A simple random sample of 49 men from a normally distributed population results in a standard deviation of 8.5 beats per minute. The normal range of pulse rates of adults is typically given as 60 to 100 beats per minute. If the range rule of thumb is applied to that normal range, the result is a standard deviation of 10 beats per minute. Use the sample results with a 0.10 significance level to test the claim that pulse rates of men have a standard deviation equal to 10 beats per minute. Identify the null and alternative hypotheses

Compute the test statistic.

Find the P-value of the test statistic

State the conclusion

Null hypothesis: The standard deviation for the pulse rates of men is 10 beats per minute.

Alternate hypothesis: The standard deviation for the pulse rates of men is not equal to 10 beats per minute.

The test statistic is -1.24

The P-value of the test statistic is 0.01

There is sufficient evidence to support the claim that the pulse rates of men have a standard deviation equal to 10 beats per minute.

Step-by-step explanation:

A null hypothesis is a statement from a population parameter which is either rejected or accepted (fail to reject) upon testing. It is expressed using the equality sign.

An alternate hypothesis is also a statement from a population parameter which negates the null hypothesis and is accepted if the null hypothesis is rejected. It is expressed using any of the inequality signs.

Test statistic (z) = (sample sd - population sd) ÷ sample sd/√n

sample sd = 8.5 beats per minute

population sd = 10 beats per minute

n = 49

z = (8.5 - 10) ÷ 8.5/√49 = -1.5 ÷ 1.214 = -1.24

The test is a two-tailed test. The cumulative area of the test statistic is 0.8925

P-value = 2(1 - 0.8925) = 2×0.1075 = 0.215

Conclusion:

Fail to reject the null hypothesis because the P-value 0.215 is greater than the significance level 0.1.

The standard deviation for the pulse rates of men is 10 beats per minute.

1 + (6 / x - 1)

step-by-step explanation:

divide using polynomial division.