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elem. stat week5 6.1.13.

# elem. stat week5 6.1.13. - (“/13 ~ ‘ Assignment Week 5...

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Unformatted text preview: (“/13 ~ ‘ Assignment: Week 5 Exercises Date: 9/24/11 Course: Math 201 Fall B10 Time: 6:42 PM Book: Larson: Elementary Statistics: Picturing the World, 5e Find the margin of error for the given values of c, s, and n. c = 0.95, s = 2.4, n: 36 Use the formula below to ﬁnd the margin of error E, where 20 is the z—score corresponding to an area of c, o is the population standard deviation and n is the sample size. When n 2 30 you can use 3 in place of 6, where s is the sample standard deviation. E c = Z _ c Vn The level of conﬁdence 0 is the area under the standard normal curve between the critical values - 2c and 20. The area outside the critical values is 1 ~ c, so the area in each tail is :(1 — 0). Substitute the value of c to solve for the area in each tail of the conﬁdence region. i i _ 2(1—c) 2(10.95) .025 While you can use a standard normal table or technology to ﬁnd the critical value, for this problem we will use a standard normal table. Use a standard normal table to ﬁnd the positive critical value that corresponds to a tail area of 0.025. 26: 1.96 The population standard deviation 0 is unknown but because n 2 30, use 5 in place of 6. Substitute the values zc= 1.960, oz 5 z 2.4, andn = 36 to ﬁnd E. o E : 26—” 2.4 1.960— V 36 .784 I! ll Therefore, the margin of error is approximately 0.784. Page 1 ...
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