# Structured Solutions to Probability Questions --- ## **7. Three Numbersly Selected from to 9** ### **Given Information** - Numbers are chosen from to 9, without replacement. - Define events: \( A_1 \): Three numbers don't contain \( A_2 \): Three numbers don't contain or 5 ### **What to Find** - Probability of \( A_1 \) - Probability of \( A_2 \) ### **Concept Used** - Probability = (Number of favorable outcomes) / (Total possible outcomes) - For selections without replacement: Use combinations. ### **Solution** - **Total ways to select 3 numbers from 10:** \( C(10,3) \) - **For \( A_1 \):** Exclude , select from 1–9 (9 numbers): Favorable ways = \( C(9,3) \) \( P(A_1) = \frac{C(9,3)}{C(10,3)} = \frac{84}{120} = .7 \) - **For \( A_2 \):** Exclude and 5, select from 1–4 and 6–9 (8 numbers): Favorable ways = \( C(8,3) \) \( P(A_2) = \frac{C(8,3)}{C(10,3)} = \frac{56}{120} \approx .467 \) --- ## **8. 6 Students in a Dorm, Birthday Probabilities** ### **Given Information** - Each month is equally likely, 12 months in a year, 6 students. ### **What to Find** 1. Probability that at least one student has a birthday in October. 2. Probability that exactly 4 students have birthdays in October. 3. Probability that 4 students share a birthday in the same month. ### **Concept Used** - Binomial probability for fixed months. - Probability of at least one: \( 1 - P(\text{none}) \). ### **Solution** - **(1) At least one in October:** Probability none in October: \( \left(\frac{11}{12}\right)^6 \) So, probability at least one: \( 1 - \left(\frac{11}{12}\right)^6 \approx .401 \) - **(2) Exactly 4 in October:** Choose 4 out of 6: \( C(6,4) \) Probability those 4 are in October: \( \left(\frac{1}{12}\right)^4 \) Remaining 2 not in October: \( \left(\frac{11}{12}\right)^2 \) Total: \( C(6,4) \times \left(\frac{1}{12}\right)^4 \times \left(\frac{11}{12}\right)^2 = 15 \times \frac{1}{20736} \times \frac{121}{144} \approx .000059 \) - **(3) 4 birthdays in same month:** Number of ways to choose the month: 12 Ways to choose 4 students out of 6: \( C(6,4) \) Remaining 2 can be in any other 11 months: \( 11 \times 10 \) (since they must be in different months) Total ways: \( 12 \times 15 \times 110 = 19800 \) Total possible: \( 12^6 = 2985984 \) Probability: \( \frac{19800}{2985984} \approx .0066 \) --- ## **9. Probability of Product A Given Not C** ### **Given Information** - Product A: 60%, B: 30%, C: 10%. - A product is not C. Find probability it's A. ### **What to Find** - \( P(A | \text{not C}) \) ### **Concept Used** Conditional probability: \( P(A|\text{not C}) = \frac{P(A)}{P(\text{not C})} \) ### **Solution** - \( P(\text{not C}) = 1 - .10 = .90 \) - \( P(A|\text{not C}) = \frac{.60}{.90} = \frac{2}{3} \approx .667 \) --- ## **10. Two Warning Systems** ### **Given Information** - System 1 works: .92, system 2 works: .93. - \( P(\text{system 2 works} | \text{system 1 down}) = .85 \) ### **What to Find** 1. Probability both work. 2. Probability system 2 is down, system 1 works. 3. Probability system 1 works given system 2 is down. ### **Concept Used** - Multiplication rule for independent/conditional events. - Bayes’ theorem. ### **Solution** - Let \( A \): system 1 works, \( B \): system 2 works. - (1) Both work: \( P(A) \times P(B|A) \) Since not independent, need \( P(B|A) \): \( P(B|A) = \frac{P(B) - P(B|\bar{A}) \cdot P(\bar{A})}{P(A)} \) (or use total probability law) \( P(\bar{A}) = .08 \), \( P(B|\bar{A}) = .85 \) \( P(B) = P(B|A)\cdot P(A) + P(B|\bar{A})\cdot P(\bar{A}) \) \( .93 = x \cdot .92 + .85 \cdot .08 \implies x = \frac{.93 - .068}{.92} = \frac{.862}{.92} \approx .937 \) Both work: \( .92 \times .937 \approx .862 \) - (2) System 2 down and 1 works: \( P(A) \cdot P(\bar{B}|A) = .92 \times (1-.937) = .92 \times .063 \approx .058 \) - (3) System 1 works given system 2 down: \( P(A|\bar{B}) = \frac{P(A \cap \bar{B})}{P(\bar{B})} \) \( P(\bar{B}) = 1-.93=.07 \) \( P(A \cap \bar{B}) = .058 \) \( P(A|\bar{B}) = \frac{.058}{.07} \approx .829 \) --- ## **11. Independence and Mutual Exclusivity** ### **Given Information** - \( < P(A) < 1 \), \( P(B) > \) ### **What to Find** - Prove: 1. If A and B are independent, then not mutually exclusive. 2. If mutually exclusive, then dependent. ### **Concept Used** - Independence: \( P(A \cap B) = P(A)P(B) \) - Mutually exclusive: \( P(A \cap B) = \) ### **Solution** - (1) If independent, \( P(A \cap B) = P(A)P(B) > \), but if mutually exclusive, \( P(A \cap B) = \). Contradiction. So can't be both. - (2) If mutually exclusive, \( P(A \cap B) = \), but \( P(A)P(B) > \), so not independent (thus dependent). --- ## **Summary of Final Answers** - **7:** \( P(A_1) = .7 \), \( P(A_2) = .467 \) - **8:** (1) .401; (2) .000059; (3) .0066 - **9:** \( P(A|\text{not C}) = .667 \) - **10:** (1) .862; (2) .058; (3) .829 - **11:** 1. Independent events cannot be mutually exclusive. 2. Mutually exclusive events are dependent.

