Answer to Question #126977 in Statistics and Probability for mengesha

Question

The average annual medical expense per family in a small city was Birr 750 in 1998. A random sample of 49 families was selected and their expenses for 1999 had a mean of Birr 800 with a standard deviation of Birr 140. Based on this information, can we conclude at α= 5% that the average annual medical expenses had increased from the 1998 average?

19. Let µ represent the population mean expenditure for 1999. Which of the following is the appropriate alternative hypothesis?
(A) µ ≠ 750 (B) µ = 750 (C) µ > 750 (D) µ < 750 (E) µ > 800

20. Which of the following is the appropriate rejection region?

(A) Z < –1.96 or Z > 1.96 (B) Z < –1.96 (C) Z < –1.645 (D) Z > 1.645
(E) –1.96 < Z < 1.96
21. Which of the following is the value of the test statistic?
(A) 50 (B) –0.36 (C) 0.36 (D) –2.5 (E) 2.5

Answer 1

$19) \mu>750\\ 20) Z>1.645\\ 21) \text{the test statistic}=\frac{800-750}{\frac{140}{\sqrt{49}}}=2.5$