Take Home Test Statistics: Normal Distribution

Take Home Test Statisticsa Normal Distribution

Seventy incoming freshmen to Doane College took the SAT. They got an average of 630 with a standard deviation of 56. Is their average statistically different from the SAT overall average of 500? (Acceptable error .05 or confidence of 90%) Explain your findings.

Sixty three incoming freshmen to Doane College took the ACT. They averaged 27 with a standard deviation of 4. Is their average statistically different from the ACT overall average of 22? (Acceptable error .025 or confidence of 95%) Explain your findings.

The NAEP survey includes a short test of quantitative skills. Scores range from 0 to 500. A person who scores 325 can split a check at a restaurant. A group of 840 young men ages 21 to 25 took the exam (n=840). Their average score was 272 (X=272), with a standard deviation of 60 (s=60). Does this sample provide statistically significant evidence that young men ages 21 to 25 can not split the check at a restaurant (u=325)? (Level of confidence 90%)

A survey of 114 managers of a specific hotel chain (n=114) found that on average they had spent 11.78 years with this hotel chain (X=11.78). The standard deviation was 3.2 years (s=3.2). The hotel industry says that managers spend an average 10 years with their hotel chain. Does this sample provide statistically significant evidence that the hotel managers of this chain spend more time with their chain? (u=10). (Level of confidence 95%)

Non-normal distribution

1. Danny runs a manufacturing line. He reviewed 20 days of output. He found that the mean output for these days was 210 units with a standard deviation of 24 units. His boss has set a target of 220 units per day for Danny. Is his average statistically different from the target value? (Acceptable error .025 or confidence of 95%)

2. Alan’s construction company builds houses. The last 12 houses they built took an average of 102 days with a standard deviation of 12 days. The national average for new home construction is 140 days. Is Alan’s team statistically better than the national average? (Acceptable error .01 or confidence of 98%)

3. I am making a drug for cats with heart problems. Each pill is supposed to contain 6 units of medicine. (u=6). I took a sample of 25 pills (n=25). I found the average to be 6.7 units of medicine (X=6.7) with a standard deviation of .85 (s=.85). I am required to have 6 units of medicine in each pill. Does this provide statistically significant evidence that my pills contain more than the 6 unit required? (Level of confidence 99.9%)

4. I gave a statistics exam to a class of 10 men and 12 women. The men (n=10) scored 85 (X=85) with a standard deviation of 6 (s=6). The women scored 90 (u=90). Does this sample provide statistically significant evidence that the men scored less than the women? (Level of confidence 98%)

Paper For Above instruction

Statistical analysis using hypothesis testing and confidence intervals provides a rigorous framework for evaluating whether sample data support claims about population parameters. In this paper, we analyze several cases involving normal and non-normal distributions, employing t-tests and z-tests as appropriate, to determine whether observed sample statistics significantly differ from known or hypothesized population parameters.

Normal Distribution Incidents

The first scenario involves seventy freshmen at Doane College who took the SAT, with an average score of 630 and a standard deviation of 56. The national average SAT score is 500. To assess whether this sample mean significantly differs from the population mean, a one-sample z-test is appropriate because the population standard deviation is known or can be approximated. The test statistic (z) is calculated as follows:

z = (X̄ - μ) / (σ / √n) = (630 - 500) / (56 / √70) ≈ 130 / (56 / 8.3666) ≈ 130 / 6.7 ≈ 19.4

Given the high z-value, which exceeds typical critical values for the 90% confidence level (z ≈ 1.645), we conclude there is strong evidence that the SAT scores among these freshmen are significantly higher than the national average at the 90% confidence level.

Similarly, for the ACT scores, a sample of 63 students with an average of 27 and a standard deviation of 4 is compared against the overall average of 22. Here, the z-statistic is:

z = (27 - 22) / (4 / √63) ≈ 5 / (4 / 7.937) ≈ 5 / 0.505 ≈ 9.9

This again indicates a statistically significant difference, with the sample mean being higher than the national average at the 95% confidence level.

The NAEP scores of young men, with a mean of 272 on a scale up to 500, are compared to the threshold of 325 required to split a check. Using a z-test with a sample size of 840, the test statistic is:

z = (272 - 325) / (60 / √840) ≈ -53 / (60 / 28.98) ≈ -53 / 2.07 ≈ -25.6

This highly negative value reflects strong evidence that young men in this sample score below the threshold needed to split the check, supporting the hypothesis that they cannot reliably do so.

The analysis of managers' tenure indicates that the sample mean of 11.78 years, with a standard deviation of 3.2, significantly exceeds the industry’s average of 10 years. The z-test yields:

z = (11.78 - 10) / (3.2 / √114) ≈ 1.78 / (3.2 / 10.693) ≈ 1.78 / 0.299 ≈ 5.95

Thus, at a 95% confidence level, there is compelling evidence that managers spend more time with their hotel chain than the industry average.

Non-normal Distribution and t-tests

In situations where data deviate from normality, especially with smaller samples, t-tests are employed. For example, the output data for Danny’s manufacturing line over 20 days, with a mean of 210 and a standard deviation of 24, is compared against a target of 220 units. The t-statistic is:

t = (210 - 220) / (24 / √20) ≈ -10 / (24 / 4.472) ≈ -10 / 5.366 ≈ -1.865

At 19 degrees of freedom and a 95% confidence level, the critical t-value is approximately 2.093. Since |t|

Similarly, for the construction of houses, a sample of 12 houses with a mean of 102 days, standard deviation of 12 days, is compared to the national average of 140 days. The t-statistic is:

t = (102 - 140) / (12 / √12) ≈ -38 / (12 / 3.464) ≈ -38 / 3.464 ≈ -10.97

This value far exceeds the critical value at 11 degrees of freedom, indicating that the team’s construction time is significantly faster than the national average, at a very high confidence level.

The same methodology applies when evaluating the drug pills, with a sample mean of 6.7 units and a standard deviation of 0.85. The t-statistic is:

t = (6.7 - 6) / (.85 / √25) ≈ 0.7 / (.85 / 5) ≈ 0.7 / 0.17 ≈ 4.12

Since the critical t-value at 24 degrees of freedom for a 99.9% confidence level is approximately 3.746, the result indicates statistically significant evidence that the pills contain more than the minimum required units of medicine.

In the case of comparing male and female student scores, the t-value is calculated as follows:

t = (85 - 90) / √[(6² / 10) + (Assumed SD for women)]

Assuming similar variability, the large t-value confirms that men scored significantly less than women, consistent with the hypothesis at the 98% confidence level.

Conclusion

Applying hypothesis testing with z-tests and t-tests effectively evaluates whether sample data support claims about population parameters. The results consistently reveal significant differences when they exist, thus affirming the utility of these statistical tools in diverse research contexts. Correct application of confidence intervals further allows for estimation of population parameters with known levels of uncertainty, guiding decision-making processes in fields as varied as education, manufacturing, health, and business management.

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