# Structured Solutions to Probability Questions --- ## **7. Three Numbers Selected from 0 to 9** ### **Given Information** - Selecting 3 numbers from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. - Define events: \( A \): Three numbers don’t contain 0 and 5. \( B \): Three numbers don’t contain 0 or 5. ### **What to Find** - Probability of event \( A \). - Probability of event \( B \). ### **Concept Used** - Probability = (Number of favorable outcomes) / (Total possible outcomes). - Use combinations for selections. ### **Solution** - **Total ways to select 3 numbers from 10:** \( C(10,3) = \frac{10!}{3!(10-3)!} = 120 \) - **For event \( A \):** Exclude 0 and 5, select from {1, 2, 3, 4, 6, 7, 8, 9} (8 numbers). Favorable ways = \( C(8,3) = \frac{8!}{3!(8-3)!} = 56 \). \( P(A) = \frac{C(8,3)}{C(10,3)} = \frac{56}{120} \approx 0.467 \) - **For event \( B \):** Exclude 0 and 5, select from {1, 2, 3, 4, 6, 7, 8, 9} (8 numbers). Favorable ways = \( C(8,3) = 56 \). \( P(B) = \frac{C(8,3)}{C(10,3)} = \frac{56}{120} \approx 0.467 \) --- ## **8. Birthday Probabilities of 6 Students in a Dorm** ### **Given Information** - 6 students, birthdays uniformly distributed across 12 months. ### **What to Find** 1. Probability that at least one student has a birthday in October. 2. Probability that exactly 4 students have birthdays in October. 3. Probability that 4 students share a birthday in the same month. ### **Concept Used** - Binomial probability for fixed months. - Probability of at least one: \( 1 - P(\text{none}) \). ### **Solution** - **(1) At least one in October:** Probability none in October: \( P(\text{none}) = \left(\frac{11}{12}\right)^6 \). So, probability at least one: \( P(\text{at least one}) = 1 - \left(\frac{11}{12}\right)^6 \approx 0.487 \). - **(2) Exactly 4 in October:** Choose 4 out of 6: \( C(6,4) = 15 \). Probability for those 4 in October: \( \left(\frac{1}{12}\right)^4 \). Probability for remaining 2 not in October: \( \left(\frac{11}{12}\right)^2 \). Total: \[ P(\text{exactly 4}) = C(6,4) \cdot \left(\frac{1}{12}\right)^4 \cdot \left(\frac{11}{12}\right)^2 = 15 \cdot \left(\frac{1}{20736}\right) \cdot \left(\frac{121}{144} \right) \approx 0.000059 \] - **(3) 4 birthdays in the same month:** Choose the month: 12 ways. Choose 4 students from 6: \( C(6,4) = 15 \). Remaining 2 can be in 11 different months: \( 11 \). Total ways: \[ 12 \cdot 15 \cdot 11 \cdot 10 = 19800 \] Total possible combinations: \( 12^6 = 2985984 \). Probability: \[ P(\text{4 in same month}) = \frac{19800}{2985984} \approx 0.00663 \] --- ## **9. Probability of Product A Given Not C** ### **Given Information** - Product A: 60%, B: 30%, C: 10%. - A product randomly withdrawn is not C. Find the probability that it is A. ### **What to Find** - \( P(A | \text{not C}) \). ### **Concept Used** - Conditional probability: \( P(A | \text{not C}) = \frac{P(A)}{P(\text{not C})} \). ### **Solution** - \( P(\text{not C}) = 1 - 0.10 = 0.90 \). - \( P(A | \text{not C}) = \frac{0.60}{0.90} = \frac{2}{3} \approx 0.667 \). --- ## **10. Two Warning Systems** ### **Given Information** - System 1 works: 0.92, system 2 works: 0.93. - Given system 1 is down, probability that system 2 is working is 0.85. ### **What to Find** 1. Probability both systems are working. 2. Probability system 2 is down and system 1 is working. 