# Comparing Two Means - Grade 9 Statistics & Probability
Section A: Multiple Choice Questions
Q1: A researcher wants to compare the mean heights of male and female students in a school. What type of samples are being compared? (a) Independent samples (b) Paired samples (c) Stratified samples (d) Systematic samples
Solution:
Ans: (a) Explanation: The samples are independent because the male and female students are two separate, unrelated groups. Each measurement in one group does not correspond to a specific measurement in the other group. Paired samples would involve the same subjects measured twice or matched pairs.
Q2: When comparing two population means with independent samples, which condition must be satisfied to use a two-sample t-test? (a) The sample sizes must be equal (b) The populations must be normally distributed or the sample sizes must be large enough (c) The population means must be equal (d) The population variances must be known
Solution:
Ans: (b) Explanation: For a two-sample t-test, we need either the populations to be approximately normally distributed or have sufficiently large sample sizes (typically \(n \geq 30\)) by the Central Limit Theorem. Equal sample sizes are not required, and we do not need to know the population variances exactly.
Q3: A 95% confidence interval for the difference between two means \(\mu_1 - \mu_2\) is calculated to be (2.5, 7.8). What can we conclude? (a) The means are equal (b) \(\mu_1\) is significantly greater than \(\mu_2\) (c) \(\mu_2\) is significantly greater than \(\mu_1\) (d) There is no significant difference between the means
Solution:
Ans: (b) Explanation: Since the entire confidence interval (2.5, 7.8) contains only positive values and does not include zero, we can conclude that \(\mu_1 - \mu_2 > 0\), meaning \(\mu_1\) is significantly greater than \(\mu_2\) at the 95% confidence level.
Q4: The pooled variance estimate is used when comparing two means under which assumption? (a) The sample sizes are equal (b) The population variances are equal (c) The samples are paired (d) The populations are not normally distributed
Solution:
Ans: (b) Explanation: The pooled variance estimate combines the variances from both samples under the assumption that the population variances are equal (homogeneity of variance). This provides a more precise estimate of the common variance.
Q5: In a paired t-test, what is the null hypothesis typically testing? (a) The mean of the first sample equals zero (b) The mean difference between paired observations equals zero (c) The two sample means are different (d) The variances of the two samples are equal
Solution:
Ans: (b) Explanation: In a paired t-test, we calculate the difference for each pair and test whether the mean difference (\(\mu_d\)) equals zero. The null hypothesis is \(H_0: \mu_d = 0\), meaning there is no average difference between the paired observations.
Q6: Two samples have means \(\bar{x}_1 = 85\) and \(\bar{x}_2 = 78\). The standard error of the difference is 2.5. What is the test statistic for testing \(H_0: \mu_1 = \mu_2\)? (a) 1.4 (b) 2.8 (c) 3.5 (d) 7.0
Solution:
Ans: (b) Explanation: The test statistic is calculated as \(t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{85 - 78}{2.5} = \frac{7}{2.5} = 2.8\). This measures how many standard errors the difference in sample means is from zero.
Q7: When should you use a paired t-test instead of an independent samples t-test? (a) When the sample sizes are small (b) When the same subjects are measured twice or observations are naturally paired (c) When the population variances are unequal (d) When the data is not normally distributed
Solution:
Ans: (b) Explanation: A paired t-test is appropriate when the data consists of matched pairs or when the same subjects are measured under two different conditions (e.g., before and after treatment). This accounts for the correlation between paired observations.
Q8: If the p-value in a two-sample t-test is 0.03 and the significance level is \(\alpha = 0.05\), what is the correct decision? (a) Fail to reject the null hypothesis (b) Reject the null hypothesis (c) Accept the alternative hypothesis as proven (d) The test is inconclusive
Solution:
Ans: (b) Explanation: Since the p-value (0.03) is less than the significance level \(\alpha = 0.05\), we reject the null hypothesis. This provides evidence that there is a significant difference between the two population means.
Section B: Fill in the Blanks
Q9: When comparing two independent means, the sampling distribution of the difference \(\bar{x}_1 - \bar{x}_2\) is approximately normal if both sample sizes are large due to the __________ theorem.
