Question

Statistics In the manual “How to Have a Number One the Easy Way,” it is stated that a song “must be no longer than three minutes and thirty seconds (210 seconds)”. A simple random sample of 40 current hit songs results in a mean length of 252.5 seconds. Assume that the standard deviation of song lengths is 54.5 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 210 seconds.

Answer 1

We conclude that the sample is from a population of songs with a mean greater than 210 seconds.

Explanation:

We are given that a simple random sample of 40 current hit songs results in a mean length of 252.5 seconds.

Assume that the standard deviation of song lengths is 54.5 sec.

Let
`\mu` = population mean length of the songs

So, Null Hypothesis,
`H_0` :
`\mu`
`\leq` 210 seconds {means that the sample is from a population of songs with a mean smaller than or equal to 210 seconds}

Alternate Hypothesis,
`H_A` :
`\mu` > 210 seconds {means that the sample is from a population of songs with a mean greater than 210 seconds}

The test statistics that will be used here is One-sample z-test statistics because we know about population standard deviation;

T.S. =
`(\bar X-\mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

where,
`\bar X` = sample mean length of songs = 252.5 seconds

`\sigma` = population standard deviation = 54.5 seconds

n = sample of current hit songs = 40

So, the test statistics =
`(252.5-210)/((54.5)/(√(40) ) )`

= 4.932

The value of z-test statistics is 4.932.

Now, at 0.05 level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since the value of our test statistics is more than the critical value of z as 4.932 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the sample is from a population of songs with a mean greater than 210 seconds.