Question

a group of third grade students is taught using a new curriculum. a control group of third grade students is taught using the old curriculum. the reading test scores for the two groups are shown in the back-to-back stem-and-leaf plot. at α=0.10, is there enough evidence to support the claim that the new method of teaching reading produces higher reading test scores than the old method does? assume the population variances are equal. complete parts (a) through (e) below. assume the samples are random and independent, and the populations are normally distributed. old curriculum new curriculum 7 3 4 0 4 8 9 9 7 6 6 3 3 0 5 1 2 9 8 6 4 2 1 0 0 6 0 0 1 2 2 3 5 5 8 8 8 9 7 0 0 1 5 6 6 8 9 8 3 8 key: 4|7|0 = 74 old and 70 new a. t > t₀ b. t < -t₀, t > t₀ c. -t₀ < t < t₀ d. t < -t₀ (c) find the standardized test statistic. t = (type an integer or decimal rounded to three decimal places as needed.)

Explanation:

Step1: Extract Data from Stem - Leaf Plot

For Old Curriculum:
- Stem 3: 7 (score: 37)
- Stem 4: 0, 4 (scores: 40, 44)
- Stem 5: 0, 3, 3, 6, 6, 7, 9, 9 (scores: 50, 53, 53, 56, 56, 57, 59, 59)
- Stem 6: 0, 0, 1, 2, 4, 6, 8, 9 (scores: 60, 60, 61, 62, 64, 66, 68, 69)
Count of Old Curriculum scores: \(n_1=1 + 2+8 + 8=19\)
Sum of Old Curriculum scores:
\(37+(40 + 44)+(50+53+53+56+56+57+59+59)+(60+60+61+62+64+66+68+69)\)
\(=37 + 84+(50\times8+3+3+6+6+7+9+9)+(60\times8+1+2+4+6+8+9)\)
\(=121+(400+34)+(480 + 30)\)
\(=121 + 434+510=1065\)
Mean of Old Curriculum (\(\bar{x}_1\)): \(\bar{x}_1=\frac{1065}{19}\approx56.0526\)

For New Curriculum:
- Stem 3: None
- Stem 4: 8 (score: 48)
- Stem 5: 1, 2 (scores: 51, 52)
- Stem 6: 0, 0, 1, 2, 2, 3, 5, 5, 8, 8, 8, 9 (scores: 60, 60, 61, 62, 62, 63, 65, 65, 68, 68, 68, 69)
- Stem 7: 0, 0, 1, 5, 6, 6, 8, 9 (scores: 70, 70, 71, 75, 76, 76, 78, 79)
- Stem 8: 3, 8 (scores: 83, 88)
Count of New Curriculum scores: \(n_2 = 1+2 + 12+8+2=25\)
Sum of New Curriculum scores:
\(48+(51 + 52)+(60+60+61+62+62+63+65+65+68+68+68+69)+(70+70+71+75+76+76+78+79)+(83 + 88)\)
\(=48+103+(60\times12+1+2+2+3+5+5+8+8+8+9)+(70\times8+1+5+6+6+8+9)+171\)
\(=151+(720 + 57)+(560+35)+171\)
\(=151+777+595+171 = 1694\)
Mean of New Curriculum (\(\bar{x}_2\)): \(\bar{x}_2=\frac{1694}{25}=67.76\)

Step2: Calculate Sample Variances

For Old Curriculum:
\(s_1^2=\frac{\sum(x_{1i}-\bar{x}_1)^2}{n_1 - 1}\)
First, calculate \((x_{1i}-\bar{x}_1)^2\) for each score:
- \(37 - 56.0526=-19.0526\), \((-19.0526)^2\approx363.00\)
- \(40 - 56.0526=-16.0526\), \((-16.0526)^2\approx257.69\)
- \(44 - 56.0526=-12.0526\), \((-12.0526)^2\approx145.26\)
- For stem 5 scores:
- \(50 - 56.0526=-6.0526\), \((-6.0526)^2\approx36.63\) (8 scores, sum of squares: \(8\times36.63 = 293.04\))
- For stem 6 scores:
- \(60 - 56.0526 = 3.9474\), \((3.9474)^2\approx15.58\) (8 scores, sum of squares: \(8\times15.58=124.64\))
Sum of squares for Old Curriculum:
\(363.00+257.69+145.26+293.04+124.64\)
\(=363.00+(257.69+145.26)+(293.04+124.64)\)
\(=363.00 + 402.95+417.68=1183.63\)
\(s_1^2=\frac{1183.63}{19 - 1}\approx65.757\)

