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jay1462
20.07.2020 •
Mathematics
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.
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Ответ:
We conclude that the sample is from a population of songs with a mean greater than 210 seconds.
Step-by-step 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
= population mean length of the songs
So, Null Hypothesis,
: ![\mu](/tpl/images/0709/8499/0746d.png)
210 seconds {means that the sample is from a population of songs with a mean smaller than or equal to 210 seconds}
Alternate Hypothesis,
:
> 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. =
~ N(0,1)
where,
= sample mean length of songs = 252.5 seconds
n = sample of current hit songs = 40
So, the test statistics =![\frac{252.5-210}{\frac{54.5}{\sqrt{40} } }](/tpl/images/0709/8499/f9bf6.png)
= 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.
Ответ: