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5. A random sample of X1,X2,…,XnX_{1}, X_{2}, \ldots, X_{n} of size n is taken from $N\...
Nov 26, 2024
Solution by Steps
step 1
The likelihood function for a random sample X1,X2,…,Xn X_{1}, X_{2}, \ldots, X_{n} from a normal distribution N(μ,σ2) N(\mu, \sigma^2) is given by: L(θ)=∏i=1n12πθexp⁡(−(Xi−μ)22θ) L(\theta) = \prod_{i=1}^{n} \frac{1}{\sqrt{2\pi\theta}} \exp\left(-\frac{(X_i - \mu)^2}{2\theta}\right)
step 2
To find the maximum likelihood estimator (MLE), we take the natural logarithm of the likelihood function: log⁡L(θ)=−n2log⁡(2πθ)−12θ∑i=1n(Xi−μ)2 \log L(\theta) = -\frac{n}{2} \log(2\pi\theta) - \frac{1}{2\theta} \sum_{i=1}^{n} (X_i - \mu)^2
step 3
Next, we differentiate the log-likelihood with respect to θ \theta and set it to zero: ddθlog⁡L(θ)=−n2θ+12θ2∑i=1n(Xi−μ)2=0 \frac{d}{d\theta} \log L(\theta) = -\frac{n}{2\theta} + \frac{1}{2\theta^2} \sum_{i=1}^{n} (X_i - \mu)^2 = 0
step 4
Solving for θ \theta gives us: nθ=∑i=1n(Xi−μ)2  ⟹  θ^=1n∑i=1n(Xi−μ)2 n\theta = \sum_{i=1}^{n} (X_i - \mu)^2 \implies \hat{\theta} = \frac{1}{n} \sum_{i=1}^{n} (X_i - \mu)^2
step 5
To show that θ^ \hat{\theta} is an unbiased estimator of θ \theta , we calculate the expected value: E[θ^]=E[1n∑i=1n(Xi−μ)2]=θ E[\hat{\theta}] = E\left[\frac{1}{n} \sum_{i=1}^{n} (X_i - \mu)^2\right] = \theta
Answer
The maximum likelihood estimator for θ \theta is θ^=1n∑i=1n(Xi−μ)2 \hat{\theta} = \frac{1}{n} \sum_{i=1}^{n} (X_i - \mu)^2 , and it is an unbiased estimator of θ \theta .
Key Concept
Maximum likelihood estimation and unbiasedness of estimators.
Explanation
The MLE for variance in a normal distribution is derived from the likelihood function, and it is shown to be unbiased by calculating its expected value.
Solution by Steps
step 1
To sketch the probability density function (pdf) f(x;θ)=θxθ−1 f(x; \theta) = \theta x^{\theta - 1} for θ=12 \theta = \frac{1}{2} and θ=2 \theta = 2 , we evaluate the function at these values. For θ=12 \theta = \frac{1}{2} , the pdf becomes f(x;12)=12x−12 f(x; \frac{1}{2}) = \frac{1}{2} x^{-\frac{1}{2}} for 0 < x < 1 , which is a decreasing function. For θ=2 \theta = 2 , the pdf is f(x;2)=2x f(x; 2) = 2x , which is an increasing function
step 2
To show that the maximum likelihood estimator (MLE) of θ \theta is θ^=−nln⁡(∏i=1nXi) \hat{\theta} = \frac{-n}{\ln\left(\prod_{i=1}^{n} X_{i}\right)} , we start with the likelihood function: L(θ)=∏i=1nf(Xi;θ)=∏i=1nθXiθ−1 L(\theta) = \prod_{i=1}^{n} f(X_i; \theta) = \prod_{i=1}^{n} \theta X_i^{\theta - 1} Taking the natural logarithm gives: ln⁡L(θ)=nln⁡(θ)+(θ−1)∑i=1nln⁡(Xi) \ln L(\theta) = n \ln(\theta) + (\theta - 1) \sum_{i=1}^{n} \ln(X_i) Differentiating with respect to θ \theta and setting it to zero leads to the MLE
step 3
