Respuesta :
Answer:
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 [tex]\mu[/tex] = population mean length of the songs
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] [tex]\leq[/tex] 210 seconds {means that the sample is from a population of songs with a mean smaller than or equal to 210 seconds}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 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. = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\bar X[/tex] = sample mean length of songs = 252.5 seconds
[tex]\sigma[/tex] = population standard deviation = 54.5 seconds
n = sample of current hit songs = 40
So, the test statistics = [tex]\frac{252.5-210}{\frac{54.5}{\sqrt{40} } }[/tex]
= 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.
