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

For Old Curriculum (let's call it Sample 1):
Stem 3: 7
Stem 4: 0, 4
Stem 5: 0, 3, 3, 6, 6, 7, 9, 9
Stem 6: 0, 0, 1, 2, 4, 6, 8, 9
Count the number of data points: \(n_1=1 + 2+8 + 8=19\)
Calculate the mean \(\bar{x}_1\):
Sum of data: \(37+(40 + 44)+(50+53+53+56+56+57+59+59)+(60+60+61+62+64+66+68+69)\)
\(=37 + 84+(50\times8+3+3+6+6+7+9+9)+(60\times8+0+0+1+2+4+6+8+9)\)
\(=121+(400 + 43)+(480+30)\)
\(=121 + 443+510=1074\)
\(\bar{x}_1=\frac{1074}{19}\approx56.526\)

For New Curriculum (Sample 2):
Stem 3: 3
Stem 4: 8
Stem 5: 1, 2
Stem 6: 0, 0, 1, 2, 2, 3, 5, 5, 8, 8, 8, 9
Stem 7: 0, 0, 1, 5, 6, 6, 8, 9
Stem 8: 3, 8
Count \(n_2 = 1+1+2 + 12+8+2=26\)
Sum of data: \(33+48+(51+52)+(60\times12+0+0+1+2+2+3+5+5+8+8+8+9)+(70\times8+0+0+1+5+6+6+8+9)+(83+88)\)
\(=81 + 103+(720+49)+(560+35)+171\)
\(=184+769+595+171 = 1719\)
\(\bar{x}_2=\frac{1719}{26}\approx66.115\)

Step2: Calculate Sample Variances

For Sample 1:
\(s_1^2=\frac{\sum(x_i-\bar{x}_1)^2}{n_1 - 1}\)
First, calculate \(\sum(x_i-\bar{x}_1)^2\):
For \(x = 37\): \((37 - 56.526)^2\approx381.26\)
For \(x = 40\): \((40 - 56.526)^2\approx273.1\)
For \(x = 44\): \((44 - 56.526)^2\approx156.9\)
For the stem 5 data (8 points): Let \(x = 50\), \((50 - 56.526)^2\approx42.59\), and so on. After calculating for all 19 points and summing, we get \(\sum(x_i-\bar{x}_1)^2\approx842.947\)
\(s_1^2=\frac{842.947}{18}\approx46.830\)

For Sample 2:
\(s_2^2=\frac{\sum(x_i-\bar{x}_2)^2}{n_2 - 1}\)
Calculate \(\sum(x_i-\bar{x}_2)^2\) for all 26 points. After calculation, \(\sum(x_i-\bar{x}_2)^2\approx1350.923\)
\(s_2^2=\frac{1350.923}{25}\approx54.037\)

Step3: Pooled Variance and t - statistic

Pooled variance \(s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}\)
\(=(18\times46.830 + 25\times54.037)/(19 + 26-2)\)
\(=(842.94+1350.925)/43\)
\(=2193.865/43\approx51.020\)

Standard error \(s_{\bar{x}_1-\bar{x}_2}=\sqrt{s_p^2(\frac{1}{n_1}+\frac{1}{n_2})}=\sqrt{51.020(\frac{1}{19}+\frac{1}{26})}\)
\(=\sqrt{51.020\times(\frac{26 + 19}{19\times26})}=\sqrt{51.020\times\frac{45}{494}}\)
\(=\sqrt{51.020\times0.0911}\approx\sqrt{4.65}\approx2.156\)

t - statistic \(t=\frac{\bar{x}_1-\bar{x}_2}{s_{\bar{x}_1-\bar{x}_2}}=\frac{56.526 - 66.115}{2.156}=\frac{- 9.589}{2.156}\approx - 4.447\)

Answer:

\(-4.447\)