Question 1a: Is A Golf Instructor Interested In Determining

Question1a Golf Instructor Is Interested In Determining If Her New Tec

Question 1a: A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She takes four new students and records their 18-hole scores before learning the new technique and then after having taken her class. She then conducts a hypothesis test at a significance level of 5%. The data are as follows. Player 1, Player 2, Player 3, Player 4. Mean score before class, mean score after class. What is the Null Hypotheses? a) The mean scores before the class and the mean scores after the class are the same. b) The mean scores before the class are greater than the mean scores after the class. Enter your answer by the appropriate identifying letter.

Use the following 4 (x, y) pairs to find the equation of the least squares regression line. round the first answer (constant) to 0 decimal places (integer) and the second answer (slope) to 1 decimal place.

Question points: Based upon material covered in class. An airline reports that its flights are on time with an average delay of 15 minutes with a variance of 150 based upon a study of 25 recent flights. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights was 22 minutes with a variance of 225. Based upon that data, was the disgruntled traveler correct in believing the variance of the delayed flights was greater than the reported 150 minutes squared? Evaluate the disgruntled traveler claim by performing a hypothesis test with a significance level of 0.05. Assume the data is normally distributed. What is the null and alternate hypothesis? a) Null hypothesis: The variance reported by the airline is not the same as the variance calculated by the disgruntled traveler. Alternate hypothesis: The variance calculated by the disgruntled traveler is the same as the variance reported by the airline. b) Null hypothesis: The variance reported by the airline is the same as the variance calculated by the disgruntled traveler. Alternate hypothesis: The variance calculated by the disgruntled traveler is not the same as the variance reported by the airline. c) Null hypothesis: The variance reported by the airline is the same as the variance calculated by the disgruntled traveler. Alternate hypothesis: The variance calculated by the disgruntled traveler is greater than the variance reported by the airline. d) Null hypothesis: The variance reported by the airline is the same as the variance calculated by the disgruntled traveler. Alternate hypothesis: The variance calculated by the disgruntled traveler is less than the variance reported by the airline. Enter the correct answer by selecting the appropriate letter.

Paper For Above instruction

The evaluation of hypotheses in statistical testing plays a crucial role in determining the validity of claims based on data. In the context of the golf instructor’s study, the primary question is whether her new teaching technique significantly improves her students’ golf scores. The null hypothesis (H₀) usually reflects no effect or no difference, serving as the default assumption unless evidence suggests otherwise.

For the first question concerning the golf instructor’s efficacy study, the null hypothesis asserts that there is no difference in the mean scores before and after the instructional technique. This is formalized as: “The mean scores before the class and the mean scores after the class are the same,” corresponding to option a. The alternative hypothesis would suggest that the technique has an effect, typically proposing that the mean scores after the class are lower (improved). The choice of null hypothesis influences the statistical test conducted, which in this case likely an average comparison such as a paired t-test due to the pretest-posttest design.

Moving to the second scenario involving the airline delays, the focus shifts to testing the variance, which reflects consistency in delays. The initial claim states that the airline’s flights are on average delayed by 15 minutes with a variance of 150. The traveler’s data show a mean delay of 22 minutes and a variance of 225 for 25 flights. The hypothesis test here evaluates whether the variance of the delays observed by the traveler exceeds the airline’s reported variance.

The null hypothesis (H₀) in this case posits that the variance claimed by the airline is accurate and matches the observed data. The alternative hypothesis (H₁) posits that the variance is greater than the airline’s claim, reflecting variability in delays that exceeds reported consistency. This corresponds to option c, which explicitly states that the null hypothesis is that the variances are equal, and the alternative claims a greater variance by the traveler.

Statistical tests for variances typically involve the chi-square distribution. The test compares the sample variance to the claimed population variance, calculating a chi-square statistic and referencing chi-square distribution tables for significance. Given the data, the traveler’s variance (225) is higher than the airline’s (150), and the larger the test statistic, the more evidence there is for greater variability.

In conclusion, hypothesis testing is fundamental across multiple disciplines to assess claims against collected data. Whether evaluating the effectiveness of a new golf teaching method or scrutinizing the consistency of airline delays, forming appropriate null and alternative hypotheses sets the framework for rigorous data analysis. Correctly specifying these hypotheses ensures the validity of conclusions drawn from statistical tests, aiding decision-making in real-world scenarios.

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