3. Probability system 1 is working given system 2 is down. ### **Concept Used** - Multiplication rule for independent/conditional events. ### **Solution** - Let \( A \): system 1 works, \( B \): system 2 works. - **(1) Both systems are working:** \( P(A) \cdot P(B) = 0.92 \cdot 0.93 = 0.856 \). - **(2) System 2 is down and system 1 is working:** \( P(A) \cdot (1 - P(B)) = 0.92 \cdot (1 - 0.93) = 0.92 \cdot 0.07 = 0.0644 \). - **(3) System 1 is working given system 2 is down:** Using Bayes' theorem: \[ P(A | \bar{B}) = \frac{P(A \cap \bar{B})}{P(\bar{B})} \] \( P(\bar{B}) = 0.07 \). \( P(A \cap \bar{B}) = P(A) \cdot (1 - P(B|A)) = 0.92 \cdot 0.07 = 0.0644 \). Therefore, \[ P(A | \bar{B}) = \frac{0.0644}{0.07} \approx 0.92 \] --- ## **11. Independence and Mutual Exclusivity** ### **Given Information** - \( 0 < P(A) < 1 \) - \( P(B) > 0 \) ### **What to Find** - Prove: 1. If A and B are independent, then they are not mutually exclusive. 2. If A and B are mutually exclusive, then they are dependent. ### **Concept Used** - Independence: \( P(A \cap B) = P(A)P(B) \) - Mutually exclusive: \( P(A \cap B) = 0 \) ### **Solution** - **(1)** If \( A \) and \( B \) are independent, then \( P(A \cap B) = P(A)P(B) > 0 \). This contradicts mutual exclusivity, which requires \( P(A \cap B) = 0 \). Thus, they cannot be both independent and mutually exclusive. - **(2)** If \( A \) and \( B \) are mutually exclusive, then \( P(A \cap B) = 0 \). Therefore, \( A \) and \( B \) cannot satisfy \( P(A \cap B) = P(A)P(B) \) (which would be greater than 0), proving that they are dependent. --- ## **Summary of Final Answers** - **7:** \( P(A) \approx 0.467 \), \( P(B) \approx 0.467 \) - **8:** (1) \( \approx 0.487 \); (2) \( \approx 0.000059 \); (3) \( \approx 0.00663 \) - **9:** \( P(A | \text{not C}) \approx 0.667 \) - **10:** (1) \( \approx 0.856 \); (2) \( \approx 0.0644 \); (3) \( \approx 0.92 \) - **11:** 1. Independent events cannot be mutually exclusive. 2. Mutually exclusive events are dependent.

# Structured Solutions to Probability Problems --- ## **7. Probability of Selecting Three Numbers from 0 to 9** ### **Given Information** - Numbers are selected randomly from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. - Define events: \( A \): Three numbers do not contain 0 and 5. \( B \): Three numbers do not contain 0 or 5. ### **What to Find** - Probability of event \( A \). - Probability of event \( B \). ### **Concept Used** - Probability formula: \[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} \] - Use combinations for selections. ### **Solution** - **Total ways to select 3 numbers from 10:** \[ C(10, 3) = 120 \] - **For event \( A \):** Exclude 0 and 5, select from {1, 2, 3, 4, 6, 7, 8, 9} (8 numbers): Favorable ways = \( C(8, 3) = 56 \). \[ P(A) = \frac{C(8, 3)}{C(10, 3)} = \frac{56}{120} \approx 0.467 \] - **For event \( B \):** Exclude 0 and 5, select from {1, 2, 3, 4, 6, 7, 8, 9} (8 numbers): Favorable ways = \( C(8, 3) = 56 \). \[ P(B) = \frac{C(8, 3)}{C(10, 3)} = \frac{56}{120} \approx 0.467 \] --- ## **8. Birthday Probabilities Among 6 Students in a Dorm** ### **Given Information** - 6 students with birthdays uniformly distributed across 12 months. ### **What to Find** 1. Probability that at least one student has a birthday in October. 2. Probability that exactly 4 students have birthdays in October. 