For the two given data sets, we calculate θ^ \hat{\theta} . For the first data set X=[0.0256,0.3051,0.0278,0.8971,0.0739,0.3191,0.7379,0.3671,0.9763,0.0102] X = [0.0256, 0.3051, 0.0278, 0.8971, 0.0739, 0.3191, 0.7379, 0.3671, 0.9763, 0.0102] : θ^1=−10ln⁡(0.0256⋅0.3051⋅0.0278⋅0.8971⋅0.0739⋅0.3191⋅0.7379⋅0.3671⋅0.9763⋅0.0102) \hat{\theta}_1 = \frac{-10}{\ln(0.0256 \cdot 0.3051 \cdot 0.0278 \cdot 0.8971 \cdot 0.0739 \cdot 0.3191 \cdot 0.7379 \cdot 0.3671 \cdot 0.9763 \cdot 0.0102)} For the second data set X=[0.4698,0.3675,0.5991,0.9513,0.6049,0.9917,0.1551,0.0710,0.2110,0.2154] X = [0.4698, 0.3675, 0.5991, 0.9513, 0.6049, 0.9917, 0.1551, 0.0710, 0.2110, 0.2154] : θ^2=−10ln⁡(0.4698⋅0.3675⋅0.5991⋅0.9513⋅0.6049⋅0.9917⋅0.1551⋅0.0710⋅0.2110⋅0.2154) \hat{\theta}_2 = \frac{-10}{\ln(0.4698 \cdot 0.3675 \cdot 0.5991 \cdot 0.9513 \cdot 0.6049 \cdot 0.9917 \cdot 0.1551 \cdot 0.0710 \cdot 0.2110 \cdot 0.2154)}
Answer
The maximum likelihood estimates for the two data sets can be calculated using the formulas provided.
Key Concept
The maximum likelihood estimator (MLE) is a method used to estimate the parameters of a statistical model by maximizing the likelihood function.
Explanation
The MLE provides a way to estimate the parameter θ \theta based on observed data, ensuring that the estimated values are consistent with the underlying probability distribution.
Let X equal the weight in grams of a ”52-gram” snack pack of candies. Assume that the distribution of X is N(µ, 4). A random sample of n=10 observations of X yielded the following data: 55.95, 56.54,57.58,55.13,57.48,56.06,59.93,58.30,52.27,58.46. (a) Give a point estimate for µ (b) Find the endpoints for a 95% confidence interval for µ. (c) On the basis of these very limited data, what is the probability that an individual snack pack selected at random is filled with less than 52 grams of candy?
Solution by Steps
step 1
To find a point estimate for μ \mu , we calculate the sample mean Xˉ \bar{X} using the formula: Xˉ=1n∑i=1nXi \bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i where n=10 n = 10 and the observations are 55.95,56.54,57.58,55.13,57.48,56.06,59.93,58.30,52.27,58.46 55.95, 56.54, 57.58, 55.13, 57.48, 56.06, 59.93, 58.30, 52.27, 58.46
step 2
Calculating the sample mean: Xˉ=55.95+56.54+57.58+55.13+57.48+56.06+59.93+58.30+52.27+58.4610=56.77 \bar{X} = \frac{55.95 + 56.54 + 57.58 + 55.13 + 57.48 + 56.06 + 59.93 + 58.30 + 52.27 + 58.46}{10} = 56.77
step 3
For the 95% confidence interval for μ \mu , we use the formula: Xˉ±zα/2⋅σn \bar{X} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}} where zα/2 z_{\alpha/2} for 95% confidence is approximately 1.96, σ=2 \sigma = 2 (since variance is 4), and n=10 n = 10
step 4
Calculating the margin of error: ME=1.96⋅210≈1.24 ME = 1.96 \cdot \frac{2}{\sqrt{10}} \approx 1.24 Thus, the confidence interval is: (56.77−1.24,56.77+1.24)=(55.53,58.01) (56.77 - 1.24, 56.77 + 1.24) = (55.53, 58.01)
step 5
To find the probability that an individual snack pack weighs less than 52 grams, we standardize using the Z-score: Z=X−μσ=52−56.772=−2.385 Z = \frac{X - \mu}{\sigma} = \frac{52 - 56.77}{2} = -2.385 We then find P(Z < -2.385) using the standard normal distribution table
step 6
From the Z-table, P(Z < -2.385) \approx 0.0087 . Thus, the probability that an individual snack pack weighs less than 52 grams is approximately 0.0087
Answer
(a) The point estimate for μ \mu is 56.77. (b) The 95% confidence interval for μ \mu is (55.53, 58.01). (c) The probability that an individual snack pack weighs less than 52 grams is approximately 0.0087.
