Respuesta :
Answer:
No, the claim is not supported by the sample evidence
Step-by-step explanation:
We will compare the mean heart rate of the boys playing Wii bowling to the heart rate of boys who ran on a treadmill.
We have:
µ = 98 (boys on treadmill)
σ = unknown or not given
x = 103 (boys playing Wii)
s = 15
n = 14
H0: µ = 98
Ha: µ ≠ 98 (claim)
We will run a hypothesis test at the 0.01 level of significance.
Since n < 30, our critical values will be a t-value.
This is a 2 tailed test, so our critical regions are
t < -3.012 and t > 3.012
If the test statistic falls in either of these regions, we will reject the null hypothesis
Using the t-distribution, it is found that since the p-value of the test is 0.2318 > 0.01, the sample does not provide convincing evidence that the mean heart rate after 15 minutes of Wii Bowling is different from the known mean heart rate after 6 minutes walking on the treadmill.
At the null hypothesis, we test if the mean heart rate is still the same, that is, of 98 bpm, hence:
[tex]H_0: \mu = 98[/tex]
At the alternative hypothesis, we test if the mean heart rate is different, that is, different of 98 bpm, hence:
[tex]H_1: \mu \neq 98[/tex]
We have the standard deviation for the sample, thus, the t-distribution is used. The test statistic is given by:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
The parameters are:
[tex]\overline{x}[/tex] is the sample mean.
[tex]\mu[/tex] is the value tested at the null hypothesis.
s is the standard deviation of the sample.
n is the sample size.
For this problem, the values of the parameters are: [tex]\overline{x} = 103, \mu = 98, s = 15, n = 14[/tex].
Hence, the value of the test statistic is:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{103 - 98}{\frac{15}{\sqrt{14}}}[/tex]
[tex]t = 1.25[/tex]
The p-value is found using a two-tailed test, as it is being tested if the mean is different of value, with t = 1.25 and 15 - 1 = 14 df.
Using a t-distribution calculator, this p-value is of 0.2318.
Since the p-value of the test is 0.2318 > 0.01, the sample does not provide convincing evidence that the mean heart rate after 15 minutes of Wii Bowling is different from the known mean heart rate after 6 minutes walking on the treadmill.
A similar problem is given at https://brainly.com/question/13873630
