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 enter the critical value(s) below. - 1.302 (type an integer or decimal rounded to three decimal places as needed. use a comma to separate answers as nee select the correct rejection region(s) below. a. t > t₀ b. t < -t₀, t > t₀ c. -t₀ < t < t₀ d. t < -t₀

Explanation:

Step1: Identify Test Type

This is a one - tailed (left - tailed? No, wait, the claim is new method produces higher scores, so we test \(H_0:\mu_{new}-\mu_{old}\leq0\), \(H_a:\mu_{new}-\mu_{old}>0\), so it's a right - tailed test? Wait, no, wait the critical value given is negative? Wait, maybe I misread. Wait, the old and new curriculum. Let's check the sample sizes.

First, find sample sizes:

Old Curriculum (left side):
- Stem 3: 7 (1)
- Stem 4: 0,4 (2)
- Stem 5: 0,3,3,6,6,7,9,9 (8)
- Stem 6: 0,0,1,2,4,6,8,9 (8) Wait, no, the old curriculum stem - leaf:

Wait the old curriculum (left of the stem) has:

Stem 3: 7 (1)
Stem 4: 0,4 (2)
Stem 5: 0,3,3,6,6,7,9,9 (8)
Stem 6: 0,0,1,2,4,6,8,9 (wait, the numbers are 0,0,1,2,4,6,8,9? Wait no, the old curriculum stem 6: 0,0,1,2,4,6,8,9? Wait the original plot: Old Curriculum has for stem 6: 0,0,1,2,4,6,8,9? Wait no, the given stem - leaf:

Old Curriculum:
3 | 7 (1)
4 | 0,4 (2)
5 | 0,3,3,6,6,7,9,9 (8)
6 | 0,0,1,2,4,6,8,9 (wait, no, the numbers are 0,0,1,2,4,6,8,9? Wait the user's plot: Old Curriculum (left) has:

Stem 3: 7 (1)
Stem 4: 0,4 (2)
Stem 5: 0,3,3,6,6,7,9,9 (8)
Stem 6: 0,0,1,2,4,6,8,9? Wait no, the old curriculum stem 6: 0,0,1,2,4,6,8,9? Wait the numbers are 0,0,1,2,4,6,8,9? Wait no, the original data: Old Curriculum (left) for stem 6: 0,0,1,2,4,6,8,9? Wait no, the user's plot:

Old Curriculum:
3 | 7 (1)
4 | 0,4 (2)
5 | 0,3,3,6,6,7,9,9 (8)
6 | 0,0,1,2,4,6,8,9? Wait no, the old curriculum stem 6: 0,0,1,2,4,6,8,9? Wait the count: 1 (stem 3) + 2 (stem 4) + 8 (stem 5) + 8 (stem 6)? Wait no, stem 6 for old: 0,0,1,2,4,6,8,9? Wait no, the numbers are 0,0,1,2,4,6,8,9? Wait that's 8 numbers? Wait no, 0,0,1,2,4,6,8,9: that's 8? Wait 0 (1), 0 (2), 1 (3), 2 (4), 4 (5), 6 (6), 8 (7), 9 (8). Yes.

New Curriculum (right side):
Stem 3: (0)
Stem 4: 8 (1)
Stem 5: 1,2 (2)
Stem 6: 0,0,1,2,2,3,5,5,8,8,8,9 (12)
Stem 7: 0,0,1,5,6,6,8,9 (8)
Stem 8: 3,8 (2)

So sample size for old (\(n_1\)): 1 + 2+8 + 8=19? Wait no, stem 3:1, stem 4:2, stem 5:8, stem 6:8? Wait no, stem 6 for old: 0,0,1,2,4,6,8,9: 8 numbers. So \(n_1 = 1+2 + 8+8=19\)? Wait no, stem 3:1, stem 4:2, stem 5:8, stem 6:8? Wait no, stem 6 for old: 0,0,1,2,4,6,8,9: 8 numbers. So \(n_1=1 + 2+8 + 8 = 19\)?

New Curriculum (\(n_2\)):
Stem 3:0
Stem 4:8 (1)
Stem 5:1,2 (2)
Stem 6:0,0,1,2,2,3,5,5,8,8,8,9 (12)
Stem 7:0,0,1,5,6,6,8,9 (8)
Stem 8:3,8 (2)
Total \(n_2=1 + 2+12 + 8+2=25\)

Degrees of freedom \(df=n_1 + n_2-2=19 + 25-2 = 42\)

Significance level \(\alpha = 0.10\), one - tailed test (since we test if new is higher, so \(H_a:\mu_2-\mu_1>0\), so the critical value is \(t_{\alpha,df}=t_{0.10,42}\). Wait, but the given critical value is - 1.302? Wait maybe I mixed up old and new. Wait the claim is new method (right) produces higher scores, so we test \(H_0:\mu_{new}\leq\mu_{old}\), \(H_a:\mu_{new}>\mu_{old}\), which is a right - tailed test. But the given critical value is - 1.302, which is left - tailed. Wait maybe the test is \(H_0:\mu_{new}-\mu_{old}\leq0\), \(H_a:\mu_{new}-\mu_{old}>0\), but if we do \(t=\frac{(\bar{x}_2-\bar{x}_1)-0}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\), and the critical region for right - tailed is \(t>t_{\alpha,df}\). But the given critical value is - 1.302, which is for left - tailed. Wait maybe the problem was set up as \(H_0:\mu_{old}\geq\mu_{new}\), \(H_a:\mu_{old}<\mu_{new}\), which is a left - tailed test. So \(t_{\alpha,df}=t_{0.10,42}\). Looking up t - table, \(t_{0.10,42}\approx1.302\), but since it's left - tailed, the critical value is \(-t_{\alpha,df}=- 1.302\).

Now, the rejection region for left - tailed test (\(H_a:\mu_1<\mu_2\) where \(\mu_1\) is old, \(\mu_2\) is new) is \(t < - t_0\), where \(t_0 = 1.302\), so \(t < - 1.302\). So the rejection region is \(t < - t_0\), which is option D.

Step2: Confirm Rejection Region

For a left - tailed test with \(\alpha = 0.10\) and \(df = 42\), the critical value is \(t_{0.10,42}\approx1.302\), so the rejection region is \(t < - 1.302\) (since it's left - tailed, we reject \(H_0\) when the test statistic is less than the negative critical value). So the correct rejection region is \(t < - t_0\), which is option D.

Answer:

D. \(t < - t_0\)