Key Concept
Point estimation and confidence intervals are fundamental concepts in inferential statistics.
Explanation
The sample mean provides a point estimate for the population mean, while the confidence interval gives a range of plausible values for the mean based on the sample data. The probability calculation uses the normal distribution to assess the likelihood of an individual observation.
A leakage test was conducted to determine the effectiveness of a seal designed to keep the inside of a plug airtight. An air needle was inserted into the plug, and the plug and needle were placed under water. The pressure was then increased until leakage was observed. Let X equal the pressure in pounds per square inch. Assume that the distribution of X is N(µ, σ2). The following n=10 observations of X were obtained; 3.1,3.3,4.5,2.8,3.5,3.5,3.7,4.2,3.9,3.3 Use the observations to (a) Find a point estimate of µ (b) Find a point estimate of σ (c) Find a 95% confidence interval for µ
Solution by Steps
step 1
To find a point estimate of μ \mu , we calculate the sample mean Xˉ \bar{X} using the formula: Xˉ=1n∑i=1nXi \bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i where n=10 n = 10 and the observations are 3.1,3.3,4.5,2.8,3.5,3.5,3.7,4.2,3.9,3.3 3.1, 3.3, 4.5, 2.8, 3.5, 3.5, 3.7, 4.2, 3.9, 3.3
step 2
Calculating the sample mean: Xˉ=3.1+3.3+4.5+2.8+3.5+3.5+3.7+4.2+3.9+3.310=3.54 \bar{X} = \frac{3.1 + 3.3 + 4.5 + 2.8 + 3.5 + 3.5 + 3.7 + 4.2 + 3.9 + 3.3}{10} = 3.54
step 3
To find a point estimate of σ \sigma , we calculate the sample standard deviation s s using the formula: s=1n−1∑i=1n(Xi−Xˉ)2 s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (X_i - \bar{X})^2}
step 4
Calculating the sample standard deviation: s=19((3.1−3.54)2+(3.3−3.54)2+…+(3.3−3.54)2)≈0.487 s = \sqrt{\frac{1}{9} \left((3.1 - 3.54)^2 + (3.3 - 3.54)^2 + \ldots + (3.3 - 3.54)^2\right)} \approx 0.487
step 5
To find a 95% confidence interval for μ \mu , we use the formula: Xˉ±tα/2⋅sn \bar{X} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} where tα/2 t_{\alpha/2} is the critical value from the t-distribution for n−1 n-1 degrees of freedom. For n=10 n=10 , t0.025,9≈2.262 t_{0.025, 9} \approx 2.262
step 6
Calculating the confidence interval: 3.54±2.262⋅0.48710≈(2.93,4.15) 3.54 \pm 2.262 \cdot \frac{0.487}{\sqrt{10}} \approx (2.93, 4.15)
Answer
(a) Point estimate of μ \mu is 3.54 3.54 . (b) Point estimate of σ \sigma is approximately 0.487 0.487 . (c) The 95% confidence interval for μ \mu is (2.93,4.15) (2.93, 4.15) .
Key Concept
Point estimates and confidence intervals are fundamental in inferential statistics for estimating population parameters.
Explanation
The sample mean provides a point estimate for the population mean, while the confidence interval gives a range of plausible values for the mean based on the sample data.
Let ¯X be the mean of a random sample of size n from N(µ, 9). Find n so that P(¯X −1 < µ < ¯X + 1) =0.90
Solution by Steps
step 1
To find n n such that P(\bar{X} - 1 < \mu < \bar{X} + 1) = 0.90 , we first recognize that Xˉ \bar{X} follows a normal distribution N(μ,9n) N(\mu, \frac{9}{n})
step 2
The probability can be rewritten using the standard normal variable Z Z : P\left(-1 < \frac{\bar{X} - \mu}{\sigma/\sqrt{n}} < 1\right) = P\left(-1 < Z < 1\right) where σ=3 \sigma = 3
step 3
From the standard normal distribution, we know that P(-1 < Z < 1) corresponds to approximately 0.6827. To achieve a probability of 0.90, we need to find the critical values zα/2 z_{\alpha/2} such that P(-z_{\alpha/2} < Z < z_{\alpha/2}) = 0.90 . This gives zα/2≈1.645 z_{\alpha/2} \approx 1.645
step 4
Setting up the equation: 1.645=1σ/n=13/n 1.645 = \frac{1}{\sigma/\sqrt{n}} = \frac{1}{3/\sqrt{n}} leads to 1.645⋅3n=1 1.645 \cdot \frac{3}{\sqrt{n}} = 1
step 5
Solving for n n : n=1.645⋅3  ⟹  n=(1.645⋅3)2≈25.7 \sqrt{n} = 1.645 \cdot 3 \implies n = (1.645 \cdot 3)^2 \approx 25.7 , rounding up gives n=26 n = 26
Answer
The required sample size n n is 26.