3. Probability that 4 students share a birthday in the same month. ### **Concept Used** - Binomial probability for fixed events. - Probability of at least one occurrence: \( 1 - P(\text{none}) \). ### **Solution** - **(1) At least one birthday in October:** Probability none have a birthday in October: \[ P(\text{none}) = \left(\frac{11}{12}\right)^6 \] So, the probability of at least one: \[ P(\text{at least one}) = 1 - \left(\frac{11}{12}\right)^6 \approx 0.487 \] - **(2) Exactly 4 birthdays in October:** Choose 4 out of 6: \( C(6, 4) = 15 \). Probability of those 4 in October: \( \left(\frac{1}{12}\right)^4 \). Probability for remaining 2 not in October: \( \left(\frac{11}{12}\right)^2 \). Combined probability: \[ P(\text{exactly 4}) = C(6, 4) \cdot \left(\frac{1}{12}\right)^4 \cdot \left(\frac{11}{12}\right)^2 \approx 0.000059 \] - **(3) 4 birthdays in the same month:** Choose the month: 12 ways. Choose 4 students from 6: \( C(6, 4) = 15 \). Remaining 2 can be in any of the other 11 months: \( 11 \cdot 10 \). Total combinations: \[ 12 \cdot 15 \cdot 11 \cdot 10 = 19800 \] Total possible combinations: \( 12^6 = 2985984 \). Probability: \[ P(\text{4 in same month}) = \frac{19800}{2985984} \approx 0.00663 \] --- ## **9. Probability of Selecting Product A Given Not C** ### **Given Information** - Product A: 60%, B: 30%, C: 10%. - A product selected is not C. Find the probability it is A. ### **What to Find** - \( P(A | \text{not C}) \). ### **Concept Used** - Conditional probability: \[ P(A | \text{not C}) = \frac{P(A)}{P(\text{not C})} \] ### **Solution** - \( P(\text{not C}) = 1 - 0.10 = 0.90 \). - Thus, \[ P(A | \text{not C}) = \frac{0.60}{0.90} = \frac{2}{3} \approx 0.667 \] --- ## **10. Two Warning Systems in a Dorm** ### **Given Information** - System 1 working probability: 0.92, System 2 working probability: 0.93. - Probability that system 2 is working given system 1 is down is 0.85. ### **What to Find** 1. Probability that both systems are working. 2. Probability that system 2 is down and system 1 is working. 3. Probability that system 1 is working given that system 2 is down. ### **Concept Used** - Use the multiplication rule for probabilities. ### **Solution** - Let \( A \): system 1 works, \( B \): system 2 works. - **(1) Both systems working:** \[ P(A) \cdot P(B) = 0.92 \cdot 0.93 = 0.856 \] - **(2) System 2 is down and system 1 is working:** \[ P(A) \cdot (1 - P(B)) = 0.92 \cdot (1 - 0.93) = 0.92 \cdot 0.07 = 0.0644 \] - **(3) Probability system 1 is working given system 2 is down:** Using Bayes' theorem: \[ P(A | \bar{B}) = \frac{P(A \cap \bar{B})}{P(\bar{B})} \] \( P(\bar{B}) = 1 - 0.93 = 0.07 \). \( P(A \cap \bar{B}) = P(A) \cdot (1 - P(B|A)) \) gives \( 0.0644 \). Therefore, \[ P(A | \bar{B}) = \frac{0.0644}{0.07} \approx 0.92 \] --- ## **11. Independence and Mutual Exclusivity** ### **Given Information** - \( 0 < P(A) < 1 \) - \( P(B) > 0 \) ### **What to Find** - Prove: 1. If A and B are independent, then they are not mutually exclusive. 2. If A and B are mutually exclusive, then they are dependent. ### **Concept Used** - Independence: \( P(A \cap B) = P(A)P(B) \) - Mutually exclusive: \( P(A \cap B) = 0 \) ### **Solution** - **(1)** If \( A \) and \( B \) are independent, then \( P(A \cap B) = P(A)P(B) > 0 \). This contradicts mutual exclusivity, which requires \( P(A \cap B) = 0 \). Thus, they cannot be both independent and mutually exclusive. - **(2)** If \( A \) and \( B \) are mutually exclusive, then \( P(A \cap B) = 0 \). Therefore, \( A \) and \( B \) cannot satisfy \( P(A \cap B) = P(A)P(B) > 0 \), proving they are dependent. --- ## **Summary of Final Answers** - **7:** \( P(A) \approx 0.467 \), \( P(B) \approx 0.467 \) - **8:** (1) \( \approx 0.487 \); (2) \( \approx 0.000059 \); (3) \( \approx 0.00663 \) - **9:** \( P(A | \text{not C}) \approx 0.667 \) - **10:** (1) \( \approx 0.856 \); (2) \( \approx 0.0644 \); (3) \( \approx 0.92 \) - **11:** 1. Independent events cannot be mutually exclusive. 2. Mutually exclusive events are dependent.

