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Creators of a new software claim that it measures the remaining notebook battery life with a standard deviation as low as 5 minutes. To test this claim, a fully charged notebook was disconnected from the power supply and continued on its battery. The experiment was repeated 50 times, and every time the predicted battery life was recorded. The sample standard deviation of these 50 normally distributed measurements was equal 5.7 minutes. At the 5% level of significance, do these data provide evidence that the actual standard deviation is greater than 5 min?

Question

Creators of a new software claim that it measures the remaining notebook battery life with a standard deviation as low as 5 minutes. To test this claim, a fully charged notebook was disconnected from the power supply and continued on its battery. The experiment was repeated 50 times, and every time the predicted battery life was recorded. The sample standard deviation of these 50 normally distributed measurements was equal 5.7 minutes. At the 5% level of significance, do these data provide evidence that the actual standard deviation is greater than 5 min?

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Solution

To test the claim that the actual standard deviation is greater than 5 minutes, we can use a Chi-Square test for a single variance.

Step 1: State the hypotheses. The null hypothesis is that the standard deviation is equal to 5 minutes. The alternative hypothesis is that the standard deviation is greater than 5 minutes.

H0: σ^2 = 5^2 = 25 H1: σ^2 > 25

Step 2: Decide the significance level. Here, we are given a 5% level of significance.

Step 3: Identify a test statistic. Since we are trying to ascertain a claim about the variance, we use the Chi-Square distribution. The test statistic is:

χ^2 = (n - 1) * s^2 / σ^2

where n is the sample size, s is the sample standard deviation, and σ is the population standard deviation.

Step 4: Calculate the test statistic.

χ^2 = (50 - 1) * 5.7^2 / 5^2 = 57.96

Step 5: Determine the critical value. The degrees of freedom is n - 1 = 50 - 1 = 49. At the 5% level of significance, the critical value of χ^2 for 49 degrees of freedom is approximately 70.222.

Step 6: Make a decision. Since the calculated value of χ^2 is less than the critical value, we do not reject the null hypothesis.

Therefore, at the 5% level of significance, the data does not provide evidence that the actual standard deviation is greater than 5 minutes.

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