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The playing time of 16 popular songs is found to have a standard deviation of 54.5 seconds. Use a 0.05 significance level to test the claim that the songs are from a population with a standard deviation less than one minute (60 seconds). State the initial and final conclusion.

A) Fail to reject the null hypothesis; there is not sufficient evidence to support the claim that the songs are from a population with a standard deviation less than one minute.
B) Reject the null hypothesis; there is not sufficient evidence to support the claim that the songs are from a population with a standard deviation less than one minute.
C) Fail to reject the null hypothesis; there is sufficient evidence to support the claim that the songs are from a population with a standard deviation less than one minute.
D) Reject the null hypothesis; there is sufficient evidence to warrant rejection of the claim that the songs are from a population with a standard deviation less than one minute.

1 answer

To test the claim that the standard deviation of song durations is less than one minute (60 seconds), we can set up our hypothesis test as follows:

  • Null hypothesis (\(H_0\)): \(\sigma \geq 60\) seconds (the standard deviation is at least one minute)
  • Alternative hypothesis (\(H_1\)): \(\sigma < 60\) seconds (the standard deviation is less than one minute)

Step 1: Identify the Sample Standard Deviation and Significance Level

  • Sample Standard Deviation (\(s\)) = 54.5 seconds
  • Hypothesized Population Standard Deviation (\(\sigma_0\)) = 60 seconds
  • Significance Level (\(\alpha\)) = 0.05

Step 2: Calculate the Test Statistic

We use the Chi-Squared (\(\chi^2\)) test for standard deviation. The test statistic is calculated as follows:

\[ \chi^2 = \frac{(n-1) \cdot s^2}{\sigma_0^2} \]

Where:

  • \(n\) = sample size = 16
  • \(s\) = sample standard deviation = 54.5 seconds
  • \(\sigma_0\) = hypothesized standard deviation = 60 seconds

Calculating \(s^2\) and \(\sigma_0^2\):

\[ s^2 = (54.5)^2 = 2970.25 \] \[ \sigma_0^2 = (60)^2 = 3600 \]

Now, substituting these values into the Chi-Squared formula:

\[ \chi^2 = \frac{(16-1) \cdot 2970.25}{3600} = \frac{15 \cdot 2970.25}{3600} = \frac{44553.75}{3600} \approx 12.38 \]

Step 3: Determine the Critical Value

Since we're conducting a one-tailed test at \(\alpha = 0.05\) with \(n - 1 = 15\) degrees of freedom, we need to find the critical value from the Chi-Squared distribution table.

Using a Chi-Squared table or calculator for \(df = 15\) and \(\alpha = 0.05\):

The critical value \(\chi^2_{0.05, 15}\) is approximately 24.996.

Step 4: Compare the Test Statistic to the Critical Value

We compare our test statistic to the critical value:

  • Test Statistic \(\chi^2 \approx 12.38\)
  • Critical Value \(\chi^2_{0.05, 15} \approx 24.996\)

Since \(12.38 < 24.996\), we fail to reject the null hypothesis.

Conclusion

Thus, we conclude:

Final Conclusion: A) Fail to reject the null hypothesis; there is not sufficient evidence to support the claim that the songs are from a population with a standard deviation less than one minute.