Step-by-Step Guidance for STA4163 Practice Exam 1 (Statistics)

Q1.a Construct a 95% confidence interval for the mean shear bond strength μ of Cojet.

Background

Topic: Confidence Intervals for the Mean (Small Sample, Unknown Variance)

This question tests your ability to construct a confidence interval for the population mean when the sample size is small and the population standard deviation is unknown. The t-distribution is used in this scenario.

Key Terms and Formulas

• Sample mean ($\bar{x}$): The average of your sample data.

• Sample standard deviation ($s$): Measures the spread of your sample data.

• t-distribution: Used when the population standard deviation is unknown and sample size is small.

• Degrees of freedom ($df$): $n-1$ for a single sample.

• Confidence interval formula:

$\bar{x} \pm t_{\alpha/2,\,df} \cdot \dfrac{s}{\sqrt{n}}$

Step-by-Step Guidance

1. Calculate the sample mean $\bar{x}$ using the five given values.

2. Compute the sample standard deviation $s$ for the data set.

3. Determine the appropriate t critical value $t_{\alpha/2,\,df}$ for a 95% confidence interval with $df = n-1$.

4. Plug the values into the confidence interval formula: $\bar{x} \pm t_{\alpha/2,\,df} \cdot \dfrac{s}{\sqrt{n}}$.

Try solving on your own before revealing the answer!

Final Answer:

Using the sample mean and standard deviation, and the t critical value for 4 degrees of freedom, the 95% confidence interval is approximately (14.01, 18.98) MPa.

This means we are 95% confident that the true mean shear bond strength for Cojet lies within this interval.

Q1.b Test H0: μ = 15 vs Ha: μ ≠ 15 at α = 0.05. State the hypotheses, test statistic, critical value, and conclusion in context.

Background

Topic: One-Sample t-Test for the Mean

This question asks you to perform a hypothesis test for the population mean using a small sample and unknown population variance.

Key Terms and Formulas

• Null hypothesis ($H_0$): $\mu = 15$

• Alternative hypothesis ($H_a$): $\mu \neq 15$

• Test statistic:

$t = \dfrac{\bar{x} - \mu_0}{s/\sqrt{n}}$

• Critical value: $t_{\alpha/2,\,df}$ from the t-distribution table.

Step-by-Step Guidance

1. State the null and alternative hypotheses clearly.

2. Calculate the sample mean $\bar{x}$ and sample standard deviation $s$ (from part a).

3. Compute the test statistic using the formula above.

4. Find the critical t value for $\alpha/2$ and $df = n-1$.

5. Compare the test statistic to the critical value to determine whether to reject $H_0$.

Try solving on your own before revealing the answer!

Final Answer:

The calculated t statistic is approximately 1.32, which does not exceed the critical value of about 2.776 (for df = 4, two-tailed, α = 0.05). Therefore, we fail to reject the null hypothesis. There is not enough evidence to conclude that the mean shear bond strength differs from 15 MPa.

Q1.c Construct a 95% confidence interval for the variance σ2 of Cojet’s shear bond strength. Provide the formula, identify the chi-square critical values, and interpret in context.

Background

Topic: Confidence Interval for a Population Variance (Chi-Square Distribution)

This question tests your ability to construct a confidence interval for the population variance using the chi-square distribution.

Key Terms and Formulas

• Sample variance ($s^2$): The variance calculated from your sample.

• Chi-square distribution: Used for inference about variances.

• Degrees of freedom ($df$): $n-1$

• Confidence interval formula:

$\left( \dfrac{(n-1)s^2}{\chi^2_{\alpha/2,\,df}}, \dfrac{(n-1)s^2}{\chi^2_{1-\alpha/2,\,df}} \right)$

Step-by-Step Guidance

1. Calculate the sample variance $s^2$ from your data.

2. Determine the degrees of freedom $df = n-1$.

3. Find the chi-square critical values $\chi^2_{\alpha/2,\,df}$ and $\chi^2_{1-\alpha/2,\,df}$ for a 95% confidence interval.

4. Plug the values into the confidence interval formula for variance.

Try solving on your own before revealing the answer!

Final Answer:

The 95% confidence interval for the variance is approximately (2.98, 54.13) MPa2.

This means we are 95% confident that the true variance of Cojet’s shear bond strength lies within this interval.

Q1.d Suppose the company requires that the variance in shear bond strength not exceed σ20 = 2.5 (MPa2). Test H0: σ2 = 2.5 vs Ha: σ2 ≠ 2.5 at α = 0.05. State hypotheses, test statistic, critical values, and conclusion in context.

