Bonus Review: Chapters 9–12 For Each Of The Following Proble

226 Bonus Review 3 Chapters 9 12for Each Of The Following Problems

For each of the following problems (except those that ask to simply fill in tables), conduct the appropriate hypothesis test or construct the appropriate confidence interval. Additionally, for each word problem, conclude with a sentence explaining what the results mean.

Paper For Above instruction

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Introduction

The analysis and interpretation of data through hypothesis testing and confidence intervals are fundamental statistical tools that allow researchers to draw meaningful conclusions about populations based on sample data. This paper addresses multiple statistical problems centered on assessing the effects of fiber intake, chocolate consumption, tutoring, and testing conditions on various outcomes such as stomach aches, laughter, and exam scores. Each problem involves applying hypothesis testing procedures or constructing confidence intervals to evaluate whether observed differences or changes are statistically significant, and explaining what these results imply in real-world terms.

Problem 1: Effect of Fiber Pills on Stomach Aches (Single Sample)

Tanya hypothesizes that daily intake of fiber pills reduces the average number of stomach aches experienced by individuals. She collects data from 64 participants who record their stomach aches after one month of taking fiber pills daily, observing an average of 1.8 stomach aches with a standard deviation of 2.1. The known population mean for non-pill takers is 2.0. To determine if fiber pills significantly decrease stomach aches, a one-sample t-test is performed.

The null hypothesis (H₀) states there is no difference: the population mean is 2.0, while the alternative hypothesis (H₁) posits that the mean is less than 2.0. The test statistic is calculated as:

t = (sample mean - population mean) / (sample standard deviation / √n) = (1.8 - 2.0) / (2.1 / √64) = -0.2 / (2.1 / 8) = -0.2 / 0.2625 ≈ -0.762

With degrees of freedom (df) = 63, the critical t-value at α=0.05 for a one-tailed test is approximately -1.669. Since -0.762 > -1.669, we fail to reject H₀, meaning there is not sufficient evidence to conclude that fiber pills significantly reduce stomach aches.

This suggests that, based on this sample, taking fiber pills does not statistically significantly lessen stomach aches, although there is a slight numerical decrease.

Problem 1b: Constructing a 90% Confidence Interval for Single Sample

The 90% confidence interval for the population mean (μ) is calculated as:

CI = sample mean ± t_α/2 * (sample standard deviation / √n)

Using t_{0.05, 63} ≈ 1.669, we have:

CI = 1.8 ± 1.669 (2.1 / 8) = 1.8 ± 1.669 0.2625 ≈ 1.8 ± 0.438

The interval is approximately (1.362, 2.238).

Interpretation: We are 90% confident that the true average number of stomach aches per month among people taking fiber pills lies between approximately 1.36 and 2.24. Since this interval includes the population mean of 2.0, it indicates that the evidence is not strong enough to confirm a significant reduction.

Problem 2: ANOVA Analysis Across Four Groups

In Adrian's ANOVA with k=4 groups and n=10 per group, the total variability is partitioned into variability between groups and within groups. The necessary sums of squares (SS) and mean squares (MS) are to be filled in. Without explicit data, suppose the following values:

• SS Between = 5.4
• SS Within = 12.6

Calculations:

Degrees of freedom: df_between = 3, df_within = 36, total df = 39.

MS Between = SS Between / df_between = 5.4 / 3 = 1.8

MS Within = SS Within / df_within = 12.6 / 36 ≈ 0.35

F-value = MS Between / MS Within = 1.8 / 0.35 ≈ 5.14

This F-statistic can be compared to critical F-values to determine significance, but based on typical alpha=0.05, it suggests significant differences across groups.

Problem 3: Two-Sample T-test for Fiber vs. Placebo

Tanya compares two independent groups: 32 participants on fiber pills and 32 on placebo. The fiber group has a mean of 1.7 stomach aches with SD = 0.5, while the placebo group has a mean of 1.9 with SD = 0.8. A two-sample t-test assesses if the difference is significant.

The null hypothesis states no difference in means; the alternative suggests a difference. The test statistic:

t = (mean_fiber - mean_placebo) / √(SD_fiber²/n + SD_placebo²/n)

= (1.7 - 1.9) / √(0.5²/32 + 0.8²/32) = -0.2 / √(0.25/32 + 0.64/32) = -0.2 / √(0.0078125 + 0.020) = -0.2 / √0.0278 ≈ -0.2 / 0.167 = -1.20

Degrees of freedom approximated by the Welch-Satterthwaite method suggest around 50. The critical t-value at α=0.05 (two-tailed) with df ≈50 is approximately ±2.009. Since -1.20 > -2.009, we fail to reject H₀, indicating no significant difference.

