A music industry producer wondered whether there is a differ

A music industry producer wondered whether there is a difference in lengths (in seconds) of rhythm & blues songs and rap songs. He obtained a random sample of 30 rhythm & blues songs and documented the song lengths. He found that the mean length time (in seconds) of this sample was 242.7 seconds with a standard deviation 26.8. He also took a random sample of 36 rap songs and found that the mean length time of this sample was 237.5 seconds with a standard deviation 28.5.

You will perform a hypothesis test at 0.05 significance level to test the claim that the length of rhythm & blues songs is different from the length of rap songs. Show your work or calculator commands in answering the following parts:

State the null and alternate hypotheses using the correct symbols.

State the type of test to be used and the requirements needed for the test.

Find the test statistic.

Find the critical value OR the p-value of the test.

Using your results from above, make a statistical decision.

Interpret your decision in the context of the problem.

Construct a 95% confidence interval for the difference between the mean time length of rhythm & blues songs and the mean time length of rap songs. Show your work or calculator command.

Explain how your conclusions from the confidence interval and the hypothesis test are the same or different.

Solution

State the null and alternate hypotheses using the correct symbols.

Let mu1 be the mean for blues songs

Let mu2 be the mean for rap songs

Null hypothesis: mu1=mu2

Alternative hypothesis: mu1 not equal to mu2

---------------------------------------------------------------------------------------------------------------

State the type of test to be used and the requirements needed for the test.

Z test

We need to assume that the popluation follows normal distribution.

---------------------------------------------------------------------------------------------------------------

Find the test statistic.

Z=(xbar1-xbar2)/sqrt(s1^2/n1+s2^2/n2)

=(242.7-237.5)/sqrt(26.8^2/30+28.5^2/36)

=0.76

---------------------------------------------------------------------------------------------------------------

Find the critical value OR the p-value of the test.

It is a two-tailed test.

So the p-value= 2*P(Z>0.76) =0.4473 (from standard normal table)

---------------------------------------------------------------------------------------------------------------

Using your results from above, make a statistical decision.

Since the p-value is larger than 0.05, we do not reject the null hypothesis.

---------------------------------------------------------------------------------------------------------------

Interpret your decision in the context of the problem.

So we can not conclude that the length of rhythm & blues songs is different from the length of rap songs.

---------------------------------------------------------------------------------------------------------------

Construct a 95% confidence interval for the difference between the mean time length of rhythm & blues songs and the mean time length of rap songs. Show your work or calculator command.

Given a=1-0.95=0.05, Z(0.025) = 1.96 (from standard normal table)

So the lower bound is

(xbar1-xbar2)-Z*sqrt(s1^2/n1+s2^2/n2)

=(242.7-237.5)- 1.96*sqrt(26.8^2/30+28.5^2/36)

=-8.165969

So the upper bound is

(xbar1-xbar2)+Z*sqrt(s1^2/n1+s2^2/n2)

=(242.7-237.5)+ 1.96*sqrt(26.8^2/30+28.5^2/36)

=18.56597

---------------------------------------------------------------------------------------------------------------

Explain how your conclusions from the confidence interval and the hypothesis test are the same or different.

Since the inerval include 0, we do not reject the null hypothesis.

So the results are the same