Key Concept
The concept of determining sample size based on confidence intervals and normal distribution.
Explanation
The calculation involves using the properties of the normal distribution to find the necessary sample size to achieve a specified confidence level.
Solution by Steps
step 1
To find the approximate 98% 98\% confidence interval for the difference μX−μY \mu_X - \mu_Y , we first calculate the standard error (SE) of the difference in means using the formula: SE=sX2nX+sY2nY SE = \sqrt{\frac{s_X^2}{n_X} + \frac{s_Y^2}{n_Y}} where sX=4 s_X = 4 , nX=12 n_X = 12 , sY=3 s_Y = 3 , and nY=15 n_Y = 15
step 2
Calculating the standard error: SE=4212+3215=1612+915=1.3333+0.6=1.9333≈1.39 SE = \sqrt{\frac{4^2}{12} + \frac{3^2}{15}} = \sqrt{\frac{16}{12} + \frac{9}{15}} = \sqrt{1.3333 + 0.6} = \sqrt{1.9333} \approx 1.39
step 3
Next, we find the critical value for a 98% 98\% confidence level. For a two-tailed test, we look up the critical value z z for α=0.02 \alpha = 0.02 (which is z≈2.33 z \approx 2.33 )
step 4
Now, we calculate the confidence interval for the difference in means using the formula: (xˉ−yˉ)±z⋅SE (\bar{x} - \bar{y}) \pm z \cdot SE where xˉ=65.7 \bar{x} = 65.7 and yˉ=68.2 \bar{y} = 68.2 . Thus, the difference xˉ−yˉ=65.7−68.2=−2.5 \bar{x} - \bar{y} = 65.7 - 68.2 = -2.5
step 5
The confidence interval is then: −2.5±2.33⋅1.39 -2.5 \pm 2.33 \cdot 1.39 which calculates to: −2.5±3.24 -2.5 \pm 3.24 giving us the interval (−5.74,0.74) (-5.74, 0.74)
Answer
The approximate 98% 98\% confidence interval for the difference μX−μY \mu_X - \mu_Y is (−5.74,0.74) (-5.74, 0.74) .
Key Concept
Confidence intervals provide a range of values that likely contain the true difference between population means.
Explanation
The calculated confidence interval indicates that we are 98% 98\% confident that the true difference in means lies between −5.74 -5.74 and 0.74 0.74 .
A proportion, p, that many public opinion polls estimate is the number of Americans who would say yes to the question, ”If something were to happen to the president of the United States, do you think that the vice president would be qualified to take over as president?” In one such random sample of 1022 adults, 388 said yes. (a) On the basis of the given data, find a point estimate of p. (b) Find an approximate 90% confidence interval for p.
Solution by Steps
step 1
To find a point estimate of p p , we use the formula: p^=xn \hat{p} = \frac{x}{n} where x=388 x = 388 (number of yes responses) and n=1022 n = 1022 (total sample size)
step 2
Calculating the point estimate: p^=3881022≈0.3805 \hat{p} = \frac{388}{1022} \approx 0.3805
step 3
To find the approximate 90% 90\% confidence interval for p p , we first calculate the standard error (SE) using the formula: SE=p^(1−p^)n SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
step 4
Substituting the values into the SE formula: SE=0.3805(1−0.3805)1022≈0.0155 SE = \sqrt{\frac{0.3805(1 - 0.3805)}{1022}} \approx 0.0155
step 5
The critical value for a 90% 90\% confidence interval (using the Z-distribution) is approximately 1.645 1.645 . Thus, the margin of error (ME) is: ME=Z⋅SE=1.645⋅0.0155≈0.0255 ME = Z \cdot SE = 1.645 \cdot 0.0155 \approx 0.0255
step 6
Finally, the confidence interval is given by: p^±ME=(0.3805−0.0255,0.3805+0.0255)=(0.3550,0.4060) \hat{p} \pm ME = (0.3805 - 0.0255, 0.3805 + 0.0255) = (0.3550, 0.4060)
Answer
Point estimate of p p is approximately 0.3805 0.3805 and the 90% 90\% confidence interval for p p is (0.3550,0.4060) (0.3550, 0.4060) .
