Presentation on theme: "5.7 SCATTER PLOTS AND TREND LINES:"— Presentation transcript:
1 5.7 SCATTER PLOTS AND TREND LINES: Scatter Plot: a graph that relates two different sets of data by displaying them as ordered pairs (x, y).Correlation: The relationship/trend found in any given data.Trend line: Line on a scatter plot, drawn near the points, that shows a correlation
2 Interpolation: The act of estimating a value between two known values. Extrapolation: The act of predicting a value outside of the range of known values.Line of Best Fit: Line that shows the most accurate relationship between two sets of data.
3 Correlation Coefficient(r): a number from -1 to 1 that tells us how closely the equation models the data.Causation: A change in one quantity causes a change in a second quantity.
10 Whenever we are given data/information/ordered pairs, we must be able to provide certain details: Ex: Make a scatter plot of the data, provide the type of relationship it represents and the approximate weight of a 7-month-old panda.Weight of a PandaAge (months)1234681012Weight (lbs)2.57.612.517.124.337.949.254.9
11 To answer the questions on the panda task we must do three procedures: Procedure 1: Create a Scatter PlotProcedure 2: Write an Equation of the Trend of the LineProcedure 3: Estimate the weight of a 7-month-old panda.
13 𝒚 =5.2𝒙 – 3.7 Procedure 2: Write an equation of the Trend Using A(4, 17.1) and B(8, 37.9), points on the positive correlation line, we find the slopeAgeWeight12.527.6312.5417.1624.3837.91049.21254.9m= 𝟑𝟕.𝟗 −𝟏𝟕.𝟏 𝟖−𝟒 = 𝟐𝟎.𝟖 𝟒 = 5.2Using one of the two points and the point-slope form equation: 𝒚- 𝒚𝟏 = m(𝒙-𝒙𝟏) we get:𝒚 =5.2(𝒙 - 4)𝒚 =5.2𝒙 – 20.8𝒚 =5.2𝒙 –𝒚 =5.2𝒙 – 3.7
14 Thus a 7-month-old panda will weight about 32.7 lbs. Procedure 3: Estimate the weight of the 7-month-old pandaUsing the found equation of the Trend Line:𝒚 =5.2𝒙 – 3.7and letting x = 7 months,we get:𝒚 =5.2(7) – 3.7𝒚 =36.4 – 𝒚 =32.7Thus a 7-month-old panda will weight about 32.7 lbs.
15 YOU TRY IT: Use the data below to create a scatter plot, provide the relationship and approximate the daily temperature in January at a latitude of 50o N.LatitudeTemp3546335230672576433240373944
16 Negative Correlation Trend Line Procedure 1: Scatter PlotLatitude(x)Temp (y)35463352306725764332403739448070605040Temperature (o F)302010Negative Correlation Trend Line20253035404550Latitude (o N)
17 𝒚 = -3𝒙+157 Procedure 2: Write an equation of the Trend Using A(30, 67) and B(40, 37), points on the negative correlation line, we find the slope(x)(y)3546335230672576433240373944m= 𝟑𝟕−𝟔𝟕 𝟒𝟎−𝟑𝟎 = −𝟑𝟎 𝟏𝟎 = - 3Using one of the two points and the point-slope form equation: 𝒚- 𝒚𝟏 = m(𝒙-𝒙𝟏) we get:𝒚- 67 = -3 (𝒙 - 30)𝒚- 𝟔𝟕 =-3𝒙+𝟗𝟎𝒚 =-3𝒙𝒚 = -3𝒙+157
18 Thus at latitude of 50o N the temperature will be about 7o F. Procedure 3: Estimate the temperature of the 50oN:Using the found equation of the Trend Line:𝒚 = -3𝒙+157and letting x = 50o N of latitudewe get:𝒚 =-3(50)+𝟏𝟓𝟕𝒚 = – 𝒚 =7Thus at latitude of 50o N the temperature will be about 7o F.