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solve 14，15 14 A confidence interval is used to estimate the population mean $\m...
May 30, 2024
solve 14，15
Generated Graph
Solution by Steps
step 1
The width of the confidence interval is inversely proportional to the square root of the sample size, nn
step 2
To decrease the width by 50%50\%, we need to increase the sample size by a factor of 44. This is because (14=12)(\frac{1}{\sqrt{4}} = \frac{1}{2})
# Part (b)
step 1
To decrease the width of the confidence interval by 20%20\%, we need to find the factor by which the sample size should be increased
step 2
Let the factor be kk. Then, 1k=0.8\frac{1}{\sqrt{k}} = 0.8
step 3
Solving for kk, we get k=(10.8)2=1.5625k = \left(\frac{1}{0.8}\right)^2 = 1.5625
step 4
The percentage increase in the sample size is (1.5625−1)×100%=56.25%(1.5625 - 1) \times 100\% = 56.25\%
# Part (c)
step 1
If the sample size is increased by a factor of 99, the width of the confidence interval will decrease by a factor of 9=3\sqrt{9} = 3
# Part (d)
step 1
If the sample size is decreased by a factor of 1616, the width of the confidence interval will increase by a factor of 16=4\sqrt{16} = 4
Answer
(a) Factor of 4, (b) 56.25% increase, (c) Decrease by a factor of 3, (d) Increase by a factor of 4
Key Concept
Sample size and confidence interval width relationship
Explanation
The width of the confidence interval is inversely proportional to the square root of the sample size.
Question 15
step 1
The formula for the margin of error (E) in a confidence interval is E=z⋅σnE = z \cdot \frac{\sigma}{\sqrt{n}}, where zz is the z-score, σ\sigma is the population standard deviation, and nn is the sample size
step 2
For a 95%95\% confidence level, the z-score is approximately 1.961.96
step 3
We need the margin of error to be 2020, so 20=1.96⋅100n20 = 1.96 \cdot \frac{100}{\sqrt{n}}
step 4
Solving for nn, we get n=(1.96⋅10020)2=96.04n = \left(\frac{1.96 \cdot 100}{20}\right)^2 = 96.04
step 5
Since the sample size must be an integer, we round up to the nearest whole number, so n=97n = 97
Answer
97
Key Concept
Margin of error in confidence intervals
Explanation
The margin of error depends on the z-score, population standard deviation, and sample size.
Solution by Steps
step 1
We need to determine the sample size nn required to estimate the mean weight μ\mu with a 95% confidence level and a margin of error of 0.5 grams. The formula for the sample size is given by n=(Z⋅σE)2n = \left( \frac{Z \cdot \sigma}{E} \right)^2, where ZZ is the z-score corresponding to the confidence level, σ\sigma is the standard deviation, and EE is the margin of error
step 2
For a 95% confidence level, the z-score ZZ is 1.96. Given σ=2.0\sigma = 2.0 grams and E=0.5E = 0.5 grams, we substitute these values into the formula: n=(1.96⋅2.00.5)2n = \left( \frac{1.96 \cdot 2.0}{0.5} \right)^2
step 3
Simplifying the expression: n=(3.920.5)2=(7.84)2=61.47n = \left( \frac{3.92}{0.5} \right)^2 = \left( 7.84 \right)^2 = 61.47
step 4
Since the sample size must be a whole number, we round up to the nearest whole number: n=62n = 62
Answer
62
Key Concept
Sample size determination for confidence intervals
Explanation
The sample size is calculated using the formula n=(Z⋅σE)2n = \left( \frac{Z \cdot \sigma}{E} \right)^2, where ZZ is the z-score for the desired confidence level, σ\sigma is the standard deviation, and EE is the margin of error.