For New Curriculum:
\(s_2^2=\frac{\sum(x_{2i}-\bar{x}_2)^2}{n_2 - 1}\)
First, calculate \((x_{2i}-\bar{x}_2)^2\) for each score:
- \(48 - 67.76=-19.76\), \((-19.76)^2\approx390.46\)
- \(51 - 67.76=-16.76\), \((-16.76)^2\approx280.89\); \(52 - 67.76=-15.76\), \((-15.76)^2\approx248.38\) (sum for stem 4 and 5: \(390.46+280.89+248.38 = 919.73\))
- For stem 6 scores (12 scores, mean deviation from 67.76: \(60 - 67.76=-7.76\) to \(69 - 67.76 = 1.24\)):
Sum of squares for stem 6:
\(\sum_{i = 1}^{12}(x_{2i}-67.76)^2\)
\(=2\times(-7.76)^2+(-6.76)^2+2\times(-5.76)^2+(-4.76)^2+2\times(-2.76)^2+(-1.76)^2+(1.24)^2\)
\(=2\times60.2176+45.6976+2\times33.1776+22.6576+2\times7.6176+3.0976+1.5376\)
\(=120.4352+45.6976+66.3552+22.6576+15.2352+3.0976+1.5376\)
\(=275.016\)
- For stem 7 scores (8 scores, mean deviation from 67.76: \(70 - 67.76 = 2.24\) to \(79 - 67.76=11.24\)):
Sum of squares for stem 7:
\(\sum_{i = 1}^{8}(x_{2i}-67.76)^2\)
\(=2\times(2.24)^2+(3.24)^2+(7.24)^2+2\times(8.24)^2+(10.24)^2+(11.24)^2\)
\(=2\times5.0176+10.4976+52.4176+2\times67.8976+104.8576+126.3376\)
\(=10.0352+10.4976+52.4176+135.7952+104.8576+126.3376\)
\(=439.9408\)
- For stem 8 scores:
\(83 - 67.76 = 15.24\), \((15.24)^2\approx232.26\); \(88 - 67.76 = 20.24\), \((20.24)^2\approx409.66\) (sum: \(232.26+409.66 = 641.92\))
Total sum of squares for New Curriculum:
\(919.73+275.016+439.9408+641.92\)
\(=919.73+(275.016+439.9408)+641.92\)
\(=919.73+714.9568+641.92=2276.6068\)
\(s_2^2=\frac{2276.6068}{25 - 1}\approx94.8586\)

Step3: Calculate Pooled Variance (\(s_p^2\))

Since \(\sigma_1^2=\sigma_2^2\) (assumed equal variances), the pooled variance is given by:
\(s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}\)
\(n_1 - 1 = 18\), \(n_2 - 1=24\)
\(s_p^2=\frac{18\times65.757+24\times94.8586}{19 + 25-2}\)
\(=\frac{1183.626+2276.6064}{42}\)
\(=\frac{3460.2324}{42}\approx82.3865\)
\(s_p=\sqrt{82.3865}\approx9.0767\)

Step4: Calculate Standardized Test Statistic (\(t\))

The formula for the two - sample \(t\) - test (equal variances) is:
\(t=\frac{\bar{x}_1-\bar{x}_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\)
\(\bar{x}_1-\bar{x}_2=56.0526 - 67.76=-11.7074\)
\(\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}=\sqrt{\frac{1}{19}+\frac{1}{25}}=\sqrt{\frac{25 + 19}{19\times25}}=\sqrt{\frac{44}{475}}\approx\sqrt{0.0926}\approx0.3043\)
\(s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}=9.0767\times0.3043\approx2.762\)
\(t=\frac{-11.7074}{2.762}\approx - 4.239\)

Answer:

\(-4.239\)