Question

R=5
please use z table for distribution question

(a) A researcher plans to carry out a study to estimate the proportion of residents in a city who have travelled abroad last year. If he wants to be 95% confident that the estimate is accurate to within (2.73 + ) %, find the minimum sample size that is required. (3 marks) A researcher plans to carry out a study to estimate the mean age of teachers in a city. If he wants to be 97% confident that the estimate is accurate to within 4+ years, find the minimum sample size that is required. Based on a previous research, it is known that the population variance of the ages of the teachers is roughly equal to 15 years. (3 marks) A survey was carried out to estimate the heights of residents in a city. A random sample of 100 residents from the city was selected. The sample mean and sample standard deviation of the heights of the selected residents were 156 cm and 15 cm respectively. If the population mean height of the residents were to be estimated by a confidence interval with a confidence level of 98.9%, find the width of the confidence interval. Correct your answer to 2 decimal places. (4 marks)

Answer 1

(a) ME = 2.73 + (5/5) = 3.73% = 0.0373

Z critical at = 0.05 is 1.96

Since nothing is known, p = 0.5 and q = 0.5

The ME = Z critical * SQRT(p * q/n).

Squaring and solving, n = (Zc/ME)² * p * q

Therefore n = (1.96 / 0.0373)² * 0.5 * 0.5 = 690.29

Therefore n = 691 (Rounding to the nearest Integer)

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(b) ME = 4 + (5 /5) = 5 years, ² = 15,

Z critical at = 0.03, is 2.17

The ME is given by :

Squaring both sides we get: (ME)² = (Z critical)² * ² / n

Therefore n = (Z critical / ME)² * ² = (2.17 * / 5)² * 15 = 2.83

Therefore n = 3 (Taking it to the next whole number)

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(c) Width of a confidence interval is given by the Margin of error, ME

Although population standard deviation is unknown, n is very large and we can assume the distribution to be normal.

The Z critical at = (100 - 98.9) / 100 = 0.011 is 2.543

ME = Z critical * s / sqrt(n) = 2.543 * 15 /sqrt(100) = 3.82

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