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Which treatment group has more variable glucose levels? Which group has more consistent outcomes? Explain why comparing standard deviations is useful. Can the treatment effectiveness be determined from variability alone? Calculate the coefficient of variation for each group.

Mathematics

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Okay, Faustina, let's continue with the rest of the questions.

Question 7: Examination Scores

a) Both classes have the same average performance.

Both Class A and Class B have an average score of 65.\boxed{\text{Both Class A and Class B have an average score of 65.}}

b) Class A is more homogeneous because it has a smaller standard deviation.

Class A (Standard deviation = 5)\boxed{\text{Class A (Standard deviation = 5)}}

c) Class B has greater score dispersion because it has a larger standard deviation.

Class B (Standard deviation = 18)\boxed{\text{Class B (Standard deviation = 18)}}

d) The mean is the average score, while the standard deviation measures the spread of scores around that average. Two classes can have the same average score, but the individual scores in one class might be much more spread out (higher standard deviation) than in the other class (lower standard deviation).

The mean is the average, while standard deviation measures spread. Classes can have the same average but different levels of score variability around that average.\boxed{\text{The mean is the average, while standard deviation measures spread. Classes can have the same average but different levels of score variability around that average.}}

e) Class B is likely to have more extreme scores due to its larger standard deviation (18) and larger range (70), indicating greater variability.

Class B\boxed{\text{Class B}}

Question 8: Standardized Scores

a) Calculate the student's z-score.

  1. Use the z-score formula

    z=X−μσz = \frac{X - \mu}{\sigma}

    z=72−606z = \frac{72 - 60}{6}
  2. Calculate the z-score.

    z=126z = \frac{12}{6} z=2z = 2 2\boxed{2}

    b) A z-score of 2 means the student's score is 2 standard deviations above the mean.

    The student’s score is 2 standard deviations above the mean.\boxed{\text{The student's score is 2 standard deviations above the mean.}}

    c) Since the z-score is positive (2), the student's score is above the mean.

    Above the mean\boxed{\text{Above the mean}}

    d) A z-score of zero represents a score that is exactly equal to the mean.

    A z-score of zero means the score is exactly at the mean.\boxed{\text{A z-score of zero means the score is exactly at the mean.}}

    e) A z-score of 2 typically corresponds to approximately the 97.7th percentile in a normal distribution.

    Approximately the 97.7th percentile\boxed{\text{Approximately the 97.7th percentile}}

    Question 9: Physical Activity and Body Mass Index

    a) The correlation coefficient is -0.612, which indicates a negative direction.

    Negative\boxed{\text{Negative}}

    b) The absolute value of the correlation coefficient is 0.612, which indicates a moderate to strong correlation.

    Moderate to strong\boxed{\text{Moderate to strong}}

    c) The p-value (Sig. (2-tailed)) is <0.001180. Since this is less than 0.05, the relationship is statistically significant.

    Yes, the relationship is statistically significant (p < 0.001180).\boxed{\text{Yes, the relationship is statistically significant (p < 0.001180).}}

    d) There is a statistically significant, moderate to strong negative correlation between weekly exercise hours and BMI. This means that as weekly exercise hours increase, BMI tends to decrease.

    There is a statistically significant, moderate to strong negative correlation. As weekly exercise hours increase, BMI tends to decrease.\boxed{\text{There is a statistically significant, moderate to strong negative correlation. As weekly exercise hours increase, BMI tends to decrease.}}

    e) No, correlation does not imply causation. This result only shows an association between exercise and BMI; it does not prove that exercise causes lower BMI. Other factors could influence both exercise habits and BMI.

    No, correlation does not imply causation. This only shows an association, not a cause-and-effect relationship.\boxed{\text{No, correlation does not imply causation. This only shows an association, not a cause-and-effect relationship.}}

    f) Calculate the coefficient of determination (R2R^2).

  3. Square the Pearson correlation coefficient.

    R2=(−0.612)2R^2 = (-0.612)^2
  4. Calculate the value.

    R2=0.374544R^2 = 0.374544 0.375\boxed{0.375}

    Question 10: Stress and Sleep Duration

    a) The correlation coefficient is -0.354. This indicates a weak to moderate negative linear relationship between stress score and sleep duration.

    There is a weak to moderate negative linear relationship between stress score and sleep duration.\boxed{\text{There is a weak to moderate negative linear relationship between stress score and sleep duration.}}

    b) The p-value (Sig. (2-tailed)) is 0.018. Since 0.018<0.050.018 < 0.05, the relationship is statistically significant at the 5% level.

    Yes, the relationship is statistically significant (p = 0.018 < 0.05).\boxed{\text{Yes, the relationship is statistically significant (p = 0.018 < 0.05).}}

    c) The null hypothesis states that there is no linear relationship between stress score and sleep duration.

    H0:ρ=0\boxed{H_0: \rho = 0}

    d) Since the p-value (0.018) is less than the significance level (α=0.05\alpha = 0.05), we reject the null hypothesis.

    Reject the null hypothesis.\boxed{\text{Reject the null hypothesis.}}

    e) A negative correlation in this context means that as stress scores increase, sleep duration tends to decrease, and conversely, as stress scores decrease, sleep duration tends to increase.

    As stress scores increase, sleep duration tends to decrease, and vice versa.\boxed{\text{As stress scores increase, sleep duration tends to decrease, and vice versa.}}

    Question 11: Age and Systolic Blood Pressure

    a) The Pearson correlation is 0.793. This indicates a strong positive relationship.

    Strong positive relationship\boxed{\text{Strong positive relationship}}

    b) The p-value is <0.001. This means there is a very low probability of observing such a strong correlation by chance if no true relationship existed. The relationship is highly statistically significant.

    The p-value (<0.001) indicates a highly statistically significant relationship.\boxed{\text{The p-value (<0.001) indicates a highly statistically significant relationship.}}

    c) Calculate the proportion of variation shared by the variables (R2R^2).

  5. Square the Pearson correlation coefficient.

    R2=(0.793)2R^2 = (0.793)^2
  6. Calculate the value.

    R2=0.628849R^2 = 0.628849 0.629\boxed{0.629}

    d) No, age cannot be described as the sole cause of increased blood pressure. While there is a strong positive correlation, correlation does not imply causation. Many other factors can contribute to increased blood pressure, and age is likely one of several contributing variables.

    No, correlation does not imply causation. Other factors also contribute to blood pressure.\boxed{\text{No, correlation does not imply causation. Other factors also contribute to blood pressure.}}
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