Q1: Suppose a random sample of 49 is selected from a population of size N - 500
ID: 3198964 • Letter: Q
Question
Q1: Suppose a random sample of 49 is selected from a population of size N - 500 witha standard deviation of 14, Ifthe sample mean is 125, the 99% confidence interval to estimate the population mean is between A. 119.85 and 130.15 B. 119.85 and 135.15 C. 118.00 and 132.00 D. 120.10 and 129.90 Q2: A manufacturer wants to purchase a certain product of foil. The foil is stored on 1534 rolls ining a varying amount of foil with a standard deviation of 12.5. In order to estimate the total number of foil on all the rolls, the manufacturer randomly selected 200 rolls and measured the number of foil on each roll. The sample mean was 48 . Then the 95% co interval to estimate the population mean of foil is between A. 46.27 and 49.73 B. 45.88 and 50.12 C. 46.64 and 49.36 D. 46.08 and 49.92 Q3: According to Gartner Inc., the largest share of the worldwide PC market is held by Hewlett- Packard with 16.0%. Suppose that a market researcher believes that Hewlett Packard holds a higher share of the market in the western region of the United States. To verify this theory, he randomly selects 428 people who purchased a personal computer in the last month in the western region of the United States. Seventy-four of these purchases were Hewlett-Packard computers. Using a 1% level of significance, test the market researcher's theory. If the market share is really 0.19 in the southwestern region of the United Siates, what is the probability of making a Type II error? he null hypothesis, the probability of committing a Type I error is 0.01, ?-07019 B. Fail to reject the null hypothesis, the probability of committing a Type I error is 0,01.p 0.7019 C. Fail to reject the null hypothesis, the probability of committing a Type I error is 0.01. p-0.1271 D. Reject the null hypothesis, the probability of committing a Type I error is 0.01. p-0.3729Explanation / Answer
#1.
CI for 99%
n = 49
mean = 125
z-value of 99% CI = 2.5758
std. dev. = 14
SE = std.dev./sqrt(n)
= 14/sqrt(49)
= 2
ME = z*SE
= 2.5758*2
= 5.1517
Lower Limit = Mean - ME = 125 - 5.1517 = 119.84834
Upper Limit = Mean + ME = 125 + 5.1517 = 130.15166
99% CI (119.88 , 130.15)
Option A is correct
#2.
CI for 95%
n = 200
mean = 48
z-value of 95% CI = 1.9600
std. dev. = 12.5
SE = std.dev./sqrt(n)
= 12.5/sqrt(200)
= 0.88388
ME = z*SE
= 1.96*0.88388
= 1.7324
Lower Limit = Mean - ME = 48 - 1.7324 = 46.26762
Upper Limit = Mean + ME = 48 + 1.7324 = 49.73238
95% CI (46.27, 49.73)
Option A is correct
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