Question

Test the following claim. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, critical​ value(s), conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. A manual states that in order to be a​ hit, a song must be no longer than three minutes and twenty seconds​ (or 200 ​seconds). A simple random sample of 40 current hit songs results in a mean length of 245.0 sec. Assume the population standard deviation of song lengths is 53.5 sec. Use a 0.05 significance level to test the claim that the sample is from a population of songs with a mean greater than 200 sec. What do these result suggest about the advice given in the​ manual? What are the null and alternative​ hypotheses? A. H0​: muequals200 sec H1​: mugreater than200 sec B. H0​: mugreater than200 sec H1​: muless than200 sec C. H0​: mugreater than200 sec H1​: muequals200 sec D. H0​: muless than200 sec H1​: mugreater than200 sec What is the value of the test​ statistic? zequals nothing ​(Round to two decimal places as​ needed.) Identify the critical​ value(s) of z. zequals nothing ​(Round to three decimal places as needed. Use a comma to separate answers as​ needed.) ▼ Reject Fail to reject H0. There ▼ is is not sufficient evidence to support the claim that the sample is from a population of songs with a mean greater than 200 sec. This result suggest that the advice given in the manual ▼ is is not sound.

Answer 1

A Null hypothesis (H0): mu equals 200 sec

Alternate hypothesis (H1): mu greater than 200 sec

Test statistic (z) is 5.31

Critical value is 1.645

Conclusion: reject the null hypothesis

Final conclusion that addresses the original claim: the result obtained suggest that the advice given in the manual is not valid

Explanation:

Null hypothesis is a statement from the population parameter subject to testing

Alternate hypothesis is also a statement from the population parameter that negates the null hypothesis

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

sample mean = 245 sec, popi mean = 200 sec, sd = 53.5 sec, n = 40

z = (245 - 200) ÷ 53.5/√40 = 45 ÷ 8.47 = 5.31

This is a one tailed test.The critical value using a significance level of 0.05 is 1.645

Conclusion: since the test statistic (5.31) is greater than the critical value (1.645), reject the null hypothesis

Final conclusion: There is sufficient evidence to support the claim that the sample is from a population of songs with a mean greater than 200 sec

This result suggest that the advice offered in the manual is not valid