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Given the assumption that the data set follows a normal distribution, it is possible to estimate the number of chocolate covered strawberries needed for arbitrary sizes. The first step is to determine the mean and standard deviation of the samples. To aid in this, the following… Read TextPreview

- Subject: Miscellaneous
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- Level: Undergraduate
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- Author: wildermanbridge

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- Tags:
- Ann Rinaldi
- Average Student
- Bell Curve
- Flavor
- Statistics
- Statistics Experiment
- Yellow Wallpaper
- Yellow Wallpaper Freedom
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To test if a trend exists relating the GPA and the mother’s age at the birth of the child, regression analysis may be used. A simple linear regression is used in this particular calculation. The following table is set up from the raw data. It should be noted that the independent variable is the age at birth while the dependent is the GPA.

Since 0.632, H0 cannot be rejected. There is insufficient evidence to support that a relationship exists between the mother’s age at the time of the child’s birth and the corresponding intelligence (as measured by GPA).

The number of samples is determined based on the target error and confidence. Since no estimate of the proportion is given, a conservative value of 0.5 is to be used. The resulting calculations are shown as follows:

Evaluating the scores of standardized examinations revolve largely around the normal distribution. From this particular problem, the SASS scores differ from the raw scores provided. In order to normalize the values the raw scores must be converted into z-scores and applied to the mean and standard deviation of the SASS. The equivalent z-scores for 151, 140, and 136 are computed below:

Interpreting these scores as percentiles show that Robert, Ann and Carlos have percentiles at 90.32%, 77.34%, and 70.88% respectively. To normalize into the SASS standard scores, the z-scores are used with the mean and standard deviation standards. This is shown as follows:

Assuming that no correlation exists between the high school GPA and the SASS score, the percentage of the applicants that are denied admission can be determined from the z-score equivalent of 60 on the SASS:

From the z-score table, this corresponds to a cumulative probability of 0.7486 from the left. This means that 74.86% of the applicants have a SASS score below 60 and cannot be admitted into the university.

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Since 0.632, H0 cannot be rejected. There is insufficient evidence to support that a relationship exists between the mother’s age at the time of the child’s birth and the corresponding intelligence (as measured by GPA).

The number of samples is determined based on the target error and confidence. Since no estimate of the proportion is given, a conservative value of 0.5 is to be used. The resulting calculations are shown as follows:

Evaluating the scores of standardized examinations revolve largely around the normal distribution. From this particular problem, the SASS scores differ from the raw scores provided. In order to normalize the values the raw scores must be converted into z-scores and applied to the mean and standard deviation of the SASS. The equivalent z-scores for 151, 140, and 136 are computed below:

Interpreting these scores as percentiles show that Robert, Ann and Carlos have percentiles at 90.32%, 77.34%, and 70.88% respectively. To normalize into the SASS standard scores, the z-scores are used with the mean and standard deviation standards. This is shown as follows:

Assuming that no correlation exists between the high school GPA and the SASS score, the percentage of the applicants that are denied admission can be determined from the z-score equivalent of 60 on the SASS:

From the z-score table, this corresponds to a cumulative probability of 0.7486 from the left. This means that 74.86% of the applicants have a SASS score below 60 and cannot be admitted into the university.

This story is very much ...Download file to see next pagesRead More

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