3.5 Answer Key Activity 3.5 Applied Statistics

Introduction

Applied statistics is the bridge between raw data and meaningful decisions, and Activity 3.5 is a cornerstone exercise that helps students practice core concepts such as descriptive measures, probability distributions, hypothesis testing, and regression analysis. This article presents a complete answer key for Activity 3.5, explains the reasoning behind each solution, and highlights the statistical principles that the activity reinforces. Whether you are a teacher preparing a grading rubric, a tutor checking student work, or a self‑studying learner, the detailed walkthrough below will clarify every step, reduce grading time, and deepen understanding of applied statistics.

Overview of Activity 3.5

Activity 3.5 typically appears in introductory textbooks on applied statistics (e.g., Applied Statistics for the Social Sciences).

Part	Topic	Typical Prompt
A	Descriptive statistics	Compute mean, median, mode, range, variance, and standard deviation for a given data set.
B	Probability & discrete distributions	Determine probabilities for a Binomial experiment and calculate expected value and variance.
C	Confidence interval for a population mean	Construct a 95 % confidence interval using the t‑distribution. That's why
D	Hypothesis testing (two‑sample t‑test)	Test whether two independent samples differ significantly at α = 0. 05.
E	Simple linear regression	Estimate the regression line, interpret the slope, and evaluate model fit (R²).

The data supplied in the original textbook are reproduced here for completeness.

Data Set for Parts A & B

Observation	Value
1	12
2	15
3	9
4	14
5	11
6	13
7	10
8	16
9	8
10	12

Binomial Scenario for Part B

A quality‑control inspector examines 20 units, each with a 0.25 probability of being defective.

Sample Statistics for Part C

Sample size (n) = 25
Sample mean ( (\bar{x}) ) = 78.4
Sample standard deviation (s) = 9.2

Two‑Sample Data for Part D

Group	n	(\bar{x})	s
A (Control)	30	82.5
B (Treatment)	28	86.Because of that, 1	7. 3

Regression Data for Part E

X (Hours studied)	Y (Test score)
2	68
4	75
6	80
8	85
10	92

Part A – Descriptive Statistics

1. Mean ( (\bar{x}) )

[ \bar{x} = \frac{\sum_{i=1}^{10} x_i}{10} = \frac{12+15+9+14+11+13+10+16+8+12}{10} = \frac{120}{10} = 12.0 ]

2. Median

Ordered data: 8, 9, 10, 11, 12, 12, 13, 14, 15, 16
With an even number of observations, the median is the average of the 5th and 6th values:

[ \text{Median} = \frac{12 + 12}{2} = 12 ]

3. Mode

The value 12 appears twice, more frequently than any other number → Mode = 12.

4. Range

[ \text{Range} = \max(x) - \min(x) = 16 - 8 = 8 ]

5. Variance ( (s^{2}) )

[ s^{2} = \frac{\sum (x_i-\bar{x})^{2}}{n-1} ]

[ \begin{aligned} \sum (x_i-\bar{x})^{2} &= (12-12)^{2}+(15-12)^{2}+…+(12-12)^{2}\ &= 0+9+9+4+1+1+4+16+16+0 = 60 \end{aligned} ]

[ s^{2} = \frac{60}{9}=6.667 ]

6. Standard Deviation (s)

[ s = \sqrt{s^{2}} = \sqrt{6.667}\approx 2.58 ]

Answer Key – Part A

Mean = 12.0
Median = 12
Mode = 12
Range = 8
Variance = 6.67 (rounded to two decimals)
Standard deviation = 2.58

Part B – Binomial Probability

The experiment: n = 20 trials, p = 0.So 25 (defective). Let X be the number of defective units.

1. Probability of exactly 5 defectives

[ P(X=5)=\binom{20}{5}p^{5}(1-p)^{15} ]

[ \binom{20}{5}=15504,; p^{5}=0.25^{5}=0.0009766,; (1-p)^{15}=0.75^{15}=0.0133 ]

[ P(X=5)=15504 \times 0.0009766 \times 0.0133 \approx 0.201 ]

2. Probability of at most 3 defectives

[ P(X\le 3)=\sum_{k=0}^{3}\binom{20}{k}p^{k}(1-p)^{20-k} ]

Calculations (rounded):

k	(\binom{20}{k})	(p^{k})	((1-p)^{20-k})	Product
0	1	1	0.0625	0.0209
2	190	0.That said, 0032
1	20	0. 0665
3	1140	0.Consider this: 75^{17}=0. Here's the thing — 75^{20}=0. Day to day, 0032	0. 75^{18}=0.Now, 75^{19}=0. 0156	0.In real terms, 0056

[ P(X\le 3)=0.0032+0.0209+0.0665+0.1334\approx 0.224 ]

3. Expected value (μ) and variance (σ²)

[ \mu = np = 20 \times 0.25 = 5 ]

[ \sigma^{2}=np(1-p)=20 \times 0.25 \times 0.75 = 3.75 ]

Answer Key – Part B

(P(X=5) \approx 0.201) (20.1 %)
(P(X\le 3) \approx 0.224) (22.4 %)
Expected number of defectives = 5
Variance = 3.75

Part C – Confidence Interval for a Population Mean

Given: n = 25, (\bar{x}=78.4), s = 9.2, confidence level = 95 %. Because σ is unknown and n < 30, use the t‑distribution.

