Question

1. Research examining the effects of preschool child care has found that children who spent time in day care, especially high-quality day care, perform better on math and language tests than children who say home with their mothers (Broberg, Wessels, Lamb, & Hwang, 1997). In a typical study, a researcher obtains a sample of n = 10 children who attended day care before starting school. Then children are given a standardized math test for which the population mean is μ = 52.

The scores for the sample are as follows:

               53,   57,   61,   49,   52,   56,   58,   62,   51,   56

On the basis of this sample, children with a history of preschool day care significantly different from the general population? Use a two-tailed test with α = .05.

1. State the null hypothesis in words and in a statistical form

2. State the alternative hypothesis in words and a statistical form

3. Compute the appropriate statistic to test the hypotheses. Sketch the distribution with the estimated standard error and locate the critical region(s) with the critical value(s) (6). [When you compute SS, you might want to use the definitional formula as it might be easier to compute SS this way].

4. State your statistical decision

5. Compute Cohen’s d. Interpret what this d really means in this context

f.    Compute 95% CI

                       

g. What is your conclusion? Interpret the result. Don’t forget to include statistical information as well (e.g., t-score, df, α, Cohen’s d)

Answer 1

Σx = 555 Σx2 = 30965 ΣΧ Mean i n 53 + 57 + 61 + 49 + 52 + 56 + 58 + 62 + 51 + 56 Π 10 555 10 = 55.5 (Σx) 2 Σχ2 n Sample Standard deviation S = η - 1 (555)2 30965 - 10 9 130965 - 30802.5 9 (162.5 - 9 = V/18.0556 = 4.2492

The provided sample mean is I = 55.5 and the sample standard deviation is s = 4.2492, and the sample size is n = 10. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u = 52 Ha: M 752 This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a two-tailed test is tc = 2.262. The rejection region for this two-tailed test is R = {t: 1t| > 2.262} (3) Test Statistics The t-statistic is computed as follows: 55.5 – 52 t = X Mo s/vn = 2.605 4.2492/V10

(4) Decision about the null hypothesis Since it is observed that t| = 2.605 > te= 2.262, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0285, and since p = 0.0285 < 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean u is different than 52, at the 0.05 significance level. Confidence Interval The 95% confidence interval is 52.46 < u < 58.54. Graphically. T-Test Results: t-stats = 2.605, p-value =0.0285 0.40 0.35 0.30 0.25 0.20 0.15 T 0.10 0.05 0.00 -4.0 -3.5 -3.0 -2.5 -2.0-1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
Thus there is sufficient evidence to conclude that children with history of preschool day care significantly different than true population.

thus we are 95% confident that the mean maths score of children who spend time in day care lies between 52.46 to 58.54

The provided sample mean is X = 55.5 and the sample standard deviation is s = 4.2492. The size of the sample is n = 10 and the required confidence level is 95%. - The number of degrees of freedom are df 10 – 1= 9, and the significance level is a = 0.05. Based on the provided information, the critical t-value for a = 0.05 and df = 9 degrees of freedom is te = 2.262. The 95% confidence for the population mean u is computed using the following expression te x s CI = X X + te x s ✓n Vn Therefore, based on the information provided, the 95% confidence for the population mean u is 2.262 x 4.2492 CI= (55.5 55.5 + 2.262 x V1 9 V10 = + (55.5 – 3.04, 55.5 +3.04) - (52.46, 58.54)

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