Question

The average pulse rate is known to be 80 beats per minutes. Suppose we take a sample of 30 adults and find that the average heart rate in the sample is 73.76 with a sample standard deviation of 7.06. We are interested in testing if the average pulse rate is actually lower than 80 beats per minute.

a. Using the sample of 30 adults, what would the 95% confidence interval be for the population mean?

b. What are the null and alternative hypotheses?

c. What is the critical value at 95% confidence?

d. Calculate the test statistic.

e. Find the p-value.

f. What conclusion would be made here at the 95% confidence level?

g. Would my conclusion change if I changed alpha to .01? Show reasoning.

Answer 1

a)

Sample size = n = 30

Sample mean = = 73.76

Standard deviation = s = 7.06

We have to construct 95% confidence interval.

Formula is

Here E is a margin of error.

Degrees of freedom = n - 1 = 30 - 1 = 29

Level of significance = 0.05

tc = 2.045 ( Using t table)

So confidence interval is ( 73.76 - 2.6362 , 73.76 + 2.6362) = > ( 71.1238 , 76.3962)

b) Claim: The average pulse rate is actually lower than 80 beats per minute.

The null and alternative hypotheses is

c) Degrees of freedom = n - 1 = 30 - 1 = 29

Level of significance = 0.05

Our test is one tailed test because alternative hypothesis has greater than sign.

Critical value = 1.699 ( Using t table)

d)

Here population standard deviation is unknown so we have to use t-test statistic.
Test statistic is

e) P-value = P(T < - 4.84) = 0.000019

P-value < 0.05 we reject null hypothesis.

Conclusion:

The average pulse rate is actually lower than 80 beats per minute.

f) The confidence interval  (71.1238, 76.3962) is less than 80, therefore, The average pulse rate is actually lower than 80 beats per minute.

g) No, because P-value < 0.01