QUESTION 1 In a random sample of employees, the number af days of medical leave

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QUESTION 1 In a random sample of employees, the number af days of medical leave for each employee was recorded as 2, 5, 0, 1, 5, 2, 3,0. Find the point estimate for the population mean. o 125 o 225 3.25 QUESTION 2 In a random sample of 24 pcople who were asked about their preference of cotfee or tea, 40% Indicated that they prefer tea. If the z value corresponding to 95% confidence levels +/-1.96, what is the 9536 cont dence interval for the population propo on? (0.104, 0.496) (0.204, 0.596) (0.304, 0.696) (0.404, 0.796) QUESTION 3 In a random sample of 100 households, it was found that themple mean income is $35,000 and the sample standard deviation is $6,000. If the z value corresponding to 90% confidence level is +/-1.645, what is te 90% confidence interval for the popular on mean income? 0 0 0 0 (25013, 26987) (34013, 35987) (36013, 37987) (38013, 39987) QUESTION 4 If the 90% cont dence interval for the population mean is(221.4, 234.7), then the 95% confidence interval would possibly be (219.8, 236.1) (222.6, 223.9) (2214, 223.9) (222.6,234.7) QUESTION S In a random sample of 11 swimmers, it was found that the sample mean training hours is 5 hours per day and the sample standard devianon is 1.2 hours per day. In calculating the 90% confidence interval, we should use the O z distribution O t distribution Binomial distributon Poisson distribution

Explanation / Answer

Q1

Point estimate of population mean = sample mean = (2 + 5 + 0 + 1 + 5 + 2 + 3 + 0)/8 = 18/8 = 2.25 ANSWER option 3

Q2

95% confidence interval for population proportion = pcap ± t/2[sq.rt{pcap(1 – pcap)/n}], where pcap = sample proportion (0.4 or 40%), n = sample size (24) and t/2 = upper 2.5% point, given to be 1.96

So, the CI is: 0.4 ± 1.96[sq.rt{(0.4 x 0.6)/24}] = 0.4 ± [1.96 x sqrt(0.24/24)]

= 0.4 ± 0.196 = (0.204, 0.596) ANSWER option 2

Q3

90% confidence interval for population mean = {Xbar ± (s/n)(t/2)}, where

Xbar = sample mean, (35000)

s = sample standard deviation, (6000)

n = sample size (100) and

t/2 = upper (/2) % point of t-Distribution with (n - 1) degrees of freedom.(1.645)

So, CI = 35000 ± 1.645(6000/100) = 35000 ± 987 = (34013, 35987) ANSWER option 2

Q4

Given 90% confidence interval for population mean = {Xbar ± (s/n)(t/2)} = (221.4, 234.7),

Xbar + (s/n)(t/2) = 234.7 ………………………………………………………(1)

Xbar - (s/n)(t/2) = 221.4 ………………………………………………………(2)

(1) + (2): 2Xbar = 456.1 or Xbar = 228.05

(1) - (2): 2(s/n)(t/2) = 13.3 or (s/n)(t/2) = 6.65

As given in Q3, t/2 = 1.645. Hence, (s/n) = 6.65/1.645 = 4.04.

As given in Q2, t/2 = 1.96 for 95% confidence interval for population mean.

So, 95% confidence interval for population mean = 228.05 ± (1.96 x 4.04) = (220.13, 235.97) ANSWER option 1

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