Question

When engaging in weight control (fitness/fat burning) types of exercise, a person is expected to attain about 60% of their maximum heart rate. For 20-year-olds, this rates is approximately 120 bpm. A simple random sample of 100 20-year-olds was taken, and the sample mean was found to be 107 bpm with a standard deviation of 45 bpm. Researchers wonder if this is evidence to conclude that the expected level is actually lower than 120 bpm. To determine this, we test the following hypothesis:
H0 : ? = 120, Ha : ? < 120

Reference: Ref 17-2

A 95% confidence interval for the population mean weight control heart rate, ?, of 20-year-olds is

Answer 1

Answer:

- There is enough evidence to support the claim that the heart rate level lower than 120 bpm  (P-value=0.002).

- The 95% confidence interval for the population mean weight control heart rate of 20-year-olds is (98.07, 115.93).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the heart rate level lower than 120 bpm

Then, the null and alternative hypothesis are:

$H_0: \mu=120\\\\H_a:\mu< 120$

The significance level is 0.05.

The sample has a size n=100.

The sample mean is M=107.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=45.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{45}{\sqrt{100}}=4.5$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{107-120}{4.5}=\dfrac{-13}{4.5}=-2.889$

The degrees of freedom for this sample size are:

$df=n-1=100-1=99$

This test is a left-tailed test, with 99 degrees of freedom and t=-2.889, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t$

As the P-value (0.002) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the heart rate level lower than 120 bpm .

Confidence interval:

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=107.

The sample size is N=100.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{45}{\sqrt{100}}=\dfrac{45}{10}=4.5$

The t-value for a 95% confidence interval is t=1.984.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=1.984 \cdot 4.5=8.929$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 107-8.929=98.07\\\\UL=M+t \cdot s_M = 107+8.929=115.93$

The 95% confidence interval for the population mean weight control heart rate of 20-year-olds is (98.07, 115.93).