Question 17
step 1
We need to determine the sample size nn required to estimate the mean number of customers per day with a 99% confidence level and a margin of error of 10. The formula for the sample size is n=(Z⋅σE)2n = \left( \frac{Z \cdot \sigma}{E} \right)^2
step 2
For a 99% confidence level, the z-score ZZ is 2.576. Given σ=50\sigma = 50 and E=10E = 10, we substitute these values into the formula: n=(2.576⋅5010)2n = \left( \frac{2.576 \cdot 50}{10} \right)^2
step 3
Simplifying the expression: n=(128.810)2=(12.88)2=165.89n = \left( \frac{128.8}{10} \right)^2 = \left( 12.88 \right)^2 = 165.89
step 4
Since the sample size must be a whole number, we round up to the nearest whole number: n=166n = 166
Answer
166
Key Concept
Sample size determination for confidence intervals
Explanation
The sample size is calculated using the formula n=(Z⋅σE)2n = \left( \frac{Z \cdot \sigma}{E} \right)^2, where ZZ is the z-score for the desired confidence level, σ\sigma is the standard deviation, and EE is the margin of error.
Question 18
step 1
We need to determine the sample size nn required to estimate the mean lifetime of light bulbs with a 90% confidence level and a margin of error of 20 hours. The formula for the sample size is n=(Z⋅σE)2n = \left( \frac{Z \cdot \sigma}{E} \right)^2
step 2
For a 90% confidence level, the z-score ZZ is 1.645. Given σ=150\sigma = 150 hours and E=20E = 20 hours, we substitute these values into the formula: n=(1.645⋅15020)2n = \left( \frac{1.645 \cdot 150}{20} \right)^2
step 3
Simplifying the expression: n=(246.7520)2=(12.3375)2=152.19n = \left( \frac{246.75}{20} \right)^2 = \left( 12.3375 \right)^2 = 152.19
step 4
Since the sample size must be a whole number, we round up to the nearest whole number: n=153n = 153
Answer
153
Key Concept
Sample size determination for confidence intervals
Explanation
The sample size is calculated using the formula n=(Z⋅σE)2n = \left( \frac{Z \cdot \sigma}{E} \right)^2, where ZZ is the z-score for the desired confidence level, σ\sigma is the standard deviation, and EE is the margin of error.
Question 19a
step 1
We need to determine the sample size nn required to estimate the mean IQ score with a 95% confidence level and a margin of error of 2 points. The formula for the sample size is n=(Z⋅σE)2n = \left( \frac{Z \cdot \sigma}{E} \right)^2
step 2
For a 95% confidence level, the z-score ZZ is 1.96. Given σ=15\sigma = 15 and E=2E = 2, we substitute these values into the formula: n=(1.96⋅152)2n = \left( \frac{1.96 \cdot 15}{2} \right)^2
step 3
Simplifying the expression: n=(29.42)2=(14.7)2=216.09n = \left( \frac{29.4}{2} \right)^2 = \left( 14.7 \right)^2 = 216.09
step 4
Since the sample size must be a whole number, we round up to the nearest whole number: n=217n = 217
Answer
217
Key Concept
Sample size determination for confidence intervals
Explanation
The sample size is calculated using the formula n=(Z⋅σE)2n = \left( \frac{Z \cdot \sigma}{E} \right)^2, where ZZ is the z-score for the desired confidence level, σ\sigma is the standard deviation, and EE is the margin of error.
Question 19b
step 1
We need to determine the sample size nn required to estimate the mean IQ score with a 99% confidence level and a margin of error of 2 points. The formula for the sample size is n=(Z⋅σE)2n = \left( \frac{Z \cdot \sigma}{E} \right)^2
step 2
For a 99% confidence level, the z-score ZZ is 2.576. Given σ=15\sigma = 15 and E=2E = 2, we substitute these values into the formula: n=(2.576⋅152)2n = \left( \frac{2.576 \cdot 15}{2} \right)^2
step 3
Simplifying the expression: n=(38.642)2=(19.32)2=373.22n = \left( \frac{38.64}{2} \right)^2 = \left( 19.32 \right)^2 = 373.22
step 4
Since the sample size must be a whole number, we round up to the nearest whole number: n=374n = 374
Answer
374
Key Concept
Sample size determination for confidence intervals
Explanation
The sample size is calculated using the formula n=(Z⋅σE)2n = \left( \frac{Z \cdot \sigma}{E} \right)^2, where ZZ is the z-score for the desired confidence level, σ\sigma is the standard deviation, and EE is the margin of error.
