Question 1 Calcium is essential to tree growth. In 1990, the concentration of calcium in precipitation in a city in New York, was 0.11 milligram per liter (mg/L). A random sample of 10 precipitation dates in 2014 results in the following data: .065 .087 .070 .262 .126 .183 .120 .234 .313 .108 A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot does not show any outliers. Does the sample evidence suggest that calcium concentrations have changed since 1990? Use a α = .05 level of significance. Answer Calcium is an essential element for the growth and development of trees, as it plays a crucial role in maintaining the structure and function of their cells. The concentration of calcium in precipitation, or rainwater, is an important factor to consider as it directly affects the amount of calcium available to trees for their growth and overall health. According to data from a city in New York, the concentration of calcium in precipitation was 0.11 mg/L in 1990. Recently, in 2014, a random sample of 10 precipitation dates resulted in the following data: 0.065 mg/L, 0.087 mg/L, 0.070 mg/L, 0.262 mg/L, 0.126 mg/L, 0.183 mg/L, 0.120 mg/L, 0.234 mg/L, 0.313 mg/L, and 0.108 mg/L. A normal probability plot of the data suggests that it could come from a population that is normally distributed. Additionally, a box plot of the data does not show any outliers. To determine if there has been a change in calcium concentrations since 1990, we can conduct a hypothesis test with a significance level of α = .05. The null hypothesis is that there has been no change in the calcium concentrations, while the alternative hypothesis is that there has been a change in the concentrations. To conduct this test, we can use a one-sample t-test since we are comparing the sample data to a known population mean from 1990. Using a calculator or statistical software, we can calculate the t-score to be -2.46 and the p-value to be 0.0353. This p-value is less than our significance level, which means we can reject the null hypothesis. In other words, there is evidence to suggest that there has been a change in the calcium concentrations since 1990. Question 2 Nexium is a drug that can be used to reduce the acid produced by the body and heal damage to the esophagus due to acid reflux. The manufacturer of Nexium claims that more than 94% of patients taking Nexium are healed within 8 weeks. In clinical trials, 213 of 224 patients suffering from acid reflux disease were healed after 8 weeks. Conduct a hypothesis test to test the manufacturer’s claim at the α = .01 level of significance. Answer Nexium, also known by its generic name esomeprazole, is a widely prescribed proton pump inhibitor (PPI) that is used to reduce the production of stomach acid and treat acid reflux disease. According to the manufacturer, AstraZeneca, more than 94% of patients who take Nexium are healed within 8 weeks. This claim has been supported by clinical trials in which 213 out of 224 acid reflux patients were reportedly healed after 8 weeks of taking Nexium. However, as a conscientious college student, it is important to critically evaluate the manufacturer's claim and conduct a hypothesis test to determine its validity. Hypothesis testing is a statistical procedure used to determine if a claim or hypothesis about a population is supported by sample evidence. In this case, the manufacturer's claim is that more than 94% of patients taking Nexium are healed within 8 weeks. This will be our alternative hypothesis, denoted by H1. The null hypothesis, denoted by H0, is the opposite of the alternative hypothesis and represents the belief that there is no difference between the sample and the population. To conduct a hypothesis test, we need to first determine the level of significance. In this case, the α level is set at .01, which means that there is only a 1% chance of rejecting the null hypothesis when it is actually true. Next, we need to choose an appropriate test statistic. Since we are interested in determining whether the proportion of healed patients (p) is significantly different from the manufacturer's claim (p0), we will use a one-sample proportion z-test. The formula for the z-test statistic is given by: z= (p- p0)/√(p0(1-p0)/n) where p is the proportion in our sample (213/224= 0.95) and n is the sample size (224). p0 is the hypothesized proportion (0.94). Substituting the values in the formula, we get: z= (0.95-0.94)/√((0.94)(1-0.94)/224) z= 1.126 Next, we need to determine the critical value (zα) at α = 0.01. From the z-table, we find that z0.01= 2.33. Since our test statistic (1.126) is less than the critical value (2.33), we fail to reject the null hypothesis. This means that there is not enough evidence to support the manufacturer's claim that more than 94% of patients taking Nexium are healed within 8 weeks. In order to further validate our findings, we can calculate the p-value, which is the probability of obtaining a test statistic as extreme as or more extreme than the one observed, assuming the null hypothesis is true. In this case, the p-value equals 0.13. This suggests that there is a 13% chance of observing a test statistic as or more extreme than 1.126 if the null hypothesis is true. Since this value is greater than the chosen level of significance (0.01), we cannot reject the null hypothesis. It is important to note that the results of this hypothesis test do not necessarily mean that Nexium is not an effective treatment for acid reflux disease. Other factors such as individual patient characteristics and compliance with medication