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Example #2: Find $z_{\frac{\alpha}{2}}$ for the confidence level 90%. Example #3: Find $z_{\frac{\alpha}{2}}$ for the confidence level 99%. A confidence interval is a specific interval estimate of a parameter determined by using data obtained from a sample and by using the specific confidence level of the estimate. Formula for the Confidence Interval of the Mean for a Specific $\alpha$ When $\sigma$ is Known $\bar{X} - z_{\frac{\alpha}{2}} \cdot \left(\frac{\sigma}{\sqrt{n}}\right) < \mu < \bar{X} + z_{\frac{\alpha}{2}} \cdot \left(\frac{\sigma}{\sqrt{n}}\right)$ Confidence Level $z_{\frac{\alpha}{2}}$ 90% 1.65 95% 1.96 99% 2.58 The term $z_{\frac{\alpha}{2}} \left(\frac{\sigma}{\sqrt{n}}\right)$ is called the margin of error E (also called the maximum error of the estimate). It is the maximum likely difference between the point estimate of a parameter and the actual value of the parameter. Assumptions for Finding a Confidence Interval for a Mean When $\sigma$ is Known 1. The sample is a random sample. 2. Either n?30 or the population is normally distributed if n < 30.

          Example #2:
Find $z_{\frac{\alpha}{2}}$ for the confidence level 90%.
Example #3:
Find $z_{\frac{\alpha}{2}}$ for the confidence level 99%.
A confidence interval is a specific interval estimate of a parameter determined by using
data obtained from a sample and by using the specific confidence level of the estimate.
Formula for the Confidence Interval of the Mean for a Specific $\alpha$ When
$\sigma$ is Known
$\bar{X} - z_{\frac{\alpha}{2}} \cdot \left(\frac{\sigma}{\sqrt{n}}\right) < \mu < \bar{X} + z_{\frac{\alpha}{2}} \cdot \left(\frac{\sigma}{\sqrt{n}}\right)$
Confidence Level	$z_{\frac{\alpha}{2}}$
90%	1.65
95%	1.96
99%	2.58
The term $z_{\frac{\alpha}{2}} \left(\frac{\sigma}{\sqrt{n}}\right)$ is called the margin of error E (also called the maximum error of
the estimate). It is the maximum likely difference between the point estimate of a
parameter and the actual value of the parameter.
Assumptions for Finding a Confidence Interval for a Mean
When $\sigma$ is Known
1. The sample is a random sample.
2. Either n?30 or the population is normally distributed if n < 30.
        
Show more…
Example #2:
Find z(α)/(2) for the confidence level 90%.
Example #3:
Find z(α)/(2) for the confidence level 99%.
A confidence interval is a specific interval estimate of a parameter determined by using
data obtained from a sample and by using the specific confidence level of the estimate.
Formula for the Confidence Interval of the Mean for a Specific α When
σ is Known
X̅ - z(α)/(2)·((σ)/(√(n))) < μ < X̅ + z(α)/(2)·((σ)/(√(n)))
Confidence Level	z(α)/(2)
90%	1.65
95%	1.96
99%	2.58
The term z(α)/(2)((σ)/(√(n))) is called the margin of error E (also called the maximum error of
the estimate). It is the maximum likely difference between the point estimate of a
parameter and the actual value of the parameter.
Assumptions for Finding a Confidence Interval for a Mean
When σ is Known
1. The sample is a random sample.
2. Either n?30 or the population is normally distributed if n < 30.

Added by Javier B.

