Question

In: Statistics and Probability

State the null and alternative hypotheses, calculate the test statistic, determine the critical value for the...

State the null and alternative hypotheses, calculate the test statistic, determine the critical value for the given alpha, determine whether to reject or retain the null, and then draw a conclusion about the original study based on the significance test results.

3. Two hundred and twenty fifth-graders in Norman took an IQ test. Students were placed in one of eight categories based on their IQ score. Determine whether this distribution of 114 fifth graders differs from the theoretical distribution based on expected IQ scores from the general population.

Observed Expected

IQ Score Frequency Frequency

135+ 24 6.6

125-134 14 17.6

115-124 22 32.4

105-114 77 68.2

95-104 38 51.6

85-94 22 25.4

75-84 9 12.8

below 75 14 4.4

a. (2 point) State the null and alternative hypotheses.

b. (1 point) State what test should be used for this problem and why.

c. (11 points) Perform this test using all the steps listed above using =.05.

d. (3 points) What conclusion can you make about the IQs of the 220 fifth-graders? If the chi-square is significant, which categories contributed most to that result?

Solutions

Expert Solution

a) The null hypothesis and alternate hypothesis can be written as:

H0: There is no significant difference between the distribution of fifth graders and the theoretical distribution based on expected IQ scores.

H1: There is a significant difference between the distribution of fifth graders and the theoretical distribution based on expected IQ scores.

b) We will be using chi-square test for this particular problem as chi-square test is used for testing the significance of the discrepancy between observed and expected frequencies.

c)

IQ Score

 Observed Frequency (Oi) Expected Frequency (Ei)
135+ 24 6.6 17.4 302.76 45.87
125-134 14 17.6 -3.6 12.96 0.74
115-124 22 32.4 -10.4 108.16 3.34
105-114 77 68.2 8.8 77.44 1.135
95-104 38 51.6 -13.6 184.96 3.58
85-94 22 25.4 -3.4 11.56 0.455
75-84 9 12.8 -3.8 14.44 1.13
below 75 14 4.4 9.6 92.16 20.94
TOTAL 220 219 77.19

Therefore,

= 77.19

Now, degrees of freedom = n - 1 = 8 - 1 = 7 and the tabulated value for 7 degree of freedom at 5% level of significance = 14.067

d) Conclusion: Since the calculated value of statistic is more than the tabulated value, the null hypothesis is rejected. Hence we conclude there is a significant difference between the distribution of fifth graders and the theoretical distribution based on expected IQ scores.

orchestra answered 2 years ago
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