solve 16，17，18，19
Solution by Steps
step 1
Identify the given values: σ=2.0\sigma = 2.0 grams, confidence level = 95%95\%, margin of error E=0.5E = 0.5 grams
step 2
For a 95%95\% confidence level, the critical value zz is approximately 1.961.96
step 3
Use the formula for sample size: n=(z⋅σE)2n = \left( \frac{z \cdot \sigma}{E} \right)^2
step 4
Substitute the values into the formula: n=(1.96⋅2.00.5)2n = \left( \frac{1.96 \cdot 2.0}{0.5} \right)^2
step 5
Calculate the sample size: n=(3.920.5)2=(7.84)2=61.47n = \left( \frac{3.92}{0.5} \right)^2 = \left( 7.84 \right)^2 = 61.47
step 6
Round up to the nearest whole number: n=62n = 62
Answer
62
Key Concept
Sample size calculation for estimating a population mean
Explanation
The sample size is determined using the formula n=(z⋅σE)2n = \left( \frac{z \cdot \sigma}{E} \right)^2, where zz is the critical value, σ\sigma is the standard deviation, and EE is the margin of error.
Problem 17
step 1
Identify the given values: σ=50\sigma = 50, confidence level = 99%99\%, margin of error E=10E = 10
step 2
For a 99%99\% confidence level, the critical value zz is approximately 2.5762.576
step 3
Use the formula for sample size: n=(z⋅σE)2n = \left( \frac{z \cdot \sigma}{E} \right)^2
step 4
Substitute the values into the formula: n=(2.576⋅5010)2n = \left( \frac{2.576 \cdot 50}{10} \right)^2
step 5
Calculate the sample size: n=(128.810)2=(12.88)2=165.89n = \left( \frac{128.8}{10} \right)^2 = \left( 12.88 \right)^2 = 165.89
step 6
Round up to the nearest whole number: n=166n = 166
Answer
166
Key Concept
Sample size calculation for estimating a population mean
Explanation
The sample size is determined using the formula n=(z⋅σE)2n = \left( \frac{z \cdot \sigma}{E} \right)^2, where zz is the critical value, σ\sigma is the standard deviation, and EE is the margin of error.
Problem 18
step 1
Identify the given values: σ=150\sigma = 150 hours, confidence level = 90%90\%, margin of error E=20E = 20 hours
step 2
For a 90%90\% confidence level, the critical value zz is approximately 1.6451.645
step 3
Use the formula for sample size: n=(z⋅σE)2n = \left( \frac{z \cdot \sigma}{E} \right)^2
step 4
Substitute the values into the formula: n=(1.645⋅15020)2n = \left( \frac{1.645 \cdot 150}{20} \right)^2
step 5
Calculate the sample size: n=(246.7520)2=(12.3375)2=152.19n = \left( \frac{246.75}{20} \right)^2 = \left( 12.3375 \right)^2 = 152.19
step 6
Round up to the nearest whole number: n=153n = 153
Answer
153
Key Concept
Sample size calculation for estimating a population mean
Explanation
The sample size is determined using the formula n=(z⋅σE)2n = \left( \frac{z \cdot \sigma}{E} \right)^2, where zz is the critical value, σ\sigma is the standard deviation, and EE is the margin of error.
