We decline to reject the null hypothesis. There is not enough evidence to suggest that the observed test scores are significantly different from the expected test scores.
H_{0}: the distribution of AIDS cases follows the ethnicities of the general population of Santa Clara County.
Graph: Check student’s solution.
Decision: Cannot accept the null hypothesis.
Reason for the Decision: Calculated value of test statistics is either in or out of the tail of the distribution.
Conclusion (write out in complete sentences): The makeup of AIDS cases does not fit the ethnicities of the general population of Santa Clara County.
Smoking level per day  African American  Native Hawaiian  Latino  Japanese Americans  White  Totals 

110  9,886  2,745  12,831  8,378  7,650  41,490 
1120  6,514  3,062  4,932  10,680  9,877  35,065 
2130  1,671  1,419  1,406  4,715  6,062  15,273 
31+  759  788  800  2,305  3,970  8,622 
Totals  18,830  8,014  19,969  26,078  27,559  10,0450 
Smoking level per day  African American  Native Hawaiian  Latino  Japanese Americans  White 

110  7777.57  3310.11  8248.02  10771.29  11383.01 
1120  6573.16  2797.52  6970.76  9103.29  9620.27 
2130  2863.02  1218.49  3036.20  3965.05  4190.23 
31+  1616.25  687.87  1714.01  2238.37  2365.49 
 Cannot accept the null hypothesis.
 Calculated value of test statistics is either in or out of the tail of the distribution.
 There is sufficient evidence to conclude that smoking level is dependent on ethnic group.
Answers will vary. Sample answer: Tests of independence and tests for homogeneity both calculate the test statistic the same way $\sum}_{(ij)}\frac{{(OE)}^{2}}{E$. In addition, all values must be greater than or equal to five.
 H_{0}: σ = 15
 H_{a}: σ > 15
 df = 42
 chisquare with df = 42
 test statistic = 26.88
 Check student’s solution.

 Alpha = 0.05
 Decision: Cannot reject null hypothesis.
 Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
 Conclusion: There is insufficient evidence to conclude that the standard deviation is greater than 15.
 H_{0}: σ ≤ 3
 H_{a}: σ > 3
 df = 17
 chisquare distribution with df = 17
 test statistic = 28.73
 Check student’s solution.

 Alpha: 0.05
 Decision: Cannot accept the null hypothesis.
 Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
 Conclusion: There is sufficient evidence to conclude that the standard deviation is greater than three.
 H_{0}: σ = 2
 H_{a}: σ ≠ 2
 df = 14
 chisquare distribution with df = 14
 chisquare test statistic = 5.2094
 Check student’s solution.

 Alpha = 0.05
 Decision: Cannot accept the null hypothesis
 Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
 Conclusion: There is sufficient evidence to conclude that the standard deviation is different than 2.
The sample standard deviation is $34.29.
H_{0} : σ^{2} = 25^{2}
H_{a} : σ^{2} > 25^{2}
df = n – 1 = 7.
test statistic: ${x}^{2}={x}_{7}^{2}=\frac{(n\u20131){s}^{2}}{{25}^{2}}=\frac{(8\u20131){(34.29)}^{2}}{{25}^{2}}=13.169$;
Alpha: 0.05
Decision: Cannot reject the null hypothesis.
Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
Conclusion: At the 5% level, there is insufficient evidence to conclude that the variance is more than 625.
Marital status  Percent  Expected frequency 

Never married  31.3  125.2 
Married  56.1  224.4 
Widowed  2.5  10 
Divorced/Separated  10.1  40.4 
 The data fits the distribution.
 The data does not fit the distribution.
 3
 chisquare distribution with df = 3
 19.27
 0.0002
 Check student’s solution.

 Alpha = 0.05
 Decision: Cannot accept null hypothesis at the 5% level of significance
 Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
 Conclusion: Data does not fit the distribution.
 H_{0}: The local results follow the distribution of the U.S. AP examinee population
 H_{a}: The local results do not follow the distribution of the U.S. AP examinee population
 df = 5
 chisquare distribution with df = 5
 chisquare test statistic = 13.4
 Check student’s solution.
 Alpha = 0.05
 Decision: Cannot accept null when a = 0.05
 Reason for Decision: Calculated value of test statistics is either in or out of the tail of the distribution.
 Conclusion: Local data do not fit the AP Examinee Distribution.
 Decision: Do not reject null when a = 0.01
 Conclusion: There is insufficient evidence to conclude that local data do not follow the distribution of the U.S. AP examinee distribution.
 H_{0}: The actual college majors of graduating females fit the distribution of their expected majors
 H_{a}: The actual college majors of graduating females do not fit the distribution of their expected majors
 df = 10
 chisquare distribution with df = 10
 test statistic = 11.48
 Check student’s solution.

