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A county with a large population need to know what the average income is for tax purposes. A simple random sample of 1000 household reveals the following information. The average household income for the sample is $21,000. The median income of the sample is $17,000. The SD of the sample is $10,000, Q1 = $11,000 and Q3 = $24,000. Find a 95% confidence interval for the average household income in this county a. $21,000 \pm 1.96 \frac{\sqrt{10,000(1-\frac{10000}{1000})}}{1000} b. $17,000 \pm 1.96 \frac{\sqrt{10,000(1-\frac{10000}{1000})}}{1000} c. $21,000 \pm 1.96 \frac{10000}{\sqrt{1000}} d. $17,000 \pm 1.96 \frac{10000}{\sqrt{1000}}

          A county with a large population need to know what the average income is for tax purposes. A simple random sample of 1000 household reveals the following information. The average household income for the sample is $21,000. The median income of the sample is $17,000. The SD of the sample is $10,000, Q1 = $11,000 and Q3 = $24,000.
Find a 95% confidence interval for the average household income in this county
a. $21,000 \pm 1.96 \frac{\sqrt{10,000(1-\frac{10000}{1000})}}{1000}
b. $17,000 \pm 1.96 \frac{\sqrt{10,000(1-\frac{10000}{1000})}}{1000}
c. $21,000 \pm 1.96 \frac{10000}{\sqrt{1000}}
d. $17,000 \pm 1.96 \frac{10000}{\sqrt{1000}}

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A county with a large population need to know what the average income is for tax purposes. A simple random sample of 1000 household reveals the following information. The average household income for the sample is 21,000. The median income of the sample is17,000. The SD of the sample is 10,000, Q1 =11,000 and Q3 = 24,000.
Find a 95% confidence interval for the average household income in this county
a.21,000 ±1.96 (√(10,000(1-(10000)/(1000))))/(1000)
b. 17,000 ± 1.96 (√(10,000(1-(10000)/(1000))))/(1000)
c.21,000 ±1.96 (10000)/(√(1000))
d. 17,000 ± 1.96 (10000)/(√(1000))

Added by Stephanie R.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert David Nguyen

08/04/2022

Step 1

Step 1: Given the sample mean, standard deviation, and sample size, we can use the formula for the confidence interval to find the range of values within which the true population mean is likely to fall. Show more…

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Texts: A county with a large population needs to know what the average income is for tax purposes. A simple random sample of 1000 households reveals the following information. The average household income for the sample is $21,000. The median income of the sample is $17,000. The standard deviation of the sample is $10,000. Q1 = $11,000 and Q3 = $24,000. Find the 95% confidence interval for the average household income in this county. Solution: To find the 95% confidence interval for the average household income, we can use the formula: Confidence Interval = Sample Mean ± (Critical Value) * (Standard Deviation / √Sample Size) Given: Sample Mean = $21,000 Standard Deviation = $10,000 Sample Size = 1000 First, we need to find the critical value. Since the sample size is large (n > 30), we can use the Z-distribution. The critical value for a 95% confidence level is 1.96. Plugging in the values into the formula: Confidence Interval = $21,000 ± (1.96) * ($10,000 / √1000) Simplifying: Confidence Interval = $21,000 ± (1.96) * ($10,000 / 31.62) Confidence Interval = $21,000 ± (1.96) * $316.23 Confidence Interval = $21,000 ± $619.20 The 95% confidence interval for the average household income in this county is approximately $20,380.80 to $21,619.20.

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Transcript

00:01 Hi, i'm david and i'm here to have yansans to your question.

00:03 In the question here we have the three questions so let me do one by one for you.

00:08 For the first one ask me to find the confident interval for the mean.

00:11 Let me remind you for the number one that the confidence interval for the mean.

00:18 Here we are given the standard division of the distribution so therefore it will be the xb plus and minus zan4 of 2, the sigma overscority to the end.

00:28 Now in this question we're given the sample size n equal to the 500 and we're given the x bar equal to the 47 ,300 and we're given the sigma equal to the 9 ,500 and want to find the 95 % confidence interval.

00:48 Look at the formula we have given the xx sigma the n the only quantity we need to find will be the z and 4 over 2.

00:56 Now at the 95 % confidence and far equal to the 5 % divided by 2 equal to the 2 .5 % which is equal to the zpon 025.

01:06 It will look up the z table this value equal to the 1 .96.

01:10 From here we have the 95 % confident interval for the mean.

01:16 It will be x bar will be the 47 300 plus and minus 1 .96 times under 9 ,500, over square root of the 500.

01:28 Now if we compute the quantity on the right, we have the 1 .96 times 9500 the equal to the 832 .71.

01:45 And now if we use the minus, we will have 473 -0 minus the answer, equal to the 4647 .67...

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