Q.15 The University of Chicago鈥檚 Ge... [FREE SOLUTION]

91影视

The Practice of Statistics for AP

David Moore,Daren Starnes,Dan Yates

$Math Studyset 91影视 Explanations$ Math

4th Edition 路 809 pages

ISBN: 9781319113339

Chapter 11: Q.15 (page 694)

The University of Chicago鈥檚 General Social Survey (GSS) is the nation鈥檚 most important social science sample survey. For reasons known only to social scientists, the GSS regularly asks a random sample of people their astrological sign. Here are the counts of responses from a recent GSS:
$12$ If births are spread uniformly across the year, we expect all 12 signs to be equally likely. Are these data inconsistent with that belief? Carry out an appropriate test to support your answer. If you find a significant result, perform a follow-up analysis

Short Answer

Expert verified

There is insufficient evidence to conclude that the data is consistent with the belief .

Step by step solution

Given information

Given the question, the General Social Survey (GSS) of the University of Chicago is the nation's most important social science sample survey. The GSS frequently asks a random sample of people their astrological sign for reasons only social scientists understand. The following are the response counts from a recent GSS:

If births are evenly distributed throughout the year, all 12 signs should be equally likely.

We must determine whether these data contradict that belief.

Explanation

The dataset is,

Count
$321$
$360$
$367$
$374$
$383$
$402$
$392$
$329$
$331$
$354$
$376$
$355$

The following formula can be used to compute the test statistic:

$x^{2} = 鈭� \frac{(O - E)^{2}}{E}$

The null and alternative hypotheses are as follows:

$H_{0} : p_{1} = 0.0833$

$H_{0} : p_{2} = 0.0833$

$H_{0} : p_{3} = 0.0833$

$H_{0} : p_{4} = 0.0833$

$H_{0} : p_{5} = 0.0833$

$H_{0} : p_{6} = 0.0833$

$H_{0} : p_{7} = 0.0833$

$H_{0} : p_{8} = 0.0833$

$H_{0} : p_{9} = 0.0833$

$H_{0} : p_{10} = 0.0833$

$H_{0} : p_{11} = 0.0833$

$H_{0} : p_{12} = 0.0833$

$H_{a} :$ At least one of $p_{i}$ is incorrect

Calculation for test statistic

The test statistic is calculated as follows:

Observed Value	Expected value	(O-E)	${(O - E)}^{2}$	$\frac{{(O - E)}^{2}}{E}$
$321$	$362$	$- 41$	$1681$	$4.46436$
$360$	$362$	$- 2$	$4$	$0.011$
$367$	$362$	$6$	$25$	$0.0691$
$374$	$362$	$12$	$144$	$0.3978$
$383$	$362$	$21$	$441$	$1.2182$
$402$	$362$	$40$	$1600$	$4.4199$
$392$	$362$	$30$	$900$	$2.4862$
$329$	$362$	$- 22$	$1089$	$3.0083$
$331$	$362$	$- 31$	$961$	$2.6547$
$354$	$362$	$- 8$	$64$	$0.1768$
$376$	$362$	$14$	$196$	$0.5414$
$355$	$362$	$- 7$	$49$	$0.1354$
				localid="1653293158775" $饾毢 = 19.7624$

Test statistic

The following is the test statistic:

$蠂^{2} = 鈭� \frac{(O - E)^{2}}{E} = 19.7624$
The degree of freedom is determined as follows:

$D e g r e e o f f r e e d o m = N u m b e r o f c a t e g o r i e s - 1 = 12 - 1 = 11$

At $11$ degrees of freedom, the chi-square table $p$ -value is $0.00054$ . The $p$ -value is not statistically significant. The null hypothesis is discovered to be incorrect.