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Unformatted text preview: Determine whether the scatter diagram indicates that a C
. . . . '1
linear relatton mav exrst between the two variables. If m “D m the relation is linear. determine whether it indicates a a 2'] positive or negative association between the variables. £3: 1'3
I] Answer both questions below. [I 2] 1]
Explanatory iIlE LlﬂLﬂ. IJ‘U‘IIlLb LlU‘ IlU‘L Ilﬂ‘r'b‘ El. [IIUL‘HI ill]: LlﬂLii PU‘ElLb LlU‘ IlU‘L Ilﬂﬁ‘L‘ a ﬂIlL‘ijl
relationship because thev do not lie relationship because thev lie mainlv in a
mainlv in a straight line. straight line. V The data points have a linear The data points have a linear
relationship because thev lie mainlv in a relationship because thev do not lie
straight line. mainlv in a straight line. Do the two variables have a positive or a negative association? V The two variables have a positive association. Determine whether the scatter diagram indicates that a ‘ . . . . '1 linear relation mav exrst between the two variables. If :1: “U i t. t . . . . . . . n U
the relauon 1s linear. determine whether It 1ndrcates a a ‘5' = g t .
positive or negative association between the variables. Dig: 1'3 I i
_ I]
Answer both questtons below.
Explanatmy
lﬂb‘ LlﬂLH. PUﬂlLb LLU' IlU'L nave 'd. DIDL‘HI iIl': LlﬂLﬂ PUElLb ﬂﬂ'Ir'l: H. IIIUL‘HI
relationship because thev lie mainlv in a relationship because thev do not lie
straight line. mainlv in a straight line.
H“ Ihe data points have a linear The data points do not have a linear relationship because thev lie mainlv in a relationship because thev do not lie
straight line. mainlv in a straight line. Do the two variables have a positive or a negative association”? The two variables have a positive association. V The two variables have a negative association. Identify the explanatory variable and the response variable. A dentist vvants to determine if the type of toothbrush used evervdav can be used to predict the
amount of cavities for a patient's next appoinunent. Ihe explanatory variable. or the independent variable. is the variable that is controlled to
determine the effect of the study: The explanatorv variable is the tvpe of toothbrush. The response variable. or the dependent variable. is the variable of interest whose value can be
determined ﬁ'om a studv. Ihe response variable is the amount of cavities. Identifv the explanatorv variable and the response variable. A golfer wants to determine if the number of lessons received everv year can be used to predict
the amount of improvement in his game. The explanatorv variable is the number of lessons The response variable is the amount of improvement in his game _ Ihe altitude (in thousands of feet} and speed of a plane (in feet per second) for various altitudes is shown
below. Cornlete arts a through c . n as
one no.4 (a) Display the data in a scatter plot. To plot the points. plot the altitude on the xaxis and the corresponding ‘1:
speeds on the '_‘."EIJ{:15. The scatter plot is shown to the right. 11m] 
1: WEI]
=1
.33 1020 .
aao . ' I] 1] 2] 3] ‘E
Altitude Eb} Calculate the correlation coefﬁcient r. Ihe correlation coefﬁcient is a measure of the strength and the direction of a linear relationship between
two variables. It is deﬁned bf; the equation ans=(Zx)(Zs') JnExl(fo where n is the number ofpairs of data. To help calculate the correlation coefﬁcient ﬁrst complete the table below. 933.6632?
1.2 32.26309
920.619.1114
1.032.138.36
990.821.16
1.01 6.46124
1.022.624.49
1.248.382.26 113
13
2D
23
3D
33 Now ﬁnd the sum of each column of the table. 931659.29
Lasamooo
soomom
LEE—3.13356 990321.16
1.016.46124
1__o?2__o?4_49
1.248.582.1'6 =s.so4__?3s_os 1:With the sums found above and n= 8. End the correlation coefﬁcient. ans=(Ex)(2v) anxI—(zxf 3045.9113— (14o) (32319} f Substitute.
iis(s__soo)—14o‘ vinasodrosaosy 823?.9‘ 14.4?4 = — Sim lifv each ex ession.
ill 3.499 ill l?=l._912.I53 p ' pr Simplify. 14.4?4 ii 3.400 ii l?4.912.o?: Therefore. the correlation coefﬁcient is r =3 0.3?8. =5 [LET'3 (c) Make a conclusion about the type of correlation. The range of the correlation coefﬁcient is  1 to 1. If x and v have a suong positive linear
correlation. r is close to 1. If x and v have a strong negative linear correlation. r is close to  1.