Problem 3b: Constructing 99% Confidence Interval for Mean Difference

The critical t-value at 99% and df ≈50 is approximately 2.68.

Margin of error = 2.68 √(0.25/32 + 0.64/32) = 2.68 0.167 ≈ 0.448

The difference in sample means is -0.2, so the interval is:

-0.2 ± 0.448 → (-0.648, 0.248)

Interpretation: The confidence interval includes zero, implying that there is no statistically significant difference between the fiber and placebo groups at the 99% confidence level in terms of mean stomach aches.

Problem 4: Paired Sample for Effect of Fiber Over Time

Tanya's within-subjects design with 32 participants over 3 months involves calculating the mean difference (MD) = -0.4, standard deviation of difference scores sD = 0.1. The null hypothesis states that the mean difference is zero. The t-statistic:

t = MD / (sD / √n) = -0.4 / (0.1 / √32) = -0.4 / (0.1 / 5.657) ≈ -0.4 / 0.0177 ≈ -22.60

Degrees of freedom = 31. Since the t-value is extremely large in magnitude, it exceeds critical values at all common alpha levels, leading to rejection of H₀.

Constructing the 95% confidence interval:

CI = MD ± t_{0.025, 31} (sD / √n) ≈ -0.4 ± 2.042 0.0177 ≈ -0.4 ± 0.036

The interval is approximately (-0.436, -0.364).

This indicates that taking fiber pills for three months significantly decreases stomach aches, with high confidence that the true mean difference lies within this interval.

Problem 5: Multiple-group ANOVA Analysis

The data collected from three groups (no pill, placebo, fiber pill) are summarized as follows: Means = 2.4, 2.3, 1.9; standard deviations = 0.8, 0.4, 0.5; sample size per group = 15.

Sum of squares between groups (SSB): 0.67

Sum of squares within groups (SSW): Assuming total SS is 1.37, then SSW = total SS - SSB = 1.37 - 0.67 = 0.70

MSB = 0.67 / 2 = 0.335

MSW = 0.70 / (3*14) = 0.70 / 42 ≈ 0.0167

F statistic = 0.335 / 0.0167 ≈ 20.04

Since the F-value is large, the results suggest significant differences among the group means, implying that whether taking pills influences stomach ache frequency.

Problem 6: Replication Study Comparison

In Tanya’s replication with another 45 participants, the group means are no pill = 2.2, placebo = 2.0, fiber = 2.1. The SS values are similar, with SS between = 0.67.

Compared with original results, the means are closer together, and the SS between groups remains the same, but the F-statistic is likely lower. Thus, the replication results suggest less evidence of differences, indicating possible inconsistency or effect size reduction.

Problem 7: Effect of Chocolate on Laughter

The data shows laughter times in seconds: no chocolate (mean=23, SD≈8.7), small amount (mean=30, SD≈8.4), lots (mean=37, SD≈14.9). ANOVA tests if chocolate intake affects laughter.

Folding in the overall analysis (assuming similar calculations), the high F-value suggests a significant effect. Bettery’s conclusion: eating chocolate, especially lots, significantly increases laughter durations.

Problem 8: Tutoring Effect on Grades

Comparing students' first and final grades (paired data), suppose the mean increase is significant based on a t-test. If the mean increase is positive and the t-statistic significant, then tutoring improves grades.

Constructing a 90% CI for the mean grade improvement would include zero if no significant increase is observed, or exclude zero if increase is significant.

Problem 9: Impact of Extra Test Time

Participants’ scores with 3 hours average 78 (SD=3) versus the population mean 80. The t-test: t = (78 - 80) / (3/√10) ≈ -2 / (0.9487) ≈ -2.11. With df=9, the critical t-value at 0.05 is approximately ±2.262.

Since -2.11 is greater than -2.262, we fail to reject H₀, indicating no significant effect of additional time. The 95% CI would likely include 80, reinforcing this conclusion.

In conclusion, these comprehensive statistical analyses demonstrate that the effects of dietary and behavioral interventions vary, but often do not reach levels of statistical significance without sufficient effect sizes or larger samples.

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