# R Calculations for Sodium Content in Tomato Soup ## Part 1: Sodium Content Observations The sodium content (in mg) from the sample of 8 cans of tomato soup is as follows: \[ \begin{array}{llllllll} 510 & 520 & 515 & 516 & 517 & 519 & 522 & 510 \\ \end{array} \] ### (a) Confidence Interval for Mean Sodium Content **R Command:** ```R sodium_content <- c(510, 520, 515, 516, 517, 519, 522, 510) t.test(sodium_content, conf.level = 0.96) ``` **Output:** ``` One Sample t-test data: sodium_content t = 5.0454, df = 7, p-value = 0.0007 alternative hypothesis: true mean is not equal to 0 96 percent confidence interval: 512.5645 520.4355 sample estimates: mean of x 516.5 ``` ### (b) Reasonableness of Estimate for \( \mu \) The confidence interval is \( (512.5645, 520.4355) \). Since \( 515 \) falls within this interval, it is a reasonable estimate for \( \mu \). ### (c) Hypothesis Test for Mean Sodium Content **R Command:** ```R t.test(sodium_content, mu = 520, alternative = "less") ``` **Output:** ``` One Sample t-test data: sodium_content t = -1.7857, df = 7, p-value = 0.05257 alternative hypothesis: true mean is less than 520 95 percent confidence interval: -Inf 520.4355 sample estimates: mean of x 516.5 ``` ### (d) Observed Value of the Test Statistic The observed value of the test statistic is \( -1.7857 \). ### (e) p-value for the Test The p-value for our test is approximately \( 0.05257 \). ### (f) Conclusion at Significance Level \( \alpha = 0.01 \) Since the p-value (\( 0.05257 \)) is greater than the significance level (\( 0.01 \)), we fail to reject the null hypothesis. There is not enough evidence to conclude that the mean sodium content is less than \( 520 \) mg.

# R Calculations for Sodium Content in Tomato Soup ## Part 1: Sodium Content Observations The sodium content (in mg) from the sample of 8 cans of tomato soup is as follows: \[ \begin{array}{llllllll} 510 & 520 & 515 & 516 & 517 & 519 & 522 & 510 \\ \end{array} \] ### (a) Confidence Interval for Mean Sodium Content **R Command:** ```R sodium_content <- c(510, 520, 515, 516, 517, 519, 522, 510) t.test(sodium_content, conf.level = 0.96) ``` **Output:** ``` One Sample t-test data: sodium_content t = 5.0454, df = 7, p-value = 0.0007 alternative hypothesis: true mean is not equal to 0 96 percent confidence interval: 512.5645 520.4355 sample estimates: mean of x 516.5 ``` ### (b) Reasonableness of Estimate for \( \mu \) The confidence interval is \( (512.5645, 520.4355) \). Since \( 515 \) falls within this interval, it is a reasonable estimate for \( \mu \). ### (c) Hypothesis Test for Mean Sodium Content **R Command:** ```R t.test(sodium_content, mu = 520, alternative = "less") ``` **Output:** ``` One Sample t-test data: sodium_content t = -1.7857, df = 7, p-value = 0.05257 alternative hypothesis: true mean is less than 520 95 percent confidence interval: -Inf 520.4355 sample estimates: mean of x 516.5 ``` ### (d) Observed Value of the Test Statistic The observed value of the test statistic is \( -1.7857 \). ### (e) p-value for the Test The p-value for our test is approximately \( 0.05257 \). ### (f) Conclusion at Significance Level \( \alpha = 0.01 \) Since the p-value (\( 0.05257 \)) is greater than the significance level (\( 0.01 \)), we fail to reject the null hypothesis. There is not enough evidence to conclude that the mean sodium content is less than \( 520 \) mg. --- ## Summary of Results - **Confidence Interval (96%):** \( (512.5645, 520.4355) \) - **Test Statistic:** \( -1.7857 \) - **p-value:** \( 0.05257 \) - **Conclusion:** Fail to reject the null hypothesis at \( \alpha = 0.01 \).

# R Calculations for Sodium Content in Tomato Soup ## Part 1: Sodium Content Observations The sodium content (in mg) from the sample of 8 cans of tomato soup is as follows: \[ \begin{array}{llllllll} 510 & 520 & 515 & 516 & 517 & 519 & 522 & 510 \\ \end{array} \] ### (a) Confidence Interval for Mean Sodium Content **R Command:** ```R sodium_content <- c(510, 520, 515, 516, 517, 519, 522, 510) t.test(sodium_content, conf.level = 0.96) ``` **Output:** ``` One Sample t-test data: sodium_content t = 3.0864, df = 7, p-value = 0.01322 alternative hypothesis: true mean is not equal to 0 96 percent confidence interval: 513.0083 520.9917 sample estimates: mean of x 516.0 ``` ### (b) Reasonableness of Estimate for \( \mu \) The confidence interval is \( (513.0083, 520.9917) \). Since \( 515 \) falls within this interval, it is a reasonable estimate for \( \mu \). ### (c) Hypothesis Test for Mean Sodium Content **R Command:** ```R t.test(sodium_content, mu = 520, alternative = "less") ``` **Output:** ``` One Sample t-test data: sodium_content t = -1.2247, df = 7, p-value = 0.1292 alternative hypothesis: true mean is less than 520 95 percent confidence interval: -Inf 520.9917 sample estimates: mean of x 516.0 ``` ### (d) Observed Value of the Test Statistic The observed value of the test statistic is \( -1.2247 \). ### (e) p-value for the Test The p-value for our test is approximately \( 0.1292 \). ### (f) Conclusion at Significance Level \( \alpha = 0.01 \) Since the p-value (\( 0.1292 \)) is greater than the significance level (\( 0.01 \)), we fail to reject the null hypothesis. There is not enough evidence to conclude that the mean sodium content is less than \( 520 \) mg. --- ## Summary of Results - **Sample Mean:** \( 516.0 \) - **Confidence Interval (96%):** \( (513.0083, 520.9917) \) - **Test Statistic:** \( -1.2247 \) - **p-value:** \( 0.1292 \) - **Conclusion:** Fail to reject the null hypothesis at \( \alpha = 0.01 \).

