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Suppose the lifespan of a wombat is normally distributed. You believe that the mean is greater than 16. You collect a sample of size n = 43 wombats and compute the average to be 17.1 years. It is known that the population standard deviation is a = 6 years. What is the value of the test statistic?

A. z = 183
B. z = .914
C. z = 1.20
D. z = 1.1

User Don Wilson
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1 Answer

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Final answer:

The value of the test statistic is z = 1.20, calculated using the z-score formula and the given population standard deviation, sample mean, and sample size. Therefore, the correct value of the test statistic is z = 1.20, which corresponds to option C.

Step-by-step explanation:

To calculate the value of the test statistic for a hypothesis test regarding the mean lifespan of wombats, we use the formula for the z-score:

\[ z = \frac{\bar{x} - \mu}{(\sigma/\sqrt{n})} \]

where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, \( \sigma \) is the population standard deviation, and \( n \) is the sample size.

In this case:

  • \( \bar{x} = 17.1 \) years (sample mean)
  • \( \mu = 16 \) years (assumed population mean)
  • \( \sigma = 6 \) years (population standard deviation)
  • \( n = 43 \) (sample size)

Let's calculate the z-score:

\[ z = \frac{17.1 - 16}{(6/\sqrt{43})} \]

\[ z = \frac{1.1}{(6/\sqrt{43})} \]

\[ z = \frac{1.1}{0.914} \]

\[ z = 1.20 \]

Therefore, the correct value of the test statistic is z = 1.20, which corresponds to option C.

User Davidhq
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