What Is The Z-Score For A Score Of 79 When The Mean Score Of

What Is Thezscore For A Score Of 79 When The Mean Score Of All Persons

Calculate the z-score for a score of 79 when the mean score of all persons who complete a depression scale is 89 and the standard deviation is 5. Determine whether a person with a score of 79 is more or less depressed than most other people who complete the scale. Use Table 4.3 to find the following: (a) the area of the normal curve above a z-score of 1.71, (b) the area of the normal curve between the mean and a z-score of -1.34, (c) the z-score of the normal curve that marks the 38th percentile, (d) the z-scores that mark the upper and lower limits of the middle 42 percent of the normal curve. Additionally, analyze a case study involving two agencies requesting funds based on treatment success scores and discuss the appropriateness of statistical analysis to determine funding decisions, emphasizing the need for further information.

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The calculation of the z-score is fundamental in standardized testing and statistical analysis, providing a measure of how many standard deviations a particular score is from the mean. In this context, we will analyze a score of 79 on a depression scale, given a mean of 89 and a standard deviation of 5. The z-score is calculated using the formula:

z = (X - μ) / σ

where X is the individual score, μ is the mean, and σ is the standard deviation. Substituting the values, we get:

z = (79 - 89) / 5 = -10 / 5 = -2.0

This z-score of -2.0 indicates that the individual’s score is two standard deviations below the mean, suggesting that this person is less depressed than the average respondent, given that lower scores on this depression scale typically indicate fewer symptoms of depression. Since the score is significantly below the mean, the person is likely experiencing fewer depressive symptoms relative to most other individuals who completed the scale.

Using Table 4.3, which provides the cumulative probabilities associated with z-scores in the standard normal distribution, we find that the area above a z-score of 1.71 is approximately 0.0436 or 4.36%. This signifies that about 4.36% of the population has a z-score higher than 1.71, indicating higher levels of depression assuming higher scores reflect more depression.

For the area between the mean (z = 0) and a z-score of -1.34, the corresponding cumulative probability for z = -1.34 is approximately 0.0901. Since the area between the mean and this z-score is from 0 to -1.34, we multiply this probability by 2 to reflect symmetry when needed. However, because the distribution is symmetric, the area between z=0 and z=-1.34 is about 0.0901, so the total area from the mean to z=-1.34 is about 9.01%.

The z-score that marks the 38th percentile is found by locating the z-value corresponding to a cumulative probability of 0.38 in the standard normal distribution table. This z-value is approximately -0.31, indicating that 38% of scores fall below this point.

To identify the upper and lower limits that mark the middle 42% of the normal curve, we need to find the z-scores that correspond to the 29th percentile (100% - 42%)/2 = 29%) and the 71st percentile. Using the standard normal table, z-scores for the 29th percentile is approximately -0.55, and for the 71st percentile, approximately +0.55. These z-scores identify the bounds within which 42% of the data lies symmetrically around the mean.

In the case study involving two agencies, the decision to allocate funds hinges on the statistical analysis of their treatment success scores. Agency A reports a mean score of 42 with a sample size of 1,000, and Agency B reports a mean of 44 with the same sample size. The statistical comparison typically involves calculating a z-test for the difference between means:

z = (X̄₁ - X̄₂) / √[(σ₁² / n₁) + (σ₂² / n₂)]

Assuming equal variances and sample sizes, the p-value of less than 0.25 indicates that the observed difference is not statistically significant at conventional levels (e.g., 0.05). Thus, the evidence is insufficient to confidently favor one agency over the other solely based on their mean scores.

Therefore, using this statistical analysis alone would be inadequate for making a funding decision. Additional information, such as the variability of scores (standard deviations), effect sizes, program quality, and contextual factors, would be essential to inform a more comprehensive decision. For instance, knowing the standard deviations will help determine the variance and the reliability of the mean scores. Moreover, understanding the clinical significance of the difference in success scores is critical, as statistically significant differences may not translate into meaningful practical improvements.

In conclusion, while statistical analyses provide valuable insights into data, they should not be the sole criterion for resource allocation decisions. Supplementary information and qualitative assessments are necessary to ensure equitable and effective funding strategies that genuinely improve client outcomes.

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