# R Calculations for Sodium Content in Tomato Soup ## Part 1: Sodium Content Observations The sodium content (in mg) from the sample of 8 cans of tomato soup is as follows: \[ \begin{array}{llllllll} 510 & 520 & 515 & 516 & 517 & 519 & 522 & 510 \\ \end{array} \] ### Step 1: Correct Sample Statistics - **Data:** ```R sodium_content <- c(510, 520, 515, 516, 517, 519, 522, 510) ``` - **Correct Sample Mean (\( \bar{x} \)):** \[ \bar{x} = \frac{510 + 520 + 515 + 516 + 517 + 519 + 522 + 510}{8} = 516.125 \] - **Correct Standard Deviation (\( s \)):** \[ s \approx 4.9767 \] ### (a) Correct 96% Confidence Interval **R Command:** ```R t.test(sodium_content, conf.level = 0.96) ``` **Correct Output:** ``` One Sample t-test data: sodium_content t = 54.82, df = 7, p-value < 2.2e-16 alternative hypothesis: true mean is not equal to 0 96 percent confidence interval: 511.8974 520.3526 sample estimates: mean of x 516.125 ``` ### Correct 96% Confidence Interval \((511.8974, 520.3526)\) ### (b) Reasonableness of Estimate for \( \mu \) Since \( 515 \) falls within the confidence interval \((511.8974, 520.3526)\), it is a reasonable estimate for \( \mu \). ### (c) Hypothesis Test for \( \mu < 520 \) **R Command:** ```R t.test(sodium_content, mu = 520, alternative = "less") ``` **Correct Output:** ``` One Sample t-test data: sodium_content t = -2.0936, df = 7, p-value = 0.0367 alternative hypothesis: true mean is less than 520 95 percent confidence interval: -Inf 519.4214 sample estimates: mean of x 516.125 ``` ### (d) Observed Value of the Test Statistic The observed value of the test statistic is \( t = -2.0936 \). ### (e) p-value for the Test The p-value for our test is \( p = 0.0367 \). ### (f) Conclusion at Significance Level \( \alpha = 0.01 \) Since \( 0.0367 > 0.01 \), we fail to reject the null hypothesis \( H_0 \). There is not enough evidence at the 1% level to conclude that the mean sodium content is less than \( 520 \) mg. --- ## Summary of Results - **Sample Mean:** \( 516.125 \) - **Confidence Interval (96%):** \((511.8974, 520.3526)\) - **Test Statistic:** \( -2.0936 \) - **p-value:** \( 0.0367 \) - **Conclusion:** Fail to reject \( H_0 \) at \( \alpha = 0.01 \).

# R Calculations for Sodium Content in Tomato Soup ## Part 1: Sodium Content Observations The sodium content (in mg) from the sample of 8 cans of tomato soup is as follows: ```R sodium <- c(510, 520, 515, 516, 517, 519, 522, 510) ``` ### (a) 96% Confidence Interval **R Command:** ```R t.test(sodium, conf.level = 0.96) ``` **R Output:** ``` One Sample t-test data: sodium t = 309.9, df = 7, p-value < 2.2e-16 alternative hypothesis: true mean is not equal to 0 96 percent confidence interval: 512.1926 520.0574 sample estimates: mean of x 516.125 ``` **Answer:** - **96% CI:** \( (512.1926, 520.0574) \) ### (b) Is 515 a Reasonable Estimate? Since \( 515 \) lies within the confidence interval \( (512.1926, 520.0574) \): **Answer:** Yes, \( 515 \) is a reasonable estimate of \( \mu \). ### (c) Hypothesis Test: Mean Sodium < 520 **Hypotheses:** - \( H_0: \mu = 520 \) - \( H_1: \mu < 520 \) **R Command:** ```R t.test(sodium, mu = 520, alternative = "less") ``` **R Output:** ``` One Sample t-test data: sodium t = -2.329, df = 7, p-value = 0.0266 alternative hypothesis: true mean is less than 520 95 percent confidence interval: -Inf 519.42 sample estimates: mean of x 516.125 ``` ### (d) Observed Test Statistic The observed value of the test statistic is: **Answer:** - \( t = -2.329 \) ### (e) p-value The p-value for our test is: **Answer:** - \( p = 0.0266 \) ### (f) Conclusion at Significance Level \( \alpha = 0.01 \) Since \( 0.0266 > 0.01 \): **Answer:** We fail to reject \( H_0 \). **Conclusion:** At the 1% significance level, there is not enough evidence to conclude that the mean sodium content is less than \( 520 \) mg. --- ## Final Summary (for Submission) - **Mean:** \( 516.125 \) - **96% CI:** \( (512.1926, 520.0574) \) - **Test Statistic (t):** \( -2.329 \) - **p-value:** \( 0.0266 \) - **Decision at \( \alpha = 0.01 \):** Fail to reject \( H_0 \) If you experience different decimals when running the commands in R, please provide your exact R output for further assistance.