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Elementary Statistics a Step by Step Approach
Elementary Statistics a Step by Step Approach
Allan G. Bluman 9th Edition
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Example #2: Find z_(alpha ) for the confidence level 90%. Example # 3: Find z_(alpha ) for the confidence level 99%. A confidence interval is a specific interval estimate of a parameter determined by using data obtained from a sample and by using the specific confidence level of the estimate. Formula for the Confidence Interval of the Mean for a Specific alpha When sigma is Known ar{x} -z_((alpha )/(2))*((sigma )/(sqrt(n)))<mu <ar{x} +z_((alpha )/(2))*((sigma )/(sqrt(n))) able[[Confidence Level,z_((alpha )/(2)) Example#2: Find zfor the confidence level 90% 2 Example#3: Find z.for the confidence level 99% A confidence interval is a specific interval estimate of a parameter determined by using data obtained from a sample and by using the specific confidence level of the estimate. ais Known Formula for the Confidence Interval of the Mean for a Specific a When x-2a12 Confidence Level Za/2 90% 1.65 95% 1.96 99% 2.58 The term Z/2 is called the margin of error Ealso called the maximum error of the estimate. It is the maximum likely difference between the point estimate of a Assumptions for Finding a Confidence Interval for a Mean When a is Known 1.The sample is a random sample 2.Either n30or the population is nomally distributed if n<30
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8-92-recall-that-confidence-interval-for-the-sample-mean-can-be-calculated-using-the-interval-i-tn-1-v-p-1-tn-1-thus-the-margin-of-error-is-tn-1-v-we-can-recover-the-margin-of-error-irom-an-62512

(S 9.2) Recall that a confidence interval for the sample mean can be calculated using the interval x̄ - tn-1 · s/∑n ≤ μ ≤ x̄ + tn-1 · s/∑n. Thus, the margin of error is tn-1 · s/∑n. We can recover the margin of error from an interval constructed on the calculator using algebra. Suppose a random sample of size 15 was taken from a normally distributed population, and the sample standard deviation was calculated to be s = 6.4. We'll assume the sample mean is 10 for convenience. a) Calculate the margin of error for a 90% confidence interval for the population mean. Round your response to at least 3 decimal places. b) Calculate the margin of error for a 95% confidence interval for the population mean. Round your response to at least 3 decimal places. NOTE both these values are over 2. Suppose we want a smaller margin of error. c) Approximately how large of a sample size is needed to construct a 90% confidence interval with a margin of error less than 1.5 given an estimate for the standard deviation of 6.4? d) Approximately How large of a sample size is needed to construct a 95% confidence interval with margine of error less than 1.5 given an estimate for the standard deviation of 6.4?

Dominador T.

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6.The boundaries of a confidence interval are called: (a) Confidence levels (b) The test statistics (c) The degrees of confidence (d) The confidence limits Answer= 7.What value of z would you use to calculate the 98% confidence interval for a population mean, given that you know the population standard deviation, the sample size and the sample mean of your sample? a) z = 1.96 b) z = 2.58 c) z = 1.28 d) z = 2.33 Answer= 8.A 95% confidence interval for the population mean is calculated to be 75.29 to 81.45. If the confidence level is reduced to 90%, the confidence interval will: a) Become narrower b) Remain the same c) Become wider d) Double in size Answer= 9.To test a hypothesis involving proportions, both np and n(1-p) should a) Be at least 30 b) Be equal or greater than 5 c) Lie in the range from 0 to 1 d) There are no specific conditions surrounding the values of n and p Answer= 10. The purpose of hypothesis testing is to: (a) Determine value of the sample mean. (b) Determine evidence in favor of a parameter. (c) Determine the appropriate value of test statistic. (d) Derive the standard error of the data.

Krishna G.

recall-that-confidencc-interva-calculated-using-the-sample-mcan-and-margin-of-error-the-margin-of-error-wlll-vary-depending-on-the-ievel-of-confidence-the-population-standard-deviatlon-0-and-11292

Recall that confidence intervals are calculated using the sample mean and margin of error. The margin of error will vary depending on the level of confidence, the population standard deviation, and the size of the sample. This value will be added to and subtracted from the sample mean, using the following formula where z/2 is the value that gives an area of ̑/2 in the upper tail of the standard normal probability distribution. The value of z is found by setting the confidence level equal to (1 - ̑) and solving for z. Here we have a 90% confidence level, so ̑ = 0.10. Common values for z are given below: Confidence Level ̑/2 z 90% 0.05 1.645 95% 0.025 1.960 99% 0.01 2.576 For a 90% confidence level, the necessary value for z/2 is 1.645.

Joanna Q.


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Transcript

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00:03 Here for our answer we have part a t critical value at 0 .10 c level with 14 degree freedom this 1 .761 we have the margin of error that is 1 .761 times 6 .6 divided by the square root of 15 we have equal to 3 .001 for part b we have t…
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