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assessing Nomality Part 2 of 4 Points: 0 of 1 Save Use the given data values (a sample of female arm circumferences in centimeters) to identify the corresponding z scores that are used for a normal quantile plot, then identify the coordinates of each point in the normal quantile plot. Construct the normal quantile plot, then determine whether the data appear to be from a population with a normal distribution. \[ 323,44.9,40.5,34.2,38.1 \] List the \( z \) scores for the normal quantile plot. \[ -1.28,-0.52,0,0.52,1.28 \] (Round to two decimal places as needed. Use ascending order.) Identify the coordinanss of each point in the normal quantile plot. Use ordered pairs of the form ( \( \mathrm{x}, \mathrm{y} \) ), where x is the sorted arm circumferences, and y is the corresponding z score. \( \square \) \( \square \) \( \square \) \( \square \) ]. ]. \( \square \) (Type ordered pairs. Round to two decimal places as needed. Use ascending order.)

Ask by Reid Gibbs. in the United States
Mar 10,2025

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**Normal Quantile Plot Coordinates:** \[ (32.3, -1.28),\ (34.2, -0.52),\ (38.1, 0.00),\ (40.5, 0.52),\ (44.9, 1.28) \] **Conclusion:** The data points on the normal quantile plot appear to follow a straight line, indicating that the arm circumference measurements likely come from a normally distributed population.

Solución

**Step 1. Sort the Data** The given arm circumferences are: \[ 32.3,\ 44.9,\ 40.5,\ 34.2,\ 38.1 \] Sort these in ascending order: \[ 32.3,\ 34.2,\ 38.1,\ 40.5,\ 44.9 \] **Step 2. List the Corresponding \( z \) Scores** The \( z \) scores to be used for the normal quantile plot are (in ascending order): \[ -1.28,\ -0.52,\ 0.00,\ 0.52,\ 1.28 \] **Step 3. Identify the Coordinates for the Normal Quantile Plot** The normal quantile plot uses ordered pairs \((\text{data value}, \text{z-score})\). Pair the sorted data with the corresponding \( z \) scores: \[ (32.3,\ -1.28) \] \[ (34.2,\ -0.52) \] \[ (38.1,\ 0.00) \] \[ (40.5,\ 0.52) \] \[ (44.9,\ 1.28) \] **Step 4. Interpretation of the Normal Quantile Plot** After constructing the normal quantile plot with these points, if the points fall approximately along a straight line, it suggests that the data come from a population that is normally distributed. Based on these coordinates, if the plotted points appear linear, then the assumption of normality is reasonable.

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To construct the normal quantile plot with the given arm circumferences and their corresponding z-scores, we first sort the arm circumference data: Sorted arm circumferences: \[ 34.2, 38.1, 40.5, 44.9, 323 \] Matching them with the z-scores: - For \( 34.2 \): it's the lowest value, corresponding to the z-score of \( -1.28 \) - For \( 38.1 \): the second value, corresponding to the z-score of \( -0.52 \) - For \( 40.5 \): the third value, corresponding to the z-score of \( 0.00 \) - For \( 44.9 \): the fourth value, corresponding to the z-score of \( 0.52 \) - For \( 323 \): the highest value, corresponding to the z-score of \( 1.28 \) Thus, the coordinates for the normal quantile plot are: - \( (34.2, -1.28) \) - \( (38.1, -0.52) \) - \( (40.5, 0.00) \) - \( (44.9, 0.52) \) - \( (323, 1.28) \) Formatted as ordered pairs, they are: (34.2, -1.28), (38.1, -0.52), (40.5, 0.00), (44.9, 0.52), (323, 1.28) To determine if the data appears to be from a population with a normal distribution, you would typically look for a linear pattern in the normal quantile plot. If the points closely follow a straight line, then the data can be considered normally distributed. Any significant deviations from linearity would suggest that the data may not come from a normal distribution.

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Question 12(Mulliple Choice Warth 5 points) \[ (04.06 \mathrm{HC}) \] A researcher wants to test the claim that the proportion of juniors who watch television regularly is greater than the proportion of seniors who watch television regularly She finds that 56 of 70 randomly selected juniors and 47 of 85 randomly selected seniors report watching television regularly. Construct \( 95 \% \) confidence intervals for each population proportion. Which of the statemente gives the correct outcome of the research or's tert of the dalim? The \( 95 \% \) confidence interval for juniors is (706, 894), and the \( 95 \% \) confidence interval for seniors is ( 447,659 ). Since the intervals overlap, there is not enough evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors. The \( 95 \% \) confidence interval for juniors is (721, 879), and the \( 95 \% \) confidence interval for seniors is (464, 642). Since the interval for juniors is higher than the interval for seniors, there is evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors. The \( 95 \% \) confidence interval for juniors is ( 706,894 ), and the \( 95 \% \) confidence interval for seniors is ( 447,659 ). Since the interval for juniors is higher than the interval for seniors, there is evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors. The \( 95 \% \) confidence interval for juniors is ( \( 721, .879 \) ), and the \( 95 \% \) confidence interval for seniors is (464, 642). Since the intervals overlap, there is not enough evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors.

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