5. Shown is the crime rate in Alaska for a period of 15 years in the late 1960s

Question

5. Shown is the crime rate in Alaska for a period of 15 years in the late 1960s and 1970s. Find the least squares linear regression equation for this data (using population and violent crime rate total as the variables) and use it to estimate the number of violent crimes in Alaska in the year 1994, when the population of Alaska reached 606,000. If the actual number of violent crimes that year was 7663, how well does the linear equation model the data? Why? Year 1960 1961 1962 1963 1964 1965 1966 1967 1968 1969 Crimes in Alaska 237357 245439 258222 263345 266245 272761 294731 295538 304597 336637 352105 368665 378939 1971 1972

Explanation / Answer

consider, x as year and y as crimes in alska

Using Excel Command, We can find Slope and Intercept

So for Slope, Command is, "=slope(y,x)"

we get b=slope= 10754.42

And for Intercept, Command is "=Intercept(y,x)"

we get a= intercept = -20849331.8

so therefore Linear Regression Equation is given by

y=a+b*x

y= -20849331.8+10754.42*x

Now, We have to us this estimate to find no. of crimes in Alaska in the year 1994

So therefore y=-20849331.8+10754.42*(1994)

so y= 594981.68

Since the model is not well fitted as compared to crimes in Alska 7663

the coefficient of determination using Excel Command is, for tthis first we have to find coefficient of correlation and we square it to get coeeficient of determination

so R=correlation using Excel Command is "=correl(y,x)"

We get R=0.984

So therefore R^2 = (0.984)^2 = 0.969

That is 96.9% Variation in the Response(Crimes in Alaska) is explained by Year(x)

x y 1960 237357 1961 245439 1962 258222 1963 263345 1964 266245 1965 272761 1966 294731 1967 295538 1968 304597 1969 314826 1970 336637 1971 352105 1972 368665 1973 378939

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