Question

a. You measure 43 watermelons' weights, and find they have a mean weight of 48 ounces. Assume the population standard deviation is 2.2 ounces. Based on this, what is the maximal margin of error associated with a 95% confidence interval for the true population mean watermelon weight?

Give your answer as a decimal, to two places ± ounces

b. You measure 25 watermelons' weights and find they have a mean weight of 44 ounces. Assume the population standard deviation is 9.4 ounces. Based on this, what is the maximal margin of error associated with a 95% confidence interval for the true population mean watermelon weight?

Give your answer as a decimal, to two places ± ounces.

c. You measure 33 turtles' weights, and find they have a mean weight of 30 ounces. Assume the population standard deviation is 9.1 ounces. Based on this, construct a 99% confidence interval for the true population mean turtle weight.

Give your answers as decimals, to two places ± ounces.

d. Assume that a sample is used to estimate a population mean μ. Find the 98% confidence interval for a sample of size 683 with a mean of 29.8 and a standard deviation of 10.8. Enter your answer accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

< μ <

The answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

Answer 1

1.96 Marginof emos + Zc: 1/6 uિa3 b) Given data I 0.66 ounced. n-25 (95.913,34.081) (25191, 34.09) 2eeemay) solutions Given doula na us mean weight fu8. population standard deviation ca 2.2 956. conficlence interval: - Z critical value at 95% confidence R1-0.45-005 Et lig6 22 a £1.96x 22 27 / Jus 7+0.6576 = I1,96 9.4 F 3.68 ounced. Given data n=33 turtles 30 Ounced. Z critical value at 997. confidence 2.58. ** Z,61 307 2.58.904 → 301 4,0840 (30–4,087 30 +4,083), 733 "Рачоци Сеи. mean werght a standard devration Ecritical value at 0-05 pot 1.96 Marginof Error HE = + Zcolon -9.4 ounce. + 3.6848 0) mean weight os 911 ounces construt 994. confidences

zentiral value at Sample Size n = 683 mcant - 29.8 Standard deviation (6) - 1018 984. confidence interval: - x=1-confidence Qo1-0-98 མ་ཡི་གནང་ L 20.02 1 20.02 is 2.33 C'I *A Zc. 9400 29.8 of 2033 X 1058 V683 E 29.8 0-9629 F (29-8.-0.9629 29+8+ 0.9629 371 30-47629 (2818 2 (28.8 cue 30.8