Answer-first summary for fast verification
Answer: z = 2.60, and reject the null hypothesis.
## Explanation Since the population size is very large (3,000,000) and the sample size is 81, we can use the z-test (normal distribution) rather than the t-test. **Hypotheses:** - H₀: μ ≤ 45 - H₁: μ > 45 (one-tailed test) **Test statistic calculation:** \[ z = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{46.3 - 45}{4.5/\sqrt{81}} = \frac{1.3}{4.5/9} = \frac{1.3}{0.5} = 2.60 \] **Critical value at 1% significance level:** For a one-tailed test at α = 0.01, the critical z-value is 2.326. **Decision:** Since z = 2.60 > 2.326, we reject the null hypothesis. Therefore, the correct answer is z = 2.60, and reject the null hypothesis.
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Bob tests the null hypothesis that the population mean is less than or equal to 45. From a population size of 3,000,000 people, 81 observations are randomly sampled. The corresponding sample mean is 46.3 and sample standard deviation is 4.5. What is the value of the most appropriate test statistic for the test of the population mean, and what is the correct decision at the 1 percent significance level?
A
z = 0.29, and fail to reject the null hypothesis.
B
z = 2.60, and reject the null hypothesis.
C
t = 0.29, and accept the null hypothesis.
D
t = 2.60, and neither reject nor fail to reject the null hypothesis.
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