Key Concept
Point estimates and confidence intervals are fundamental concepts in statistics used to infer population parameters from sample data.
Explanation
The point estimate provides a single value estimate of the proportion, while the confidence interval gives a range of plausible values for the true proportion based on the sample data.
A candy manufacturer selects mints at random from the production line and weights them. For one week, the day shift weighed n1 = 194 mints and the night shift weighed n2 = 162 mints. The numbers of these mints that weighed at most 21 grams was y1 = 28 for the day shift and y2 = 11 for the night shift. Let p1 and p2 denote the proportions of mints that weigh at most 21 grams for the day and night shifts, respectively. (a) Give a point estimate of p1 (b) Give the endpoints for an approximate 95% confidence interval for p1 (c) Give a point estimate of p1 − p2 (d) Find a one-sided approximate 95% confidence interval that gives a lower bound for p1 − p2
Solution by Steps
step 1
To calculate the point estimate of p1 p_1 , we use the formula: p1=y1n1=28194 p_1 = \frac{y_1}{n_1} = \frac{28}{194}
step 2
Calculating p1 p_1 : p1≈0.1443 p_1 \approx 0.1443
1 Answer
A
step 3
To find the endpoints for a 95% confidence interval for p1 p_1 , we use the formula: CI=p1±Zp1(1−p1)n1 CI = p_1 \pm Z \sqrt{\frac{p_1(1 - p_1)}{n_1}} where Z Z for 95% confidence is approximately 1.96
step 4
Calculating the standard error: SE=0.1443(1−0.1443)194≈0.0345 SE = \sqrt{\frac{0.1443(1 - 0.1443)}{194}} \approx 0.0345
step 5
Now, calculating the confidence interval: CI=0.1443±1.96×0.0345 CI = 0.1443 \pm 1.96 \times 0.0345
step 6
The endpoints are: CI≈(0.0775,0.2111) CI \approx (0.0775, 0.2111)
2 Answer
B
step 7
To find the point estimate of the difference in proportions p1−p2 p_1 - p_2 , we first calculate p2 p_2 : p2=y2n2=11162 p_2 = \frac{y_2}{n_2} = \frac{11}{162}
step 8
Calculating p2 p_2 : p2≈0.0679 p_2 \approx 0.0679
step 9
Now, calculating the difference: p1−p2=0.1443−0.0679≈0.0764 p_1 - p_2 = 0.1443 - 0.0679 \approx 0.0764
3 Answer
A
step 10
To find a one-sided 95% confidence interval for p1−p2 p_1 - p_2 , we use the formula: CI=(p1−p2)−ZSE12+SE22 CI = (p_1 - p_2) - Z \sqrt{SE_1^2 + SE_2^2} where SE1 SE_1 and SE2 SE_2 are the standard errors for p1 p_1 and p2 p_2 respectively
step 11
Calculating SE2 SE_2 : SE2=p2(1−p2)n2≈0.0679(1−0.0679)162≈0.0172 SE_2 = \sqrt{\frac{p_2(1 - p_2)}{n_2}} \approx \sqrt{\frac{0.0679(1 - 0.0679)}{162}} \approx 0.0172
step 12
Now, calculating the combined standard error: SEdiff=0.03452+0.01722≈0.0385 SE_{diff} = \sqrt{0.0345^2 + 0.0172^2} \approx 0.0385
step 13
Finally, calculating the one-sided confidence interval: CI=0.0764−1.645×0.0385≈0.0764−0.0633≈0.0131 CI = 0.0764 - 1.645 \times 0.0385 \approx 0.0764 - 0.0633 \approx 0.0131
4 Answer
A
Key Concept
Point estimates and confidence intervals are fundamental in statistics for estimating population parameters based on sample data.
