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Dottie Kleman is the "Cookie Lady." She bakes and sells cookies at locations in...
May 17, 2024
Solution by Steps
step 1
Calculate the mean (xˉ\bar{x}) of the sample. The formula for the mean is: xˉ=∑xin \bar{x} = \frac{\sum x_i}{n} where xix_i are the sample values and nn is the number of values
step 2
Sum the sample values: 4+1+2+2+1+2+2+1+0+3=18 4 + 1 + 2 + 2 + 1 + 2 + 2 + 1 + 0 + 3 = 18
step 3
Divide the sum by the number of values (n=10n = 10): xˉ=1810=1.8 \bar{x} = \frac{18}{10} = 1.8
step 4
Calculate the standard deviation (ss) using the formula: s=∑(xi−xˉ)2n−1 s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}
step 5
Compute each (xi−xˉ)2(x_i - \bar{x})^2: (4−1.8)2,(1−1.8)2,(2−1.8)2,(2−1.8)2,(1−1.8)2,(2−1.8)2,(2−1.8)2,(1−1.8)2,(0−1.8)2,(3−1.8)2 (4-1.8)^2, (1-1.8)^2, (2-1.8)^2, (2-1.8)^2, (1-1.8)^2, (2-1.8)^2, (2-1.8)^2, (1-1.8)^2, (0-1.8)^2, (3-1.8)^2 =4.84,0.64,0.04,0.04,0.64,0.04,0.04,0.64,3.24,1.44 = 4.84, 0.64, 0.04, 0.04, 0.64, 0.04, 0.04, 0.64, 3.24, 1.44
step 6
Sum these squared differences: 4.84+0.64+0.04+0.04+0.64+0.04+0.04+0.64+3.24+1.44=11.6 4.84 + 0.64 + 0.04 + 0.04 + 0.64 + 0.04 + 0.04 + 0.64 + 3.24 + 1.44 = 11.6
step 7
Divide by n−1n-1: 11.69=1.29 \frac{11.6}{9} = 1.29
step 8
Take the square root to find the standard deviation: s=1.29≈1.14 s = \sqrt{1.29} \approx 1.14
Answer
Mean: 1.8, Standard Deviation: 1.14
(b) What is the population mean? What is the best estimate of that value?
step 1
The population mean (μ\mu) is unknown. The best estimate of the population mean is the sample mean (xˉ\bar{x})
Answer
Best estimate of the population mean: 1.8
(c) Develop a 95%95\% confidence interval for the population mean. Assume that the population distribution is normal.
step 1
Use the formula for the confidence interval: xˉ±tα/2(sn) \bar{x} \pm t_{\alpha/2} \left( \frac{s}{\sqrt{n}} \right) where tα/2t_{\alpha/2} is the critical value from the tt-distribution for 95%95\% confidence and n−1n-1 degrees of freedom
step 2
For n−1=9n-1 = 9 degrees of freedom, t0.025≈2.262t_{0.025} \approx 2.262
step 3
Calculate the margin of error: 2.262(1.1410)≈0.816 2.262 \left( \frac{1.14}{\sqrt{10}} \right) \approx 0.816
step 4
Construct the confidence interval: 1.8±0.816 1.8 \pm 0.816 =(0.984,2.616) = (0.984, 2.616)
Answer
95%95\% confidence interval: (0.984, 2.616)
(d) Explain why the tt distribution is used as a part of the confidence interval.
step 1
The tt distribution is used because the sample size is small (n < 30) and the population standard deviation is unknown. The tt distribution accounts for the additional variability introduced by estimating the population standard deviation from the sample
Answer
The tt distribution is used due to small sample size and unknown population standard deviation.
(e) Is it reasonable to conclude that the typical worker does not miss any days during a pay period?
step 1
Examine the confidence interval (0.984, 2.616). Since 0 is not within this interval, it is not reasonable to conclude that the typical worker does not miss any days during a pay period
Answer
It is not reasonable to conclude that the typical worker does not miss any days during a pay period.
Key Concept
Confidence intervals provide a range of values within which the population parameter is expected to lie.
Explanation
The confidence interval for the mean does not include 0, indicating that the average number of days missed is greater than 0.
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