1. a. large p value b. small p value 2. a. can be rejected b. cannot be rejected

ID: 1134743 • Letter: 1

Question

1. a. large p value b. small p value

2. a. can be rejected b. cannot be rejected

3. a. there is no b. there is

Data on fifth-grade test scores (reading and mathematics) for 404 school districts in California yield = 697.9 and standard deviation sY_ 21.1 The 95% confidence interval for the mean test score in the population is ( When the districts were divided into districts with small classes ( 20 students per teacher) and large classes ( 20 students per teacher), the following results were found: . (Round your responses to two decimal places) Class Size Small Large Average Score (Y) 710.0 702.0 Standard Deviation (sy) 21.0 19.3 243 184 Is there statistically significant evidence that the districts with smaller classes have higher average test scores? The t-statistco esting the null hypothesis is(Round your response to two decimal places.) The p-value for the test is Round your response to six decimal places.) Hint: Use the Excel function normsdist to help answer this question. The test is one-sided test. Is there statistically significant evidence that the districts with smaller classes have higher average test scores? The smaller classes have hiaher averaoe test scores Enter your answer in the edit fields and then click Check Answer suggests that the null hypothesis with a high degree of confidence. Hence, Vstatistically significant evidence that the districts with

Explanation / Answer

Part 1) The 95% confidence interval for the mean test score can be calculated by using the following formula

(mean test score) ± 1.96(/ n)

Mean test score = 697.9

= standard deviation = 21.1

n = total school districts = 404

(697.9) ± 1.96(21.1/ 404)

(697.9) ± 1.96(21.1/20.1)

(697.9) ± 2.057

695.8 – 699.9

Part 2) We have the following information

Class Size

Average Score

Standard Deviation

Small

710.0

21.0

243

Large

702.0

19.3

184

Null Hypothesis: There is no significant evidence that the districts with smaller classes have higher average test scores.

Alternate Hypothesis: There is significant evidence that the districts with smaller classes have higher average test scores.

t-statistic for testing the null hypothesis can be calculated from the following formula

t = [(mean of small sample) – (mean of large sample)/S] × [n1n2)/(n1+n2)]

Mean of small sample = 710

Mean of large sample = 702

n1 = 243

n2 = 184

S = Combined Standard Deviation

S = [(n1 – 1)Ss2 + (n2 – 1)sl2]/(n1+n2 – 2)

Ss2 = Standard deviation of small sample

Sl2 = Standard deviation of large sample

S = [(243 – 1)(21)2 + (184 – 1)(19.3)2]/(243 + 184 – 2)

S = [(242)(441) + (183)(372.49)]/(425)

S = (106722 + 68165.67)/(425)

S = (411.50)

S = 20.29

t = [(710 – 702)/20.29] × (243×184)/(243+184)

t = (8/20.29) × 211.45/427

t = (8/20.29) × 0.495

t = 0.195

Degree of freedom in the present case is (243+184 – 2) = 425

The table value of t statistics at 5% significance level and 425 degrees of freedom is 1.645

Since, the estimated value is less than the table value, so the null hypothesis is true.

The p value in the case of right tailed test is the area to the right of the computed value of the test statistic under the null hypothesis.

So, the p value for t = 0.195 and degree of freedom being 425 at 5% level of significance is 0.422743.

The large p value suggests that the null hypothesis cannot be rejected with a high degree of confidence. Hence, there is no statistically significant evidence that the districts with smaller cases have higher average test scores.