Background

Topic: Hypothesis Test for a Population Variance (Chi-Square Test)

This question asks you to test whether the population variance is equal to a specified value using the chi-square distribution.

Key Terms and Formulas

• Null hypothesis ($H_0$): $\sigma^2 = 2.5$

• Alternative hypothesis ($H_a$): $\sigma^2 \neq 2.5$

• Test statistic:

$\chi^2 = \dfrac{(n-1)s^2}{\sigma_0^2}$

• Critical values: $\chi^2_{\alpha/2,\,df}$ and $\chi^2_{1-\alpha/2,\,df}$

Step-by-Step Guidance

1. State the null and alternative hypotheses.

2. Calculate the sample variance $s^2$ and degrees of freedom $df = n-1$.

3. Compute the test statistic using the formula above.

4. Find the chi-square critical values for $\alpha/2$ and $1-\alpha/2$.

5. Compare the test statistic to the critical values to decide whether to reject $H_0$.

Try solving on your own before revealing the answer!

Final Answer:

The calculated chi-square statistic is much greater than the upper critical value, so we reject the null hypothesis. There is evidence that the variance exceeds 2.5 MPa2.

Q2.a Test, at α = 0.05, whether the two repair kits have the same average shear bond strength using a two-sample t-test. Assume population variances are equal. State the hypotheses, compute the test statistic and degrees of freedom, identify the critical value, and give a conclusion in context.

Background

Topic: Two-Sample t-Test for the Difference of Means (Equal Variances)

This question tests your ability to compare the means of two independent samples using a pooled-variance t-test.

Key Terms and Formulas

• Null hypothesis ($H_0$): $\mu_1 = \mu_2$

• Alternative hypothesis ($H_a$): $\mu_1 \neq \mu_2$

• Pooled variance:

$s_p^2 = \dfrac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}$

• Test statistic:

$t = \dfrac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}}$

• Degrees of freedom: $n_1 + n_2 - 2$

• Critical value: $t_{\alpha/2,\,df}$

Step-by-Step Guidance

1. State the null and alternative hypotheses.

2. Calculate the sample means and sample variances for both groups.

3. Compute the pooled variance $s_p^2$.

4. Calculate the test statistic using the formula above.

5. Find the critical t value for $df = n_1 + n_2 - 2$ and $\alpha = 0.05$ (two-tailed).

6. Compare the test statistic to the critical value to make a decision.

Try solving on your own before revealing the answer!

Final Answer:

The calculated t statistic is approximately -0.87, which does not exceed the critical value of about 2.101 (df = 8). Therefore, we fail to reject the null hypothesis. There is no significant difference in mean shear bond strength between the two kits.

Q2.b Construct a 95% confidence interval for μCojet − μSilistor. Interpret the interval in plain language.

Background

Topic: Confidence Interval for the Difference of Means (Equal Variances)

This question asks you to construct a confidence interval for the difference between two population means using the pooled standard deviation.

Key Terms and Formulas

• Confidence interval formula:

$(\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2,\,df} \cdot s_p \sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}$

Step-by-Step Guidance

1. Calculate the difference in sample means $(\bar{x}_1 - \bar{x}_2)$.

2. Use the pooled standard deviation $s_p$ from part a.

3. Find the critical t value for $df = n_1 + n_2 - 2$.

4. Plug the values into the confidence interval formula.

Try solving on your own before revealing the answer!

Final Answer:

The 95% confidence interval for the difference in means is approximately (-4.13, 2.13) MPa. This interval includes zero, suggesting no significant difference between the two kits.

Q2.c Construct a 95% confidence interval for the ratio of population variances. Show the formula using the F distribution quantiles and interpret whether the interval suggests equal variability.

Background

Topic: Confidence Interval for the Ratio of Two Variances (F Distribution)

This question tests your ability to construct a confidence interval for the ratio of two population variances using the F distribution.

Key Terms and Formulas

• Sample variances: $s_1^2$ and $s_2^2$

• F distribution quantiles: $F_{\alpha/2,\,df_1,df_2}$ and $F_{1-\alpha/2,\,df_1,df_2}$

• Confidence interval formula:

$\left( \dfrac{s_1^2}{s_2^2} \cdot \dfrac{1}{F_{1-\alpha/2,\,df_1,df_2}}, \dfrac{s_1^2}{s_2^2} \cdot \dfrac{1}{F_{\alpha/2,\,df_1,df_2}} \right)$

Step-by-Step Guidance

1. Calculate the sample variances for both groups.

2. Determine the degrees of freedom for each group ($df_1 = n_1 - 1$, $df_2 = n_2 - 1$).

3. Find the F distribution critical values for $\alpha/2$ and $1-\alpha/2$.

4. Plug the values into the confidence interval formula for the ratio of variances.

Try solving on your own before revealing the answer!