Solution:
Ans: Central Limit Explanation: The Central Limit Theorem states that the sampling distribution of the sample mean (and the difference between sample means) approaches a normal distribution as the sample size increases, typically when \(n \geq 30\).
Q10: The standard error of the difference between two sample means measures the __________ of the difference \(\bar{x}_1 - \bar{x}_2\).
Solution:
Ans: variability (or standard deviation) Explanation: The standard error quantifies the variability or spread of the sampling distribution of the difference between two sample means. It indicates how much the difference might vary from sample to sample.
Q11: In a paired t-test, we analyze the __________ between each pair of observations rather than the individual values.
Solution:
Ans: differences Explanation: A paired t-test focuses on the differences within each matched pair. By calculating \(d_i = x_{1i} - x_{2i}\) for each pair, we reduce the problem to a one-sample test on these differences.
Q12: The degrees of freedom for a two-sample t-test with independent samples of sizes \(n_1\) and \(n_2\), when population variances are assumed equal, is __________.
Solution:
Ans: \(n_1 + n_2 - 2\) Explanation: When using the pooled variance method with equal variances assumed, the degrees of freedom is calculated as \(df = n_1 + n_2 - 2\), which accounts for estimating parameters from both samples.
Q13: A confidence interval for the difference between two means that contains zero suggests there is __________ significant difference between the population means.
Solution:
Ans: no Explanation: If a confidence interval for \(\mu_1 - \mu_2\) includes zero, it means zero is a plausible value for the difference, indicating no significant difference between the two population means at that confidence level.
Q14: When population variances are not assumed to be equal, we use the __________ t-test (also called the Welch's t-test).
Solution:
Ans: unequal variance (or Welch's) Explanation:Welch's t-test is an adaptation of the two-sample t-test that does not assume equal population variances. It uses a different formula for the standard error and degrees of freedom.
Section C: Word Problems
Q15: A teacher wants to compare the effectiveness of two teaching methods. She randomly assigns 15 students to Method A and 12 students to Method B. After the course, the mean test score for Method A is 82 with a standard deviation of 6, and for Method B is 78 with a standard deviation of 5. Assuming equal population variances, calculate the pooled variance estimate.
Solution:
Ans: The pooled variance is calculated using: \[s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}\] Substituting the values: \[s_p^2 = \frac{(15 - 1)(6)^2 + (12 - 1)(5)^2}{15 + 12 - 2}\] \[s_p^2 = \frac{14 \times 36 + 11 \times 25}{25}\] \[s_p^2 = \frac{504 + 275}{25} = \frac{779}{25} = 31.16\] Final Answer: The pooled variance is 31.16
Q16: A nutritionist measures the weight loss (in pounds) of 8 participants before and after a diet program. The differences (before - after) are: 3, 5, 2, 4, 6, 3, 5, 4. Calculate the mean difference and determine if there is evidence of weight loss at \(\alpha = 0.05\) level. (Use that the standard deviation of differences is approximately 1.31)
Solution:
Ans: First, calculate the mean difference: \[\bar{d} = \frac{3 + 5 + 2 + 4 + 6 + 3 + 5 + 4}{8} = \frac{32}{8} = 4\] The standard error is: \[SE = \frac{s_d}{\sqrt{n}} = \frac{1.31}{\sqrt{8}} = \frac{1.31}{2.828} \approx 0.463\] The test statistic is: \[t = \frac{\bar{d}}{SE} = \frac{4}{0.463} \approx 8.64\] With \(df = n - 1 = 7\) and \(\alpha = 0.05\) (one-tailed), the critical value is approximately 1.895. Since \(t = 8.64 > 1.895\), we reject the null hypothesis. Final Answer: Mean difference = 4 pounds; there is significant evidence of weight loss (t ≈ 8.64, p <>
Q17: Two brands of batteries are tested for longevity. Brand X has a sample of 20 batteries with a mean life of 45 hours and standard deviation of 4 hours. Brand Y has a sample of 25 batteries with a mean life of 42 hours and standard deviation of 5 hours. Assuming unequal variances, calculate the standard error of the difference between the two means.