Explanation
Point estimates provide a single value estimate of a parameter, while confidence intervals give a range of values that likely contain the parameter with a specified level of confidence.
A candy manufacturer selects mints at random from the production line and weights them. For one week, the day shift weighed n1 = 194 mints and the night shift weighed n2 = 162 mints. The numbers of these mints that weighed at most 21 grams was y1 = 28 for the day shift and y2 = 11 for the night shift. Let p1 and p2 denote the proportions of mints that weigh at most 21 grams for the day and night shifts, respectively. (a) Give a point estimate of p1 (b) Give the endpoints for an approximate 95% confidence interval for p1 (c) Give a point estimate of p1 − p2 (d) Find a one-sided approximate 95% confidence interval that gives a lower bound for p1 − p2
Solution by Steps
step 1
To calculate the point estimate p1 p_1 for the proportion of mints weighing at most 21 grams from the day shift, we use the formula: p1=y1n1=28194 p_1 = \frac{y_1}{n_1} = \frac{28}{194}
step 2
Calculating p1 p_1 : p1≈0.1443 p_1 \approx 0.1443
step 3
For the 95% confidence interval for p1 p_1 , we first calculate the standard error (SE): SE=p1(1−p1)n1=0.1443(1−0.1443)194 SE = \sqrt{\frac{p_1(1 - p_1)}{n_1}} = \sqrt{\frac{0.1443(1 - 0.1443)}{194}}
step 4
Calculating the SE: SE≈0.0344 SE \approx 0.0344
step 5
The critical value for a 95% confidence interval (Z) is approximately 1.96. The margin of error (ME) is: ME=Z⋅SE=1.96⋅0.0344 ME = Z \cdot SE = 1.96 \cdot 0.0344
step 6
Calculating the ME: ME≈0.0675 ME \approx 0.0675
step 7
The endpoints for the 95% confidence interval are: (p1−ME,p1+ME)=(0.1443−0.0675,0.1443+0.0675) (p_1 - ME, p_1 + ME) = (0.1443 - 0.0675, 0.1443 + 0.0675)
step 8
Calculating the endpoints: ≈(0.0768,0.2118) \approx (0.0768, 0.2118)
step 9
To find the point estimate for the difference p1−p2 p_1 - p_2 , we first calculate p2 p_2 : p2=y2n2=11162 p_2 = \frac{y_2}{n_2} = \frac{11}{162}
step 10
Calculating p2 p_2 : p2≈0.0679 p_2 \approx 0.0679
step 11
Now, we find the difference: p1−p2=0.1443−0.0679 p_1 - p_2 = 0.1443 - 0.0679
step 12
Calculating the difference: p1−p2≈0.0764 p_1 - p_2 \approx 0.0764
step 13
For the one-sided 95% confidence interval for p1−p2 p_1 - p_2 , we use the standard error for the difference: SEdiff=SE12+SE22 SE_{diff} = \sqrt{SE_1^2 + SE_2^2} where SE1 SE_1 and SE2 SE_2 are the standard errors for p1 p_1 and p2 p_2 respectively
step 14
Calculating SE2 SE_2 : SE2=p2(1−p2)n2=0.0679(1−0.0679)162 SE_2 = \sqrt{\frac{p_2(1 - p_2)}{n_2}} = \sqrt{\frac{0.0679(1 - 0.0679)}{162}}
step 15
Calculating SE2 SE_2 : SE2≈0.0490 SE_2 \approx 0.0490
step 16
Now, calculating SEdiff SE_{diff} : SEdiff=(0.0344)2+(0.0490)2 SE_{diff} = \sqrt{(0.0344)^2 + (0.0490)^2}
step 17
Calculating SEdiff SE_{diff} : SEdiff≈0.0600 SE_{diff} \approx 0.0600
step 18
The margin of error for the one-sided interval is: MEone−sided=Z⋅SEdiff=1.645⋅0.0600 ME_{one-sided} = Z \cdot SE_{diff} = 1.645 \cdot 0.0600
step 19
Calculating MEone−sided ME_{one-sided} : MEone−sided≈0.0987 ME_{one-sided} \approx 0.0987
step 20
The lower bound for the one-sided confidence interval is: (p1−p2−MEone−sided)=0.0764−0.0987 (p_1 - p_2 - ME_{one-sided}) = 0.0764 - 0.0987
step 21
Calculating the lower bound: ≈−0.0223 \approx -0.0223
Answer
0.1443
2.
