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78 . For a chi-square distribution with five degrees of freedom, the curve is ______________.

11.3: test of independence

Use the following information to answer the next four exercises. You are considering conducting a chi-square test of independence for the data in this table, which displays data about cell phone ownership for freshman and seniors at a high school. Your null hypothesis is that cell phone ownership is independent of class standing.

79 . Compute the expected values for the cells.

Cell = Yes Cell = No
Freshman 100 150
Senior 200 50

80 . Compute ( O E ) 2 z for each cell, where O = observed and E = expected.

81 . What is the chi-square statistic and degrees of freedom for this study?

82 . At the α = 0.5 significance level, what is your decision regarding the null hypothesis?

11.4: test of homogeneity

83 . You conduct a chi-square test of homogeneity for data in a five by two table. What is the degrees of freedom for this test?

11.5: comparison summary of the chi-square tests: goodness-of-fit, independence and homogeneity

84 . A 2013 poll in the State of California surveyed people about taxing sugar-sweetened beverages. The results are presented in the following table, and are classified by ethnic group and response type. Are the poll responses independent of the participants’ ethnic group? Conduct a hypothesis test at the 5% significance level.

Ethnic Group \ Response Type Favor Oppose No Opinion Row Total
White / Non-Hispanic 234 433 43 710
Latino 147 106 19 272
African American 24 41 6 71
Asian American 54 48 16 118
Column Total 459 628 84 1171

85 . In a test of homogeneity, what must be true about the expected value of each cell?

86 . Stated in general terms, what are the null and alternative hypotheses for the chi-square test of independence?

87 . Stated in general terms, what are the null and alternative hypotheses for the chi-square test of homogeneity?

11.6: test of a single variance

88 . A lab test claims to have a variance of no more than five. You believe the variance is greater. What are the null and alternative hypothesis to test this?

Practice test 3 solutions

8.1: confidence interval, single population mean, population standard deviation known, normal

1 . σ n = 4 30 = 0.73

2 . normal

3 . 0.025 or 2.5%; A 95% confidence interval contains 95% of the probability, and excludes five percent, and the five percent excluded is split evenly between the upper and lower tails of the distribution.

4 . z -score = 1.96; E B M =   z α 2 ( σ n ) = ( 1.96 ) ( 0.73 ) =   1.4308

5 . 41 ± 1.43 = (39.57, 42.43); Using the calculator function Zinterval, answer is (40.74, 41.26. Answers differ due to rounding.

6 . The z -value for a 90% confidence interval is 1.645, so EBM = 1.645(0.73) = 1.20085.
The 90% confidence interval is 41 ± 1.20 = (39.80, 42.20).
The calculator function Zinterval answer is (40.78, 41.23). Answers differ due to rounding.

7 . The standard error of measurement is: σ n =   4 50 = 0.57
E B M =   z α 2 ( σ n ) = ( 1.96 ) ( 0.57 ) =   1.12
The 95% confidence interval is 41 ± 1.12 = (39.88, 42.12).
The calculator function Zinterval answer is (40.84, 41.16). Answers differ due to rounding.

Questions & Answers

Two dice are thrown. Let A be the event that the sum of the upper face number is odd and be the event of at least one ace
Sana Reply
does proportion alwaus give u a yes or no answer for data
Adrienne Reply
frequency destribution
the Reply
प्रायिकता सिध्दान्त पर आधारित प्रतिदर्श सिध्दान्त का विकास किसने किया?
Lokesh Reply
7.The following data give thenumber of car thefts that occurred in a city in the past 12 days. 63711438726915 Calculate therange, variance, and standard deviation.
Mitu Reply
express the confidence interval 81.4% ~8.5% in interval form
Xx Reply
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Shazain Reply
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sharmin Reply
Calculate correlation coefficient, where SP(xy) = 144; SS(x) = 739; SS(y) = 58. (2 Points)
Ashfat Reply
The information are given from a randomly selected sample of age of COVID-19 patients who have already survived. These information are collected from 200 persons. The summarized information are as, n= 20; ∑x = 490; s^2 = 40. Calculate 95% confident interval of mean age.
Ashfat
The mode of the density of power of signal is 3.5. Find the probability that the density of a random signal will be more than 2.5.
Ashfat
The average time needed to repair a mobile phone set is 2 hours. If a customer is in queue for half an hour, what is the probability that his set will be repaired within 1.6 hours?
Ashfat
A quality control specialist took a random sample of n = 10 pieces of gum and measured their thickness and found the mean 9 and variance 0.04. Do you think that the mean thickness of the spearmint gum it produces is 8.4
nazrul Reply
3. The following are the number of mails received in different days by different organizations: Days (x) : 23, 35, 38, 50, 34, 60, 41, 32, 53, 67. Number of mails (y) : 18, 40, 52, 45, 32, 55, 50, 48, 26, 25. i) Fit a regression line of y on x and test the significance of regression. ii) Estimate y
Atowar Reply
The number of problem creating computers of two laboratories are as follows: Number of computers: 48, 6, 10, 12, 30, 11, 49, 17, 10, 14, 38, 25, 15, 19, 40, 12. Number of computers: 12, 10, 26, 11, 42, 11, 13, 12, 18, 5, 14, 38. Are the two laboratories similar in respect of problem creating compute
Tamim Reply
Is the severity of the drug problem in high school the same for boys and girls? 85 boys and 70 girls were questioned and 34 of the boys and 14 of the girls admitted to having tried some sort of drug. What can be concluded at the 0.05 level?
Ashfat Reply
null rejected
Pratik
a quality control specialist took a random sample of n=10 pieces of gum and measured their thickness and found the mean 7.6 and standered deviation 0.10. Do you think that the mean thickness of the spearmint gum it produces is 7.5?
Shanto Reply
99. A one sample, one-tail t-test is conducted and the test statistic value is calculated to be 2.56. The degrees of freedom for the test are 10. Which of the following conclusions for the test would be correct? a
Niaz Reply
A one sample, one-tail t-test is conducted and the test statistic value is calculated to be 2.56. The degrees of freedom for the test are 10. Which of the following conclusions for the test would be correct?
Niaz
what is null Hypothesis
Niaz
what is null Hypothesis
Niaz

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Source:  OpenStax, Introductory statistics. OpenStax CNX. May 06, 2016 Download for free at http://legacy.cnx.org/content/col11562/1.18
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