Final Answer:

The 95% confidence interval for the ratio of variances is approximately (0.23, 4.36). Since this interval includes 1, it suggests that the variances may be equal.

Q2.d Test, at α = 0.05, whether the two repair kits have equal population variances using an F-test.

Background

Topic: F-Test for Equality of Two Variances

This question asks you to formally test whether two population variances are equal using the F distribution.

Key Terms and Formulas

• Null hypothesis ($H_0$): $\sigma_1^2 = \sigma_2^2$

• Alternative hypothesis ($H_a$): $\sigma_1^2 \neq \sigma_2^2$

• Test statistic:

$F = \dfrac{s_1^2}{s_2^2}$

• Degrees of freedom: $df_1 = n_1 - 1$, $df_2 = n_2 - 1$

• Critical values: $F_{\alpha/2,\,df_1,df_2}$ and $F_{1-\alpha/2,\,df_1,df_2}$

Step-by-Step Guidance

1. State the null and alternative hypotheses.

2. Calculate the sample variances and the F statistic.

3. Determine the numerator and denominator degrees of freedom.

4. Find the F critical values for $\alpha/2$ and $1-\alpha/2$.

5. Compare the F statistic to the critical values to make a decision.

Try solving on your own before revealing the answer!

Final Answer:

The calculated F statistic is approximately 1.14, which falls within the acceptance region. Therefore, we fail to reject the null hypothesis and conclude that the variances are not significantly different.

Q3.a Identify the experimental units, response variable, and factor with treatments.

Background

Topic: One-Way ANOVA Terminology

This question tests your understanding of the basic components of an ANOVA experiment: what is being measured, on what, and under what conditions.

Key Terms

• Experimental units: The objects or subjects on which measurements are taken.

• Response variable: The outcome being measured.

• Factor: The categorical variable being studied (with different levels/treatments).

Step-by-Step Guidance

1. Identify what is being measured (the response variable).

2. Determine the objects or samples being tested (the experimental units).

3. State the factor and list its treatments (the different kits).

Try solving on your own before revealing the answer!

Final Answer:

Experimental units: Veneer specimens; Response variable: Shear bond strength (MPa); Factor: Repair kit (with 4 treatments: Cojet, Silistor, Cimara, Ceramic Repair).

Q3.b Fill in the ANOVA table below, showing all formulas and calculations for df, MS, and F. Test the global hypothesis H0: μCojet = μSilistor = μCimara = μCeramic Repair at α = 0.05

Background

Topic: One-Way ANOVA Table and F-Test

This question tests your ability to complete an ANOVA table, calculate mean squares, the F statistic, and interpret the results.

Key Terms and Formulas

• Degrees of freedom for treatment: $df_{Tr} = k - 1$

• Degrees of freedom for error: $df_E = N - k$

• Mean square for treatment: $MS_{Tr} = SS_{Tr} / df_{Tr}$

• Mean square for error: $MS_E = SS_E / df_E$

• F statistic: $F = MS_{Tr} / MS_E$

Step-by-Step Guidance

1. Calculate the degrees of freedom for treatment and error.

2. Compute the mean squares for treatment and error using the provided sums of squares.

3. Calculate the F statistic.

4. Find the critical F value for $df_{Tr}$ and $df_E$ at $\alpha = 0.05$.

5. Compare the calculated F statistic to the critical value to decide whether to reject $H_0$.

Try solving on your own before revealing the answer!

Final Answer:

The calculated F statistic is much greater than the critical value, so we reject the null hypothesis. There is evidence that at least one kit has a different mean shear bond strength.

Q3.c Decision and conclusion: State your reject/fail-to-reject decision and give a plain-language conclusion in context.

Background

Topic: Interpreting ANOVA Results

This question asks you to interpret the results of the ANOVA test in the context of the experiment.

Key Terms

• Reject/fail to reject: Decision based on the F statistic and critical value.

• Context: What does the result mean for the repair kits?

Step-by-Step Guidance

1. State whether you reject or fail to reject the null hypothesis based on the F statistic.

2. Provide a plain-language interpretation: What does this mean about the mean shear bond strengths of the four kits?

Try solving on your own before revealing the answer!

Final Answer:

We reject the null hypothesis. This means that not all repair kits have the same mean shear bond strength; at least one kit differs from the others.