Jack and jill exercise in a 25.0-m-long swimming pool. jack swims nine lengths of the pool in 2 minutes and 34.5 seconds, while

The P-value to evaluate the claim that the mean length of pencils produced in this factory equals 18.0 cm is 0.00736. Step-by-step explanation: In this case, a quality control specialist extracted a random sample of 45 pencils from the assembly line, which exhibited a mean length of 17.9 cm. With a known population standard deviation of 0.25 cm, we denote by the population mean length for pencils produced in the factory. Thus, Null Hypothesis: = 18.0 cm (indicating that the population mean length equals 18.0 cm). Alternate Hypothesis: ≠ 18.0 cm (suggesting different from 18.0 cm). We apply the one-sample z-test since the population standard deviation is known. The test statistic yields: T.S. ~ N(0,1), with the sample mean length 17.9 cm and population standard deviation 0.25 cm for a sample size of 45. Hence, the calculated test statistic is -2.68. The corresponding P-value is derived from P(Z < -2.68) = 1 - P(Z > 2.68), equating to 1- 0.99632 = 0.00368. For a two-tailed test, the resulting P-value computes to 2 * 0.00368 = 0.00736.

Response: 4.4

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