Fran’s store spent $64,000 on expenses last year. Rent of the store was 35% of those expenses .How much did Fran spend on rent.
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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 = <u><em>population mean length of the songs</em></u>
So, Null Hypothesis, :
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 <u>One-sample z-test statistics </u>because we know about population standard deviation;
T.S. = ~ N(0,1)
where, = sample mean length of songs = 252.5 seconds
= population standard deviation = 54.5 seconds
n = sample of current hit songs = 40
So, <u><em>the test statistics</em></u> =
= 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 <u><em>we have sufficient evidence to reject our null hypothesis</em></u> 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.