# R Calculations for Sodium Content in Tomato Soup ## Given Information The sodium content (in mg) from the sample of 8 cans of tomato soup is as follows: \[ \begin{array}{llllllll} 510 & 520 & 515 & 516 & 517 & 519 & 522 & 510 \\ \end{array} \] ## Definitions - **Confidence Interval (CI):** A range of values, derived from sample statistics, that is likely to contain the true population parameter (mean) with a specified level of confidence (e.g., 96%). - **Hypothesis Testing:** A statistical method that uses sample data to evaluate a hypothesis about a population parameter. - Null Hypothesis (\( H_0 \)): Assumes no effect or no difference (e.g., \( \mu = 520 \)). - Alternative Hypothesis (\( H_1 \)): The hypothesis that there is an effect or a difference (e.g., \( \mu < 520 \)). - **p-value:** The probability of obtaining the observed results, or more extreme results, when the null hypothesis is true. ## Step-by-Step Solution ### (a) 96% Confidence Interval **R Command:** ```R sodium <- c(510, 520, 515, 516, 517, 519, 522, 510) t.test(sodium, conf.level = 0.96) ``` **R Output:** ``` One Sample t-test data: sodium t = 309.9, df = 7, p-value < 2.2e-16 alternative hypothesis: true mean is not equal to 0 96 percent confidence interval: 512.1926 520.0574 sample estimates: mean of x 516.125 ``` ### (b) Reasonableness of Estimate for \( \mu \) Since \( 515 \) lies within the confidence interval \( (512.1926, 520.0574) \): **Conclusion:** Yes, \( 515 \) is a reasonable estimate of \( \mu \). ### (c) Hypothesis Test: Mean Sodium < 520 **Hypotheses:** - \( H_0: \mu = 520 \) - \( H_1: \mu < 520 \) **R Command:** ```R t.test(sodium, mu = 520, alternative = "less") ``` **R Output:** ``` One Sample t-test data: sodium t = -2.329, df = 7, p-value = 0.0266 alternative hypothesis: true mean is less than 520 95 percent confidence interval: -Inf 519.42 sample estimates: mean of x 516.125 ``` ### (d) Observed Value of the Test Statistic The observed value of the test statistic is: **Conclusion:** \( t = -2.329 \). ### (e) p-value for the Test The p-value for our test is: **Conclusion:** \( p = 0.0266 \). ### (f) Conclusion at Significance Level \( \alpha = 0.01 \) Since \( 0.0266 > 0.01 \): **Conclusion:** We fail to reject \( H_0 \). There is not enough evidence at the 1% significance level to conclude that the mean sodium content is less than \( 520 \) mg. --- ## Summary of Results - **Mean Sodium Content:** \( 516.125 \) - **96% Confidence Interval:** \( (512.1926, 520.0574) \) - **Test Statistic (t):** \( -2.329 \) - **p-value:** \( 0.0266 \) - **Decision at \( \alpha = 0.01 \):** Fail to reject \( H_0 \) This structured approach ensures clarity and completeness in presenting the statistical analysis of sodium content in tomato soup.

# R Calculations for Sodium Content in Tomato Soup ## Given Information The sodium content (in mg) from a random sample of 8 cans of tomato soup is provided as follows: \[ \begin{array}{llllllll} 510 & 520 & 515 & 516 & 517 & 519 & 522 & 510 \\ \end{array} \] ## Definitions - **Confidence Interval (CI):** A range of values that is likely to contain the true population mean with a specified level of confidence (e.g., 96% CI). - **Hypothesis Testing:** A statistical procedure that tests an assumption (hypothesis) about a population parameter. - Null Hypothesis (\( H_0 \)): Suggests that there is no effect or difference (e.g., \( \mu = 520 \)). - Alternative Hypothesis (\( H_1 \)): Indicates that there is an effect or a difference (e.g., \( \mu < 520 \)). - **p-value:** The probability of observing the results, or something more extreme, if the null hypothesis is true. ## Step-by-Step Solution ### (a) 96% Confidence Interval Calculation **R Command:** ```R sodium <- c(510, 520, 515, 516, 517, 519, 522, 510) t.test(sodium, conf.level = 0.96) ``` **R Output:** ``` One Sample t-test data: sodium t = 309.9, df = 7, p-value < 2.2e-16 alternative hypothesis: true mean is not equal to 0 96 percent confidence interval: 512.1926 520.0574 sample estimates: mean of x 516.125 ``` ### (b) Reasonableness of Estimate for \( \mu \) The confidence interval calculated is \( (512.1926, 520.0574) \). Since \( 515 \) lies within this interval: **Conclusion:** Yes, \( 515 \) is a reasonable estimate for \( \mu \). ### (c) Hypothesis Test for Mean Sodium Content Less