Solution:
Ans: For unequal variances, the standard error is: \[SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\] Substituting the values: \[SE = \sqrt{\frac{4^2}{20} + \frac{5^2}{25}}\] \[SE = \sqrt{\frac{16}{20} + \frac{25}{25}}\] \[SE = \sqrt{0.8 + 1.0} = \sqrt{1.8} \approx 1.34\] Final Answer: The standard error of the difference is approximately 1.34 hours
Q18: A researcher conducts a study comparing the reaction times (in milliseconds) of 30 participants under two conditions. The mean reaction time in Condition 1 is 285 ms with a standard deviation of 25 ms, and in Condition 2 is 295 ms with a standard deviation of 30 ms. Construct a 95% confidence interval for the difference in means \(\mu_1 - \mu_2\) assuming equal variances and using a critical value of 2.002.
Solution:
Ans: First, calculate the pooled variance: \[s_p^2 = \frac{(30-1)(25)^2 + (30-1)(30)^2}{30+30-2} = \frac{29 \times 625 + 29 \times 900}{58}\] \[s_p^2 = \frac{18125 + 26100}{58} = \frac{44225}{58} \approx 762.5\] \[s_p = \sqrt{762.5} \approx 27.6\] The standard error is: \[SE = s_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}} = 27.6\sqrt{\frac{1}{30} + \frac{1}{30}} = 27.6\sqrt{\frac{2}{30}} = 27.6 \times 0.258 \approx 7.12\] The difference in means is: \(\bar{x}_1 - \bar{x}_2 = 285 - 295 = -10\) The confidence interval is: \[CI = -10 \pm 2.002 \times 7.12 = -10 \pm 14.25\] \[CI = (-24.25, 4.25)\] Final Answer: The 95% confidence interval is (-24.25, 4.25) ms
Q19: A fitness instructor wants to determine if a new exercise program improves flexibility. She measures the sit-and-reach distance (in cm) for 10 participants before and after the program. The mean difference (after - before) is 3.2 cm with a standard deviation of 1.5 cm. Test at \(\alpha = 0.01\) whether the program significantly improves flexibility. Use a critical value of 3.25 for a one-tailed test with 9 degrees of freedom.
Solution:
Ans: Calculate the standard error: \[SE = \frac{s_d}{\sqrt{n}} = \frac{1.5}{\sqrt{10}} = \frac{1.5}{3.162} \approx 0.474\] Calculate the test statistic: \[t = \frac{\bar{d}}{SE} = \frac{3.2}{0.474} \approx 6.75\] Since \(t = 6.75 > 3.25\) (critical value), we reject the null hypothesis. Final Answer: Yes, there is significant evidence at \(\alpha = 0.01\) that the program improves flexibility (t ≈ 6.75)
Q20: A company tests two assembly methods. Method 1 is used on 40 products with a mean assembly time of 12.5 minutes and standard deviation of 2.1 minutes. Method 2 is used on 35 products with a mean of 13.8 minutes and standard deviation of 2.4 minutes. If the pooled standard deviation is approximately 2.24 minutes, calculate the test statistic to compare the two methods assuming equal variances.
Solution:
Ans: The standard error is: \[SE = s_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}} = 2.24\sqrt{\frac{1}{40} + \frac{1}{35}}\] \[SE = 2.24\sqrt{0.025 + 0.0286} = 2.24\sqrt{0.0536} = 2.24 \times 0.232 \approx 0.52\] The difference in means is: \[\bar{x}_1 - \bar{x}_2 = 12.5 - 13.8 = -1.3\] The test statistic is: \[t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{-1.3}{0.52} \approx -2.50\] Final Answer: The test statistic is approximately -2.50
The document Worksheet (with Solutions): Comparing Two Means is a part of Grade 9 category.
Exam, Free, practice quizzes, Worksheet (with Solutions): Comparing Two Means, mock tests for examination, Previous Year Questions with Solutions, ppt, Important questions, Sample Paper, pdf , study material, shortcuts and tricks, Objective type Questions, Extra Questions, MCQs, Semester Notes, video lectures, Worksheet (with Solutions): Comparing Two Means, Summary, Viva Questions, past year papers, Worksheet (with Solutions): Comparing Two Means;
