A Random Sample Of 12 Lunch Orders At Noodles And Company
A Random Sample Of 12 Lunch Orders At Noodles And Company Showed A Mea
A random sample of 12 lunch orders at Noodles and Company showed a mean bill of $12.99 with a standard deviation of $4.6. Find the 98 percent confidence interval for the mean bill of all lunch orders. (Round your answers to 4 decimal places.)
A random sample of 140 items is drawn from a population whose standard deviation is known to be σ = 50. The sample mean is 780. The 98 percent confidence interval is from?
Find a confidence interval for μ assuming that each sample is from a normal population. (Round the value of t to 3 decimal places and your final answers to 2 decimal places.)
(a) Sample mean = 23, standard deviation = 2, sample size = 7, 90 percent confidence.
(b) Sample mean = 40, standard deviation = 4, sample size = 16, 99 percent confidence.
(c) Sample mean = 116, standard deviation = 11, sample size = 26, 95 percent confidence.
The margin of error for a poll, assuming that 95% confidence and p = 0.3, with:
(a) n = 75
(b) n = 300
(c) n = 750
(d) n = 3,000
is to be calculated, rounding answers to four decimal places.
Paper For Above instruction
Introduction
Confidence intervals are fundamental tools in inferential statistics, providing a range within which we expect the true population parameter to lie with a certain level of confidence. They allow researchers to make educated guesses about the population based on sample data, and are used extensively across various fields such as marketing, healthcare, and social sciences. The following paper addresses multiple problems related to confidence intervals, including those for population means and proportions, with known and unknown variances, as well as different confidence levels.
Confidence Interval for the Mean Bill of Lunch Orders at Noodles and Company
The initial problem involves constructing a 98% confidence interval for the mean cost of lunch orders at Noodles and Company based on a sample of 12 orders. The sample's mean is $12.99, with a standard deviation of $4.6. When the sample size is small (n
Given data:
- Sample mean (\( \bar{x} \)) = 12.99
- Sample standard deviation (s) = 4.6
- Sample size (n) = 12
- Confidence level = 98%
The degrees of freedom (df) = n - 1 = 11. The critical t-value for 98% confidence and df = 11, found from t-tables or statistical software, is approximately 2.718. The margin of error (ME) is calculated as:
\[
ME = t_{\alpha/2, df} \times \frac{s}{\sqrt{n}}
\]
\[
ME = 2.718 \times \frac{4.6}{\sqrt{12}} \approx 2.718 \times 1.327 \approx 3.6054
\]
Thus, the 98% confidence interval is:
\[
(12.99 - 3.6054,\, 12.99 + 3.6054) = (9.3846,\, 16.5954)
\]
Rounded to four decimal places:
(9.3846, 16.5954)
This interval suggests that, with 98% confidence, the mean lunch bill at Noodles and Company lies between approximately $9.38 and $16.60.
Confidence Interval When Population Standard Deviation Is Known
The second scenario involves a sample of 140 items with a known population standard deviation \( \sigma = 50 \). The sample mean is 780.
Given:
- \( \sigma = 50 \)
- \( \bar{x} = 780 \)
- n = 140
- Confidence level = 98%
The z-critical value for a 98% confidence interval (because \( \sigma \) is known) is approximately 2.326, obtained from standard normal distribution tables. The margin of error:
\[
ME = z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} = 2.326 \times \frac{50}{\sqrt{140}} \approx 2.326 \times 4.226 \approx 9.835
\]
Therefore, the confidence interval:
\[
(780 - 9.835,\, 780 + 9.835) = (770.165,\, 789.835)
\]
Rounded to two decimal places:
(770.17, 789.84)
This interval indicates that the true population mean is likely between approximately $770.17 and $789.84 at 98% confidence.
Confidence Intervals for Various Sample Sizes and Confidence Levels
For the next three cases assuming normal populations, the formula involves the t-distribution, especially when the sample size is small or the standard deviation is unknown. The general form:
\[
\text{CI} = \bar{x} \pm t_{\alpha/2, df} \times \frac{s}{\sqrt{n}}
\]
Where:
- \( \bar{x} \) = sample mean
- \( s \) = sample standard deviation
- \( n \) = sample size
- \( t_{\alpha/2, df} \) = t-value for given confidence and degrees of freedom.
(a) For \( \bar{x} = 23 \), s=2, n=7, 90% confidence:
Degrees of freedom \( df = 6 \). The t-value for 90% confidence and df=6 is approximately 1.943. The standard error:
\[
SE = \frac{2}{\sqrt{7}} \approx 0.7559
\]
Margin of error:
\[
ME = 1.943 \times 0.7559 \approx 1.470
\]
Confidence interval:
\[
(23 - 1.470, 23 + 1.470) = (21.530, 24.470)
\]
(b) For \( \bar{x} = 40 \), s=4, n=16, 99% confidence:
degrees of freedom = 15. The t-value for 99% confidence is approximately 2.947. Standard error:
\[
SE= \frac{4}{\sqrt{16}}= 1
\]
Margin of error:
\[
ME=2.947 \times 1= 2.947
\]
Confidence interval:
\[
(40 - 2.947, 40 + 2.947) = (37.053, 42.947)
\]
(c) For \( \bar{x} = 116 \), s=11, n=26, 95% confidence:
Degrees of freedom = 25. The t-value is approximately 2.060. Standard error:
\[
SE= \frac{11}{\sqrt{26}} \approx 2.151
\]
Margin of error:
\[
ME= 2.060 \times 2.151 \approx 4.439
\]
Confidence interval:
\[
(116 - 4.439, 116 + 4.439) = (111.561, 120.439)
\]
Margin of Error for Poll Proportions
The margin of error (ME) when estimating a population proportion p at a confidence level \( 1-\alpha \) is:
\[
ME = z_{\alpha/2} \times \sqrt{\frac{p(1-p)}{n}}
\]
Where:
- \( p=0.3 \)
- \( n \) = sample size
- \( z_{\alpha/2} \) = critical value from the standard normal distribution corresponding to the confidence level.
Using approximate z-values:
- 95% confidence: \( z_{0.025} \approx 1.960 \)
Calculations:
(a) n=75
\[
ME=1.960 \times \sqrt{\frac{0.3 \times 0.7}{75}} \approx 1.960 \times \sqrt{\frac{0.21}{75}} \approx 1.960 \times 0.0529 \approx 0.1038
\]
(b) n=300
\[
ME=1.960 \times \sqrt{\frac{0.21}{300}} \approx 1.960 \times 0.0265 \approx 0.0520
\]
(c) n=750
\[
ME=1.960 \times \sqrt{\frac{0.21}{750}} \approx 1.960 \times 0.0167 \approx 0.0328
\]
(d) n=3,000
\[
ME=1.960 \times \sqrt{\frac{0.21}{3000}} \approx 1.960 \times 0.0084 \approx 0.0165
\]
Conclusion
Constructing confidence intervals requires understanding the nature of the data, whether standard deviations are known, the sample sizes, and the targeted confidence levels. These calculations exemplify how variability and sample size influence the precision of estimates. Small samples typically require t-distribution, whereas large samples or known population parameters warrant the z-distribution. Additionally, margin of error calculations for proportions are vital in survey research, providing bounds within which the true population proportion is expected to fall with specified confidence. Understanding these principles is fundamental for accurate statistical inference across varied fields.
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