 Alpha = 0.05
 Decision: Cannot reject null when a = 0.05 and a = 0.01
 Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
 Conclusion: There is insufficient evidence to conclude that the distribution of actual college majors of graduating females fits the distribution of their expected majors.
The hypotheses for the goodnessoffit test are:
 H_{0}: Surveyed obese fit the distribution of expected obese
 H_{a}: Surveyed obese do not fit the distribution of expected obese
Use a chisquare distribution with df = 4 to evaluate the data.
The test statistic is X^{2} = 9.85
The Pvalue = 0.0431
At 5% significance level, α = 0.05. For this data, P < α. Reject the null hypothesis.
At the 5% level of significance, from the data, there is sufficient evidence to conclude that the surveyed obese do not fit the distribution of expected obese.
 H_{0}: Car size is independent of family size.
 H_{a}: Car size is dependent on family size.
 df = 9
 chisquare distribution with df = 9
 test statistic = 15.8284
 Check student’s solution.

 Alpha: 0.05
 Decision: Cannot reject the null hypothesis.
 Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
 Conclusion: At the 5% significance level, there is insufficient evidence to conclude that car size and family size are dependent.
 H_{0}: Honeymoon locations are independent of bride’s age.
 H_{a}: Honeymoon locations are dependent on bride’s age.
 df = 9
 chisquare distribution with df = 9
 test statistic = 15.7027
 Check student’s solution.

 Alpha: 0.05
 Decision: Cannot reject the null hypothesis.
 Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
 Conclusion: At the 5% significance level, there is insufficient evidence to conclude that honeymoon location and bride age are dependent.
 H_{0}: The types of fries sold are independent of the location.
 H_{a}: The types of fries sold are dependent on the location.
 df = 6
 chisquare distribution with df = 6
 test statistic =18.8369
 Check student’s solution.

 Alpha: 0.05
 Decision: Cannot accept the null hypothesis.
 Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
 Conclusion: At the 5% significance level, There is sufficient evidence that types of fries and location are dependent.
 H_{0}: Salary is independent of level of education.
 H_{a}: Salary is dependent on level of education.
 df = 12
 chisquare distribution with df = 12
 test statistic = 255.7704
 Check student’s solution.

Alpha: 0.05
Decision: Cannot accept the null hypothesis.
Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
Conclusion: At the 5% significance level, there is sufficient evidence to conclude that salary and level of education are dependent.
 H_{0}: Age is independent of the youngest online entrepreneurs’ net worth.
 H_{a}: Age is dependent on the net worth of the youngest online entrepreneurs.
 df = 2
 chisquare distribution with df = 2
 test statistic = 1.76
 Check student’s solution.

 Alpha: 0.05
 Decision: Cannot reject the null hypothesis.
 Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
 Conclusion: At the 5% significance level, there is insufficient evidence to conclude that age and net worth for the youngest online entrepreneurs are dependent.
 H_{0}: The distribution for personality types is the same for both majors
 H_{a}: The distribution for personality types is not the same for both majors
 df = 4
 chisquare with df = 4
 test statistic = 3.01
 Check student’s solution.

 Alpha: 0.05
 Decision: Cannot reject the null hypothesis.
 Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
 Conclusion: There is insufficient evidence to conclude that the distribution of personality types is different for business and social science majors.
 H_{0}: The distribution for fish caught is the same in Green Valley Lake and in Echo Lake.
 H_{a}: The distribution for fish caught is not the same in Green Valley Lake and in Echo Lake.
 3
 chisquare with df = 3
 11.75
 Check student’s solution.

 Alpha: 0.05
 Decision: Cannot accept the null hypothesis.
 Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
 Conclusion: There is evidence to conclude that the distribution of fish caught is different in Green Valley Lake and in Echo Lake
 H_{0}: The distribution of average energy use in the USA is the same as in Europe between 2005 and 2010.
 H_{a}: The distribution of average energy use in the USA is not the same as in Europe between 2005 and 2010.
 df = 4
 chisquare with df = 4
 test statistic = 2.7434
 Check student’s solution.

 Alpha: 0.05
 Decision: Cannot reject the null hypothesis.
 Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
 Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the average energy use values in the US and EU are not derived from different distributions for the period from 2005 to 2010.
 H_{0}: The distribution for technology use is the same for community college students and university students.
 H_{a}: The distribution for technology use is not the same for community college students and university students.
 2
 chisquare with df = 2
 7.05
 pvalue = 0.0294
 Check student’s solution.

 Alpha: 0.05
 Decision: Cannot accept the null hypothesis.
 Reason for decision: pvalue < alpha
 Conclusion: There is sufficient evidence to conclude that the distribution of technology use for statistics homework is not the same for statistics students at community colleges and at universities.
 The test statistic is always positive and if the expected and observed values are not close together, the test statistic is large and the null hypothesis will be rejected.
 Testing to see if the data fits the distribution “too well” or is too perfect.