may also play a role in the healing process. In conclusion, as a responsible and critical thinking student, it is important to question the validity of claims made by manufacturers and support them with evidence. The hypothesis test conducted in this answer provides a quantitative approach to determine the significance of the manufacturer's claim about the healing rate of Nexium. For further reading on hypothesis testing, I recommend consulting reputable sources such as the textbook "Statistics for Business and Economics" by James T. McClave. Question 3 Seat belt use in a state was estimated at 96%, which means 96% of people use their seat belts. Suppose two independent drivers have been randomly selected. a. What is the probability that both of them are using a seatbelt? b. What is the probability that neither of them is using a seatbelt? c. What is the probability that at least one is using a seatbelt? Answer a. The probability that both drivers are using a seatbelt can be calculated using the formula for independent events, which states that the probability of two independent events occurring together is equal to the product of their individual probabilities. In this case, the probability of the first driver using a seatbelt is 96%, and the probability of the second driver using a seatbelt is also 96%. Therefore, the probability of both drivers using a seatbelt is 96% x 96% = 0.96 x 0.96 = 0.9216, or 92.16%. b. The probability that neither of the drivers is using a seatbelt can be calculated by taking the complement of the probability that at least one of them is using a seatbelt. This can be expressed as P(neither) = 1 - P(at least one). Using the same formula as before, the probability of at least one driver using a seatbelt is 1 - (0.96 x 0.96) = 1 - 0.9216 = 0.0784, or 7.84%. Therefore, the probability of neither driver using a seatbelt is 7.84%. c. The probability that at least one driver is using a seatbelt is 1 - P(neither) = 1 - 0.0784 = 0.9216, or 92.16%. This is the same as the probability calculated in part a, as the events are mutually exclusive (i.e. if one driver is using a seatbelt, the other driver cannot also be using a seatbelt). Question 4 Consider a basketball player that is a 79% free-throw shooter. Assume that free throw shots are independent. Suppose, over the course of a season, the player attempts 550 free throws. Complete parts a and b below. b. Would it be surprising if the player made 453 of his free throws? Why or why not? Since 453 is than 2 from the it be surprising. Answer According to the given information, the basketball player has a 79% accuracy rate in making free throw shots and attempts 550 free throws over the course of a season. Based on this, the expected number of successful free throws would be 550 x 0.79 = 434.5. Therefore, it would not be surprising if the player made 453 of his free throws as this is only a slight deviation from the expected value. To further understand the likelihood of this occurrence, we can calculate the standard deviation of the number of successful free throws using the formula σ= √(np(1-p)), where n is the number of trials and p is the probability of success. In this case, the standard deviation would be equal to √(550x0.79x0.21) = 9.05. This means that there is approximately a 68% chance that the player would make between 425.5 (434.5-9.05) and 443.5 (434.5+9.05) free throws, which includes the value of 453. Therefore, it is not considered a surprising outcome. Question 5 Consider a basketball player that is a 79% free-throw shooter. Assume that free throw shots are independent. Suppose, over the course of a season, the player attempts 550 free throws. Complete parts a and b below. Question content area bottom Part 1 a. Find the mean and the standard deviation for the expected number of free throws we expect the player to make. The mean number of free throws we expect the player to make is Answer According to statistical analysis, a 79% free-throw shooter is considered to be a highperforming basketball player. Therefore, over the course of a season, we can expect this player to make a significant number of free throws. In this scenario, the player is expected to make 79% of 550 free throw attempts. This results in an expected number of 434.5 successful free throws. To find the mean, we simply multiply the probability of making a free throw (79%) by the total number of attempts (550). This gives us a mean of 434.5 successful free throws. To calculate the standard deviation, we can use the formula: √npq, where n is the number of trials (550), p is the probability of success (0.79), and q is the probability of failure (1 p). Plugging these values into the formula, we get a standard deviation of √(550 x 0.79 x 0.21) = 10.37 successful free throws. Therefore, we can expect the player to make an average of 435 successful free throws with a standard deviation of 10.37 over the course of the season. It is important to note that these calculations are based on the assumption that free throw shots are independent. This means that the outcome of one free throw does not affect the outcome of the next. However, in reality, players may experience streaks of successful or unsuccessful shots, affecting the overall mean and standard deviation. In conclusion, the mean and standard deviation for the expected number of successful free throws for a 79% free-throw shooter attempting 550 free throws can be determined by multiplying the probability of success by the number of attempts and calculating the square root of npq, respectively.