical model.

The given information will be shown below.

The first sample size n₁ = 36

The second sample size n₂ = 34

The mean of the first sample x₁ = 68.5

The mean of second sample x₂ = 70

Then the test statistic will be given as

[tex]\rm t = \dfrac{x_1 - x_2}{\sqrt{S^2 (\dfrac{1}{n_1}+\dfrac{1}{n_1})}} \\\\\\\t = \dfrac{68.5 - 70}{\sqrt{2 (\dfrac{1}{36} + \dfrac{1}{34})}} \\\\\\t = \dfrac{-1.5}{0.337}\\\\\\t = -4.45[/tex]

The test statistic | t | = |–4.45| = 4.45

More about the test statistic link is given below.

https://brainly.com/question/15236063

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