Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. In a manual on how to have a number one song, it is stated that a song must be no longer than 210 seconds. A simple random sample of 40 current hit songs results in a mean length of 234.1 sec and a standard deviation of 54.52 sec. Use a 0.05 significance level and the accompanying Minitab display to test the claim that the sample is from a population of songs with a mean greater than 210 sec. What do these results suggest about the advice given in the manual?

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ChiKesselman ChiKesselman · Answer 1 · 2019-12-23T08:16:12+01:00

Answer:

There is enough evidence to conclude that the sample is from a population of songs with a mean greater than 210 sec.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 210 seconds

Sample mean, [tex]\bar{x}[/tex] = 234.1 sec

Sample size, n = 40

Sample standard deviation, s = 54.52 sec

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu \leq 210\text{ seconds}\\H_A: \mu > 210\text{ seconds}[/tex]

We use one-tailed t test to perform this hypothesis.

Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{234.1 - 210}{\frac{54.52}{\sqrt{40}} } = 2.795[/tex]

Now, [tex]t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.684[/tex]

We calculate the p-value with the help of standard normal table.

P-value = 0.004005

Since,

[tex]t_{stat} > t_{critical}[/tex]

We fail to accept the null hypothesis and accept the alternate hypothesis.

Thus, there is enough evidence to conclude that the sample is from a population of songs with a mean greater than 210 sec.