1. Degrees of freedom (df)

(df = n-1 = 24)

2. Critical t‑value

(t_{0.025,24} \approx 2.064) (from t‑table).

3. Standard error (SE)

[ SE = \frac{s}{\sqrt{n}} = \frac{9.2}{\sqrt{25}} = \frac{9.2}{5}=1.84 ]

4. Margin of error (ME)

[ ME = t^{*}\times SE = 2.064 \times 1.84 \approx 3.

5. Confidence interval

[ \bar{x} \pm ME = 78.4 \pm 3.80 ]

[ \boxed{(74.6,; 82.2)} ]

Interpretation: We are 95 % confident that the true population mean lies between 74.6 and 82.2.

Answer Key – Part C

95 % CI for μ: (74.6, 82.2)

Part D – Two‑Sample t‑Test (Independent Samples)

Goal: Test (H_{0}: \mu_{A} = \mu_{B}) against (H_{a}: \mu_{A} \neq \mu_{B}) at α = 0.05.

1. Pooled standard deviation (Sp)

[ S_{p} = \sqrt{\frac{(n_{A}-1)s_{A}^{2}+(n_{B}-1)s_{B}^{2}}{n_{A}+n_{B}-2}} ]

If you found this helpful, you might also enjoy why my vision is getting worse or wind is an example of.

[ \begin{aligned} S_{p} &= \sqrt{\frac{(30-1)7.5^{2}+(28-1)6.8^{2}}{30+28-2}}\ &= \sqrt{\frac{29 \times 56.25 + 27 \times 46.24}{56}}\ &= \sqrt{\frac{1631.25 + 1248.Think about it: 48}{56}}\ &= \sqrt{\frac{2879. Which means 73}{56}}\ &= \sqrt{51. 42}\approx 7.

2. Test statistic (t)

[ t = \frac{\bar{x}{B}-\bar{x}{A}}{S_{p}\sqrt{\frac{1}{n_{A}}+\frac{1}{n_{B}}}} = \frac{86.3-82.1}{7.17\sqrt{\frac{1}{30}+\frac{1}{28}}} ]

[ \sqrt{\frac{1}{30}+\frac{1}{28}} = \sqrt{0.0333+0.0357}= \sqrt{0.069}=0.262 ]

[ t = \frac{4.2}{7.17 \times 0.262} = \frac{4.In real terms, 2}{1. 88} \approx 2.

3. Degrees of freedom

(df = n_{A}+n_{B}-2 = 56)

4. Critical t‑value (two‑tailed, α/2 = 0.025)

(t_{0.025,56} \approx 2.004)

5. Decision

t statistic	Critical value	Decision
2.23	2.004	Reject (H_{0})

Since |t| > 2.004, the difference between groups is statistically significant at the 5 % level.

6. Effect size (Cohen’s d) – optional but often requested

[ d = \frac{\bar{x}{B}-\bar{x}{A}}{S_{p}} = \frac{4.2}{7.17} \approx 0.59 ]

Interpretation: a medium effect size.

Answer Key – Part D

t = 2.23, df = 56
Critical t ≈ 2.004 → Reject (H_{0}) (significant difference)
Cohen’s d ≈ 0.59 (medium effect).

Part E – Simple Linear Regression

Data: (X = hours studied, Y = test score)

X	Y
2	68
4	75
6	80
8	85
10	92

1. Compute sums

[ \begin{aligned} \sum X &= 2+4+6+8+10 = 30\ \sum Y &= 68+75+80+85+92 = 400\ \sum XY &= 2\cdot68 + 4\cdot75 + 6\cdot80 + 8\cdot85 + 10\cdot92 = 136 + 300 + 480 + 680 + 920 = 2516\ \sum X^{2} &= 2^{2}+4^{2}+6^{2}+8^{2}+10^{2}=4+16+36+64+100=220 \end{aligned} ]

2. Slope (b)

[ b = \frac{n\sum XY - (\sum X)(\sum Y)}{n\sum X^{2} - (\sum X)^{2}} = \frac{5(2516) - 30(400)}{5(220) - 30^{2}} ]

[ b = \frac{12580 - 12000}{1100 - 900}= \frac{580}{200}=2.90 ]

3. Intercept (a)

[ a = \bar{Y} - b\bar{X} ]

[ \bar{X}= \frac{30}{5}=6,\qquad \bar{Y}= \frac{400}{5}=80 ]

[ a = 80 - 2.Now, 90(6) = 80 - 17. 4 = 62.

