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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?1)What are the null and alternative hypothesis?2) What is the value of the test statistic? z=?3) Identify the critical value(s) of z?4)......... H0. There...... Sufficient evidence to support the claim that the sample is from a population of songs with a mean greater than 200 sec. This result suggests that the advice given in the manual..... sound.

User Yoni Doe
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Answer:

(1) Null hypothesis (H0): The sample is from a population of songs with a mean equal to 200 sec

Alternate hypothesis (H1): The sample is from a population of songs with a mean length greater than 200 sec

(2) Test statistic (z) = 5.31

(3) Critical value = 1.645

(4) Reject. is not, is not

Explanation:

(1) 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

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

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

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

(3) The critical value for a one tailed test with a significance level of 0.05 is 1.645

(4) Reject H0 because the test statistic (5.31) is greater than the critical value (1.645).

There 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 suggests that the advice given in the manual is not sound

User Prasad Gayan
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