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 the null hypothesis, h₀, is μ₁ ≥ μ₂. the alternative hypothesis, hₐ, is μ₁ < μ₂. which hypothesis is the claim? the null hypothesis, h₀ the alternative hypothesis, hₐ (b) find the critical value(s) and identify the rejection region(s). enter the critical value(s) below. (type an integer or decimal rounded to three decimal places as needed. use a comma to separate

Explanation:

Step1: Determine sample sizes

First, we find the number of data points for the old curriculum (sample 1) and new curriculum (sample 2) from the stem - and - leaf plot.
For the old curriculum:

• Stem 3: 1 value (37)
• Stem 4: 2 values (40, 44)
• Stem 5: 8 values (50, 53, 53, 56, 56, 57, 59, 59)
• Stem 6: 9 values (60, 60, 61, 62, 64, 66, 68, 69, 69)
• Total for old curriculum: \(1 + 2+8 + 9=20\)

For the new curriculum:

• Stem 3: 1 value (33)
• Stem 4: 1 value (48)
• Stem 5: 2 values (51, 52)
• Stem 6: 10 values (60, 60, 61, 62, 62, 63, 65, 65, 68, 68, 68, 69)
• Stem 7: 8 values (70, 70, 71, 75, 76, 76, 78, 79)
• Stem 8: 2 values (83, 88)
• Total for new curriculum: \(1+1 + 2+10+8 + 2=24\)

The degrees of freedom for a two - sample t - test with equal variances is \(df=n_1 + n_2-2=20 + 24-2 = 42\)

Step2: Identify the test type and significance level

We are performing a left - tailed two - sample t - test (since \(H_a:\mu_1<\mu_2\)) with \(\alpha = 0.10\)

Step3: Find the critical value

Using a t - distribution table or a t - distribution calculator, for a left - tailed test with \(df = 42\) and \(\alpha=0.10\), the critical value \(t_{\alpha,df}\) is approximately \(- 1.302\) (we can use the t - table or a statistical software. For \(df = 40\) (closest value in the table to 42) and one - tailed \(\alpha = 0.10\), the critical value is \(- 1.303\), and for \(df=45\), it is \(- 1.301\). Using a calculator for \(df = 42\) and \(\alpha=0.10\) (left - tailed), we get \(t=- 1.302\))

Answer:

The critical value is \(-1.302\)