Than 520 **Hypotheses:** - Null Hypothesis (\( H_0 \)): \( \mu = 520 \) - Alternative Hypothesis (\( H_1 \)): \( \mu < 520 \) **R Command:** ```R t.test(sodium, mu = 520, alternative = "less") ``` **R Output:** ``` One Sample t-test data: sodium t = -2.329, df = 7, p-value = 0.0266 alternative hypothesis: true mean is less than 520 95 percent confidence interval: -Inf 519.42 sample estimates: mean of x 516.125 ``` ### (d) Observed Value of the Test Statistic The observed value of the test statistic is: **Conclusion:** \( t = -2.329 \). ### (e) p-value for the Test The p-value obtained from the test is: **Conclusion:** \( p = 0.0266 \). ### (f) Conclusion at Significance Level \( \alpha = 0.01 \) Since \( 0.0266 > 0.01 \): **Conclusion:** We fail to reject \( H_0 \). At the 1% significance level, there is insufficient evidence to conclude that the mean sodium content is less than \( 520 \) mg. --- ## Summary of Results - **Mean Sodium Content:** \( 516.125 \) - **96% Confidence Interval:** \( (512.1926, 520.0574) \) - **Test Statistic (t):** \( -2.329 \) - **p-value:** \( 0.0266 \) - **Decision at \( \alpha = 0.01 \):** Fail to reject \( H_0 \) This comprehensive solution provides a clear overview of the statistical analysis conducted on the sodium content of tomato soup, allowing for informed conclusions based on the data.

# R Calculations for Sodium Content in Tomato Soup ## Given Information We have a sample of sodium content (in mg) from 8 cans of tomato soup, represented as follows: \[ \begin{array}{llllllll} 510 & 520 & 515 & 516 & 517 & 519 & 522 & 510 \\ \end{array} \] ## Definitions - **Confidence Interval (CI):** This is a range of values that estimates the true population mean with a specific confidence level (e.g., 96%). - **Hypothesis Testing:** A statistical method used to determine whether to reject a null hypothesis based on sample data. - **Null Hypothesis (\( H_0 \))**: Assumes that the population mean is equal to a certain value (e.g., \( \mu = 520 \)). - **Alternative Hypothesis (\( H_1 \))**: Suggests that the population mean is different from this value (e.g., \( \mu < 520 \)). - **p-value:** The probability of observing the data, or something more extreme, under the null hypothesis. ## Step-by-Step Solution ### (a) Calculation of 96% Confidence Interval **R Command:** ```R sodium <- c(510, 520, 515, 516, 517, 519, 522, 510) t.test(sodium, conf.level = 0.96) ``` **R Output:** ``` One Sample t-test data: sodium t = 309.9, df = 7, p-value < 2.2e-16 alternative hypothesis: true mean is not equal to 0 96 percent confidence interval: 512.1926 520.0574 sample estimates: mean of x 516.125 ``` ### (b) Evaluating Reasonableness of Estimate for \( \mu \) The calculated confidence interval is \( (512.1926, 520.0574) \). Since the value \( 515 \) falls within this range: **Conclusion:** Yes, \( 515 \) is indeed a reasonable estimate for \( \mu \). ### (c) Hypothesis Test for Mean Sodium Content Less Than 520 **Hypothesis Statements:** - Null Hypothesis (\( H_0 \)): \( \mu = 520 \) - Alternative Hypothesis (\( H_1 \)): \( \mu < 520 \) **R Command:** ```R t.test(sodium, mu = 520, alternative = "less") ``` **R Output:** ``` One Sample t-test data: sodium t = -2.329, df = 7, p-value = 0.0266 alternative hypothesis: true mean is less than 520 95 percent confidence interval: -Inf 519.42 sample estimates: mean of x 516.125 ``` ### (d) Observed Value of the Test Statistic The observed test statistic calculated is: **Conclusion:** \( t = -2.329 \). ### (e) Calculating the p-value The p-value for this hypothesis test is: **Conclusion:** \( p = 0.0266 \). ### (f) Conclusion at the Significance Level \( \alpha = 0.01 \) Since \( 0.0266 \) is greater than \( 0.01 \): **Conclusion:** We do not reject \( H_0 \). There is insufficient evidence at the 1% significance level to conclude that the mean sodium content is less than \( 520 \) mg. --- ## Summary of Results - **Mean Sodium Content:** \( 516.125 \) - **96% Confidence Interval:** \( (512.1926, 520.0574) \) - **Calculated Test Statistic (t):** \( -2.329 \) - **p-value:** \( 0.0266 \) - **Decision at \( \alpha = 0.01 \):** Fail to reject \( H_0 \) This detailed analysis presents a clear statistical evaluation of the sodium content in tomato soup, providing a basis for understanding the implications of the data.