Problem 19a
step 1
Identify the given values: σ=15\sigma = 15, confidence level = 95%95\%, margin of error E=2E = 2
step 2
For a 95%95\% confidence level, the critical value zz is approximately 1.961.96
step 3
Use the formula for sample size: n=(z⋅σE)2n = \left( \frac{z \cdot \sigma}{E} \right)^2
step 4
Substitute the values into the formula: n=(1.96⋅152)2n = \left( \frac{1.96 \cdot 15}{2} \right)^2
step 5
Calculate the sample size: n=(29.42)2=(14.7)2=216.09n = \left( \frac{29.4}{2} \right)^2 = \left( 14.7 \right)^2 = 216.09
step 6
Round up to the nearest whole number: n=217n = 217
Answer
217
Key Concept
Sample size calculation for estimating a population mean
Explanation
The sample size is determined using the formula n=(z⋅σE)2n = \left( \frac{z \cdot \sigma}{E} \right)^2, where zz is the critical value, σ\sigma is the standard deviation, and EE is the margin of error.
Problem 19b
step 1
Identify the given values: σ=15\sigma = 15, confidence level = 99%99\%, margin of error E=2E = 2
step 2
For a 99%99\% confidence level, the critical value zz is approximately 2.5762.576
step 3
Use the formula for sample size: n=(z⋅σE)2n = \left( \frac{z \cdot \sigma}{E} \right)^2
step 4
Substitute the values into the formula: n=(2.576⋅152)2n = \left( \frac{2.576 \cdot 15}{2} \right)^2
step 5
Calculate the sample size: n=(38.642)2=(19.32)2=373.22n = \left( \frac{38.64}{2} \right)^2 = \left( 19.32 \right)^2 = 373.22
step 6
Round up to the nearest whole number: n=374n = 374
Answer
374
Key Concept
Sample size calculation for estimating a population mean
Explanation
The sample size is determined using the formula n=(z⋅σE)2n = \left( \frac{z \cdot \sigma}{E} \right)^2, where zz is the critical value, σ\sigma is the standard deviation, and EE is the margin of error.
solve 19ab
Solution by Steps
step 1
To determine the sample size required for a 95% confidence level with a margin of error of 2 points, we use the formula for sample size in estimating a population mean: n=(Z⋅σE)2n = \left( \frac{Z \cdot \sigma}{E} \right)^2
step 2
For a 95% confidence level, the Z-score (Z) is 1.96. The standard deviation (σ\sigma) is 15, and the margin of error (E) is 2
step 3
Plugging in the values: n=(1.96⋅152)2n = \left( \frac{1.96 \cdot 15}{2} \right)^2
step 4
Simplifying inside the parentheses: n=(29.42)2=(14.7)2n = \left( \frac{29.4}{2} \right)^2 = \left( 14.7 \right)^2
step 5
Calculating the square: n=216.09n = 216.09
step 6
Since the sample size must be a whole number, we round up to the nearest whole number: n=217n = 217
Part (b)
step 1
To determine the sample size required for a 99% confidence level with a margin of error of 2 points, we use the same formula: n=(Z⋅σE)2n = \left( \frac{Z \cdot \sigma}{E} \right)^2
step 2
For a 99% confidence level, the Z-score (Z) is 2.576. The standard deviation (σ\sigma) is 15, and the margin of error (E) is 2
step 3
Plugging in the values: n=(2.576⋅152)2n = \left( \frac{2.576 \cdot 15}{2} \right)^2
step 4
Simplifying inside the parentheses: n=(38.642)2=(19.32)2n = \left( \frac{38.64}{2} \right)^2 = \left( 19.32 \right)^2
step 5
Calculating the square: n=373.22n = 373.22
step 6
Since the sample size must be a whole number, we round up to the nearest whole number: n=374n = 374
Answer
Part (a): 217
Part (b): 374
Key Concept
Sample Size Calculation for Confidence Intervals
Explanation
The sample size required to estimate a population mean with a given confidence level and margin of error can be calculated using the formula n=(Z⋅σE)2n = \left( \frac{Z \cdot \sigma}{E} \right)^2. The Z-score corresponds to the desired confidence level, σ\sigma is the population standard deviation, and E is the margin of error.
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