Regression equation:

[ \boxed{\hat{Y}=62.6 + 2.9X} ]

4. Interpretation of slope

For each additional hour studied, the predicted test score increases by 2.9 points.

5. Coefficient of determination (R²)

First compute total sum of squares (SST) and regression sum of squares (SSR).

[ SST = \sum (Y_i - \bar{Y})^{2} ]

[ \begin{aligned} (Y_i-\bar{Y})^{2}:&\ (68-80)^{2}=144\ &(75-80)^{2}=25\ &(80-80)^{2}=0\ &(85-80)^{2}=25\ &(92-80)^{2}=144\ \end{aligned} ]

[ SST = 144+25+0+25+144 = 338 ]

Predicted values (\hat{Y}_i) using the line:

X	(\hat{Y})
2	62.0
8	62.Now, 4
4	62. 9·6 = 80.8
10	62.9·4 = 74.2
6	62.Now, 6 + 2. On the flip side, 6 + 2. 6 + 2.Consider this: 9·8 = 85. 6 + 2.6 + 2.9·2 = 68.9·10 = 91.

[ SSR = \sum (\hat{Y}_i - \bar{Y})^{2} ]

[ \begin{aligned} (68.4-80)^{2}=136.9\ (74.2-80)^{2}=33.6\ (80.0-80)^{2}=0\ (85.8-80)^{2}=33.6\ (91.6-80)^{2}=136.9\ \end{aligned} ]

[ SSR = 341.0 \ (\text{rounded}) ]

[ R^{2}= \frac{SSR}{SST}= \frac{341}{338}\approx 1.009 ]

Because of rounding, R² slightly exceeds 1; using more precise calculations yields R² ≈ 0.99, indicating an excellent fit.

6. Residual standard error (optional)

[ SE = \sqrt{\frac{SST-SSR}{n-2}} \approx \sqrt{\frac{-3}{3}} \ (\text{negative due to rounding})\rightarrow 0 ]

In practice, the residual variance is essentially zero, confirming the linear model captures almost all variability.

Answer Key – Part E

Slope (b) = 2.90
Intercept (a) = 62.6
Regression equation: (\hat{Y}=62.6+2.9X)
Interpretation: each extra study hour raises the predicted score by 2.9 points.
R² ≈ 0.99 (very strong linear relationship).

Frequently Asked Questions (FAQ)

1. Why do we use the t‑distribution in Part C instead of the normal distribution?

The population standard deviation (σ) is unknown and the sample size (n = 25) is less than 30. Under these conditions, the sampling distribution of the mean follows a t‑distribution, which has heavier tails and yields a slightly wider confidence interval—reflecting greater uncertainty.

2. Can the binomial expected value be non‑integer, even though the variable counts successes?

Yes. The expected value (np) is a theoretical average over many repetitions. It need not be an integer; it simply indicates the long‑run mean number of successes.

3. What assumptions underlie the two‑sample t‑test in Part D?

Independence of observations within and between groups.
Normality of each group’s distribution (reasonable with n > 30).
Equal variances (homoscedasticity); we used the pooled variance, so we assume σ₁ ≈ σ₂. If variances differ markedly, a Welch’s t‑test would be more appropriate.

4. Why does the regression R² in Part E appear slightly above 1?

Rounding errors in intermediate calculations (especially SSR) can push the ratio over 1. Using full‑precision values or statistical software eliminates this artifact, giving an R² just under 1.

5. How can I extend Activity 3.5 for a more advanced class?

Replace the simple linear regression with multiple regression (add a second predictor, e.g., prior GPA).
Introduce non‑parametric alternatives for the two‑sample test (Mann‑Whitney U).
Use bootstrapping to construct confidence intervals for the median instead of the mean.

Conclusion

The answer key for Activity 3.Also, 5 – Applied Statistics not only supplies the correct numerical results but also walks through the logical steps that transform raw data into statistical insight. By mastering these calculations—descriptive measures, binomial probabilities, confidence intervals, hypothesis testing, and regression analysis—students gain a solid foundation for real‑world data analysis. Practically speaking, teachers can employ this key as a grading template, while learners can use it to verify their work and deepen conceptual understanding. Remember that statistics is as much about interpretation as it is about computation; always relate the numbers back to the substantive context, whether it be quality control, educational performance, or any field where data drive decisions.