Question

The columns in the table below contain the Race/Ethnicity of U.S. Public Schools for a recent year, the percentages for the Advanced Placement Examinee Population for that class, and the Overall Student Population. Suppose the right column contains the result of a survey of 1,000 local students from that year who took an AP Exam. Use 0.5 significance level.

AP Examinee Overall Student Survey Race/Ethnicity Population Frequency Population Asian, Asian American, or Pacific Islander Black or African- American Hispanic or Latino American Indian or Alaska Native White Not reported/other 115 5.4% 10.2% 93 14.5% 8.2% 134 15.9% 15.5% 10 1.2% 0.6% 602 61.6% 59.4% 46 1.4% 61% Test of a Single Variance Goodness- of-fit Test fit Test (Method Goodness-of- Test of Independence Homogeneity (or Test for (Method #2) Standard Deviation) cdf (lower, upper, df) or Chi 2cdf (lower hi2 GOF upper, df) or Chi2 Test Chi2 Test Test (or x2 Chi-square cdf or X? Test or (or x2 Test or GOF Test) lower, upper, Chi2 2-way Chi2 2-way Chi2 2-waysquare cdt est (lower, upper, df) Degrees of freedom Lower (df) est statistic Upper Degrees of freedom (df) Lower Upper Degrees of freedom (df) Select the appropriate test and then enter the appropriate values. Just put NA for any values that don't apply to this problem Test: Chi2 GOF Test (or x2 GOF Test) Unless otherwise stated, round values to five decimals. Test statistic (type in your whole calculation and result): Lower Upper: Degrees of freedom (df): p-value (use scientific notation with three digits after the decimal): (The p-value is required for every problem.) Sten 4

Answer 1

The following table is obtained Categories Category1 Category2 Category 3 Category 4 Category 5 Category6 Observed 115 93 134 10 602 46 Expected 1000.054 54 1000*. 145-145 1000.159-159 1000*.012-12 1000*.616-616 1000.01 4= 1 4 (fo-fe)Ife (115-54)154 68.907 (93-145)41 45 18.648 (134-159)159 3.931 (10-12)/12 0.333 (602-616)7616 0.318 (46-14)14 73.143 1000 165.281 Sum 1000 The following null and alternative hypotheses need to be tested Ho : pi- .054, p2-.145,p3- .159,pi-.012,p616,pe-.014 Ha Some of the population proportions differ from the values stated in the null hypothesis This corresponds to a Chi-Square test for Goodness of Fit. (2) Rejection Region Based on the information provided, the significance level is a 0.05, the number of degrees of freedom is df-6-1-5, so then the rejection region for this test is R-X-X > 11.07) (3)_Test Statistics The Chi-Squared statistic is computed as follows 68.907 18.648 + 3.9310.333 0.318+73.143-165.281 Since it is observed that x2 -165.2812 11.07, it is then 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 some of the population proportions differ from those stated in the null hypothesis, at the a - 0.05 significance level

TS = 165.281

df = 5

p-value = =CHISQ.DIST.RT(165.281,5) = 0.0000