(8 points) The National Assessment of Educational Progress (NAEP) tested a sample of students who had used a computer in their mathematics classes, and another sample of students who had not used a computer. The mean score for students using a computer was 309, with a standard deviation of 29. For students not using a computer the mean was 303, with a standard deviation of 32. Assume there were 60 students in the computer sample, and 40 students in the sample that hadn’t used a computer. Can you conclude that the population mean scores differ?

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. it is a two-tailed test. Let c b the subscript for students using computer and n be the subscript for students not using computer.

Therefore, the population means would be μc and μn.

The random variable is xc - xn = difference in the sample mean scores of students who used computers and those who didn't use computers

We would set up the hypothesis.

The null hypothesis is

H0 : μc = μn H0 : μc - μn = 0

The alternative hypothesis is

Ha : μc ≠ μn Ha : μc - μn ≠ 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(xc - xn)/√(sc²/nc + sn²/nn)

From the information given,

xc = 309

xn = 303

sc = 29

sn = 32

nc = 60

nn = 40

t = (309 - 303)/√(29²/60 + 32²/40)

t = 0.953

The formula for determining the degree of freedom is

df = [sc²/nc + sn²/nn]²/(1/nc - 1)(sc²/nc)² + (1/nn - 1)(sn²/nn)²

df = [29²/60 + 32²/40]²/(1/60 - 1)(29²/60)² + (1/40 - 1)(32²/40)² = 1569.48/20.13

df = 78

We would determine the probability value from the t test calculator. It becomes

p value = 0.344

Assuming a level of significance of 0.05, we would not reject the null hypothesis because the p value, 0.344 is > 0.05

Therefore, we cannot conclude that the population mean scores differ.

the data of the detroit pistons in numerical order from least to greatest is 0.30, 0.37, 0.55, 0.62, 0.69, 0.72, 0.73, 0.75, 0.77, 0.78, 0.80, 0.81, 0.81, 0.83, 0.84, 0.89

the data of the miami heat in numerical order from least to greatest is 0.50, 0.50, 0.56, 0.57, 0.67, 0.72, 0.72, 0.72, 0.75, 0.75, 0.76, 0.81, 0.82, 0.90, 1.00, 1.00

the detroit pistons median : 1.52

the miami heat median: 1.47

the iqr of this data is 0.05