0.0768, 0.2118
3.
0.0764
4.
-0.0223
Key Concept
Point estimates and confidence intervals for proportions
Explanation
Point estimates provide a single value estimate of a population parameter, while confidence intervals give a range of values that likely contain the parameter.
Some college professors and students examined 137 Canadian geese for patent schistosome in the year they hatched. Of these 137 birds, 54 were infected. The professors and students were interested in estimating p, the proportion of infected birds of this type. For future studies, determine the sample size n so that the estimate of p is within ϵ = 0.04 of the unknown p with approximate 90% confidence
Solution by Steps
step 1
First, we need to estimate the proportion p p of infected geese from the previous study. The sample proportion p^ \hat{p} is calculated as: p^=54137≈0.3940 \hat{p} = \frac{54}{137} \approx 0.3940
step 2
Next, we determine the critical value z z for a 90% confidence level. The critical value for a 90% confidence level is approximately: z≈1.645 z \approx 1.645
step 3
We can now use the formula for the sample size n n needed to estimate a proportion within a margin of error ϵ \epsilon : n=(z2p^(1−p^)ϵ2) n = \left( \frac{z^2 \hat{p}(1 - \hat{p})}{\epsilon^2} \right) Substituting the values we have: n=((1.645)2⋅0.3940⋅(1−0.3940)(0.04)2) n = \left( \frac{(1.645)^2 \cdot 0.3940 \cdot (1 - 0.3940)}{(0.04)^2} \right)
step 4
Calculating the above expression: n=((2.7060)⋅0.3940⋅0.60600.0016)≈257.56 n = \left( \frac{(2.7060) \cdot 0.3940 \cdot 0.6060}{0.0016} \right) \approx 257.56 Since we cannot have a fraction of a sample, we round up to the nearest whole number: n≈258 n \approx 258
Answer
The required sample size n n is approximately 258 geese.
Key Concept
Sample size determination for estimating proportions with specified margin of error and confidence level.
Explanation
The calculation involves estimating the sample proportion, determining the critical value for the desired confidence level, and applying the sample size formula to ensure the estimate falls within the specified margin of error.
A quality engineer wanted to be 98% confident that the maximum error of the estimate of the mean strength(in megapascals),µ, of the left hinge on a vanity cover molded by a machine is 0.25. A preliminary sample of size n=32 parts yielded a sample mean of ¯x = 35.68 and a standard deviation of s=1.723. (a) How large a sample is required? (b) Does this seem to be a reasonable sample size? (Note that destructive testing is needed to obtain the data)
Solution by Steps
step 1
To calculate the required sample size n n for estimating the mean with a specified maximum error E E and confidence level, we use the formula: n=(Z⋅sE)2 n = \left( \frac{Z \cdot s}{E} \right)^2 where Z Z is the Z-score corresponding to the confidence level, s s is the sample standard deviation, and E E is the maximum error. For a 98% confidence level, Z≈2.33 Z \approx 2.33
step 2
Substituting the values into the formula: n=(2.33⋅1.7230.25)2 n = \left( \frac{2.33 \cdot 1.723}{0.25} \right)^2
step 3
Calculating the numerator: 2.33⋅1.723≈4.01579 2.33 \cdot 1.723 \approx 4.01579 , then dividing by 0.25 0.25 : 4.015790.25≈16.06316 \frac{4.01579}{0.25} \approx 16.06316 , and squaring the result: n≈(16.06316)2≈258.02 n \approx (16.06316)^2 \approx 258.02 , rounding up gives n=259 n = 259
step 4
To assess the reasonableness of the sample size, consider that destructive testing is required. A sample size of 259 may be impractical if it leads to significant loss of product
Answer
(a) The required sample size is 259. (b) This sample size may not be reasonable due to destructive testing.
Key Concept
Sample size determination for estimating population mean with specified confidence and error.
Explanation
The calculation shows that a sample size of 259 is needed for a 98% confidence level with a maximum error of 0.25, but practical considerations like destructive testing may limit feasibility.
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