An experiment in teaching introductory chemistry was recently conducted at a large university. One section was taught by lecture-lab; one section was taught by lab only (no lectures); and a third section was taught online exclusively. All three sections were held at the same time and days of the week. Students were randomly assigned to each of the three sections. Each section held over 500 students. The same final exam was used for each section. Random samples of each final exam were selected. Statistics are reported below. Is there a difference in final exam grades by method? What hypotheses would be used to test this?

Lecture-Lab Lab only Online

𝑥
¯
=
75

𝑥
¯
=
70

𝑥
¯
=
62

𝑠
=
11.3

𝑠
=
14.8

𝑠
=
17.0

𝑛
=
11

𝑛
=
11

𝑛
=
11

Question

An experiment in teaching introductory chemistry was recently conducted at a large university. One section was taught by lecture-lab; one section was taught by lab only (no lectures); and a third section was taught online exclusively. All three sections were held at the same time and days of the week. Students were randomly assigned to each of the three sections. Each section held over 500 students. The same final exam was used for each section. Random samples of each final exam were selected. Statistics are reported below. Is there a difference in final exam grades by method? What hypotheses would be used to test this?

Lecture-Lab	Lab only	Online

𝑥
¯
=
75
	
𝑥
¯
=
70
	
𝑥
¯
=
62

𝑠
=
11.3
	
𝑠
=
14.8
	
𝑠
=
17.0

𝑛
=
11
	
𝑛
=
11
	
𝑛
=
11

BlackTom AI · Accepted Answer

One-way ANOVA is the appropriate method because we compare three independent group means (Lecture-Lab, Lab only, Online) from randomized samples and want to know if any group mean differs from the others. The null and alternative hypotheses for a one-way ANOVA are: H0: μ_LL = μ_L = μ_O HA: at least one mean differs 1) Relevant knowledge points and concepts - One-way ANOVA tests equality of k population means by partitioning total variability into “between-group” and “within-group” components. - H0 states all group means are equal; HA states that not all means are equal (i.e., at least one differs). - Assumptions: independent samples, approximately normal within each group, and (approximately) equal variances. 2) Step-by-step reasoning - We have k = 3 groups, each with n = 11, sample means x̄ = 75, 70, 62, and sample standard deviations s = 11.3, 14.8, 17.0. - Use ANOVA: compute treatment (between) sum of squares SST, error (within) sum of squares SSE, then mean squares and F statistic to test H0. 3) Calculations (all intermediate steps) - Grand mean: x̄_G = (75 + 70 + 62) / 3 = 207 / 3 = 69. - SST (between): n * Σ (x̄_i − x̄_G)^2 = 11 * [(75−69)^2 + (70−69)^2 + (62−69)^2] = 11 * [36 + 1 + 49] = 11 * 86 = 946. - SSE (within): Σ (n−1) s_i^2 s1^2 = 11.3^2 = 127.69; s2^2 = 14.8^2 = 219.04; s3^2 = 17.0^2 = 289.00 SSE = 10 * (127.69 + 219.04 + 289.00) = 10 * 635.73 = 6357.30. - Degrees of freedom: df_between = k−1 = 2; df_within = N−k = 33−3 = 30. - Mean squares: MST = SST/2 = 946/2 = 473.00; MSE = SSE/30 ≈ 6357.30/30 ≈ 211.91. - F = MST / MSE ≈ 473.00 / 211.91 ≈ 2.231. For F(2,30) this gives p > 0.05 (critical F ≈ 3.32 at α=0.05), so we would not reject H0 at the 5% level. 4) Why the correct answer is correct - The correct hypothesis form for comparing three means is H0: all means equal and HA: at least one differs. That is exactly the ANOVA null and alternative and matches the provided correct statement. 5) Why other options are incorrect - Directional alternatives (μ_LL > μ_L > μ_O or μ_LL < μ_L < μ_O) are too specific and not implied by the data; ANOVA uses a non-directional HA. - The alternative written as μ_LL ≠ μ_L ≠ μ_O is ambiguous and incorrectly suggests pairwise inequality for every pair; HA should state “at least one mean differs,” not that all three are pairwise unequal. 6) Conclusion - The appropriate hypotheses are H0: μ_LL = μ_L = μ_O and HA: at least one mean differs, which matches the provided correct answer.

类似问题

Using the same dataset, Flaye now decides to investigate whether concert attendance differs by the day of the concert (Friday, Saturday, or Sunday). Assuming all assumptions are met, what statistical test would you suggest they run?

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