If there is a weak linear correlation r is close to [L If there is no linear correlation. r is equal to El
or extremer close to U (with an absolute value less than ELI}. Let an absolute value of r greater than III.5 indicate a strong linear relationship. and an absolute
value of r less than [LS and greater than III.1 indicate a weak linear relationship. Using the
information found above and recalling that r = [LE—ES. we see that the correlation is a vveak positive linear correlation. Ihe altitude (in thousands of feet) and speed of a plane (in feet per sec ond) for various altitudes
is shown below. Cornlete arts a through c . “ A
.— H"
1130
v31 w: '31
E Hi E
s a s
d: in “1
ﬁll]
—E ‘El
_i'_l[i[u:13 Sp eed
Eb) Calculate the correlation coefﬁcient r.
r = £1.38? (Round to three decimal places as needed.)
(c) Make a conclusion about the type of correlation.
The correlation is a weak positive linear correlation. T The number of hours 10 students spent studying for a test and their scores on that test are
shown in the table. Is there enough evidence to conclude that there is a signiﬁcant linear
correlation between the data? Use or = [1.01. nun mm Is there enough evidence to conclude that there is a signiﬁcant linear correlation? H" Yes there is enough evidence at the 1% level to conclude that there is a signiﬁcant
linear correlation. No. there is not enough evidence at the 1% level to conclude that there is no signiﬁcant
linear correlation No. there is not enough evidence at the 1% level to conclude that there is a signiﬁcant
linear correlation. Tfest there is enough evidence at the 1% level to conclude that there is no signiﬁcant
linear correlation. The descriptions in the left cohnnn match the descriptions in the right cohrrnn. Slope n1
yinter'c ept b
The savalue of a data point corresponding to xr. vr. Ihe }=s'alue for a point on the regression line
corresponding to xr. .1 Find the equation of the regression line for the given data. Ihen construct a scatter plot of the data and draw the
regression line. (Ihe pair of variables have a signiﬁcant correlation.) Ihen use the regression equation to predict
the value of y for each of the given xvalues if meaningful. The number of hours r5 students spent for a test and
their scores on that test are shown below. I: (willows (hm—shows
m chx=11hours (mm hm Ihe equation of a regression line for an independent variable x and a dependent variable y is mx b._ w
where y is the predicted yvalue for a given xvalue. The slope m and yintercept b are given by the formulas below where is the mean of the yvalues in the data set and E is the mean of the xvalues. To ﬁnd the
equation of the regression line for the given data. use technology or the formulas below.  ewe meg—m: anl{ijl n n 1iFir“bile you can ﬁnd the equation of the regression line for me given data using formulas or technology in this
problem we will use technology. L'se technology to ﬁnd the regression equation. §= 5.242x— 31.511
Now consnuct a scatter plot of the data and draw the regression line. In a scatter plot the ordered pairs (xy)
are graphed as points in a coordinate plane. The independent variable x is measured by the horizontal axis
and the dependent variable y is measured by the vertical axis.
Now determine the ordered pairs to be plotted. (1.35). (2.44). (3.51). (4.43}. (6.63}. (1:59) To draw the regression line. use any two points within the range of data and calculate their corresponding
yval'ues ﬁom the regression line. I'hen draw a line through the two points. Ihus the scatter plot of the data and the regression line is shown to Q:
the right. 5'3 _ iii 36313
I Hours studying (a) Nov: use the regression equation to predict the value of v for each of the given xvalues if meaningﬁJl.
Prediction values are meaningﬁJl onlv for xval'ues in the range of the data. Begin v.ith x = 5. Since x = 5 is in the range of the original data it is meaningﬁJl to predict the value of v for
x = 5. Substitute 5 for x into the regression equation. then calculate the predicted vvalue. n j; = 5.242x—31.5_.I'1
= 5.242(5)31.5'.I'1 Substitute 5 for x.
=5 511.8 Calculate. Eb) Nov: predict the value of v for x = 2.5. Since x: 2.5 is in the range of the original data it is meaningﬁJl to
predict the value off; for x = 2.5. Now substitute 2.5 for x into the regression equation. then calculate the predicted vvalue. n 3.. = 5.242x315'1‘1
= 5.242(2.5)31.5'.I‘1 Subs1itute 2.5 for x.
=5 44.? Calculate. (c) Next predict the value of v for x = 1 1. Since x = 1 1 is not in the range of the original data it is not
meaningful to predict the value off; for x = 1 1. (d) Finallv predict the value of v for x = 3.5. Since x = 3.5 is in the range of the original data it is meaningﬁJl to
predict the value off; for x = 3.5. Since it is meaningﬁJl to predict the value of substitute 3.5 for x into the regression equation. then calculate the predicted vvalue. n j; = 524231—3157“
= 5.242(3.5)—31.5'.I‘1 Substitute 3.5 for x.
=5 49.9 Calculate. ...
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This note was uploaded on 12/30/2009 for the course MATH 1410 taught by Professor Vergo during the Spring '09 term at Metropolitan Community College Omaha.
 Spring '09
 vergo
 Statistics

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