{"id":112,"date":"2021-01-12T22:19:54","date_gmt":"2021-01-12T22:19:54","guid":{"rendered":"https:\/\/textbooks.jaykesler.net\/introstats\/chapter\/prediction\/"},"modified":"2021-05-11T20:14:02","modified_gmt":"2021-05-11T20:14:02","slug":"prediction","status":"publish","type":"chapter","link":"https:\/\/textbooks.jaykesler.net\/introstats\/chapter\/prediction\/","title":{"rendered":"Prediction"},"content":{"raw":"\r\n[latexpage]\r\n<\/span>\r\n
\r\n

Recall the third exam\/final exam example<\/a>.<\/p>\r\n

We examined the scatterplot and showed that the correlation coefficient is significant. We found the equation of the best-fit line for the final exam grade as a function of the grade on the third-exam. We can now use the least-squares regression line for prediction.<\/p>\r\n

Suppose you want to estimate, or predict, the mean final exam score of statistics students who received 73 on the third exam. The exam scores (x<\/em>-values)<\/strong> range from 65 to 75. Since 73 is between the x<\/em>-values 65 and 75<\/strong>, substitute x<\/em> = 73 into the equation. Then:<\/p>\r\n$$\\hat y=\u2212173.51+4.83(73)=179.08$$\r\n

We predict that statistics students who earn a grade of 73 on the third exam will earn a grade of 179.08 on the final exam, on average.<\/p>\r\n\r\n

\r\n

Example <\/span>12.11<\/span><\/h3>\r\n<\/header>
\r\n
\r\n

Recall the third exam\/final exam example<\/a>.<\/p>\r\n \r\n

<\/header>
\r\n
\r\n
\r\n

a. What would you predict the final exam score to be for a student who scored a 66 on the third exam?<\/p>\r\n\r\n

\r\n
\r\n
\r\n\r\n \r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n
<\/header>
\r\n
\r\n
\r\n

b. What would you predict the final exam score to be for a student who scored a 90 on the third exam?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
<\/div>\r\n
\r\n

Solution <\/span>12.11<\/span><\/h4>\r\n
\r\n

a. 145.27<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n

\r\n
\r\n
\r\n

b. The x<\/em> values in the data are between 65 and 75. Ninety is outside of the domain of the observed x<\/em> values in the data (independent variable), so you cannot reliably predict the final exam score for this student. (Even though it is possible to enter 90 into the equation for x<\/em> and calculate a corresponding y<\/em> value, the y<\/em> value that you get will not be reliable.)\r\n\r\n<\/span>\r\n<\/span><\/p>\r\nTo understand really how unreliable the prediction can be outside of the observed x<\/em> values observed in the data, make the substitution x<\/em> = 90 into the equation.\r\n\r\n$$\\hat y=\u2013173.51+4.83(90)=261.19$$\r\n\r\nThe final-exam score is predicted to be 261.19. The largest the final-exam score can be is 200.\r\n\r\n<\/span>\r\n

<\/header>
\r\n
\r\n\r\nNote<\/span>\r\n

The process of predicting inside of the observed x<\/em> values observed in the data is called interpolation<\/span>. The process of predicting outside of the observed x<\/em> values observed in the data is called extrapolation<\/span>.<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n

\r\n

Try It <\/span>12.11<\/span><\/h3>\r\n<\/header>
\r\n
\r\n
\r\n
\r\n

Data are collected on the relationship between the number of hours per week practicing a musical instrument and scores on a math test. The line of best fit is as follows:<\/p>\r\n

\u0177<\/em> = 72.5 + 2.8x<\/em>\r\n\r\n<\/span>Assuming you have confirmed that the correlation coefficient is significant, what would you predict the score on a math test would be for a student who practices a musical instrument for five hours a week?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n

If you are trying to predict an outcome but the correlation coefficient is not significant, we can simply use the mean of the $y$-data as our predicted value, since the $x$-data and the $y$-data are not well correlated in this case.<\/div>\r\n \r\n\r\n<\/div>","rendered":"


\n[latexpage]
\n<\/span><\/p>\n

\n

Recall the third exam\/final exam example<\/a>.<\/p>\n

We examined the scatterplot and showed that the correlation coefficient is significant. We found the equation of the best-fit line for the final exam grade as a function of the grade on the third-exam. We can now use the least-squares regression line for prediction.<\/p>\n

Suppose you want to estimate, or predict, the mean final exam score of statistics students who received 73 on the third exam. The exam scores (x<\/em>-values)<\/strong> range from 65 to 75. Since 73 is between the x<\/em>-values 65 and 75<\/strong>, substitute x<\/em> = 73 into the equation. Then:<\/p>\n

$$\\hat y=\u2212173.51+4.83(73)=179.08$$<\/p>\n

We predict that statistics students who earn a grade of 73 on the third exam will earn a grade of 179.08 on the final exam, on average.<\/p>\n

\n
\n

Example <\/span>12.11<\/span><\/h3>\n<\/header>\n
\n
\n

Recall the third exam\/final exam example<\/a>.<\/p>\n

 <\/p>\n

\n
<\/header>\n
\n
\n
\n

a. What would you predict the final exam score to be for a student who scored a 66 on the third exam?<\/p>\n

\n
\n
\n
\n
\n

 <\/p>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n

\n
<\/header>\n
\n
\n
\n

b. What would you predict the final exam score to be for a student who scored a 90 on the third exam?<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/div>\n

\n
<\/div>\n
\n

Solution <\/span>12.11<\/span><\/h4>\n
\n

a. 145.27<\/p>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n

\n
\n
\n
\n
\n

b. The x<\/em> values in the data are between 65 and 75. Ninety is outside of the domain of the observed x<\/em> values in the data (independent variable), so you cannot reliably predict the final exam score for this student. (Even though it is possible to enter 90 into the equation for x<\/em> and calculate a corresponding y<\/em> value, the y<\/em> value that you get will not be reliable.)
\n
\n<\/span>
\n<\/span><\/p>\n

To understand really how unreliable the prediction can be outside of the observed x<\/em> values observed in the data, make the substitution x<\/em> = 90 into the equation.<\/p>\n

$$\\hat y=\u2013173.51+4.83(90)=261.19$$<\/p>\n

The final-exam score is predicted to be 261.19. The largest the final-exam score can be is 200.
\n
\n<\/span><\/p>\n

\n
<\/header>\n
\n
\n

Note<\/span><\/p>\n

The process of predicting inside of the observed x<\/em> values observed in the data is called interpolation<\/span>. The process of predicting outside of the observed x<\/em> values observed in the data is called extrapolation<\/span>.<\/p>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n

\n
\n

Try It <\/span>12.11<\/span><\/h3>\n<\/header>\n
\n
\n
\n
\n
\n

Data are collected on the relationship between the number of hours per week practicing a musical instrument and scores on a math test. The line of best fit is as follows:<\/p>\n

\u0177<\/em> = 72.5 + 2.8x<\/em>
\n
\n<\/span>Assuming you have confirmed that the correlation coefficient is significant, what would you predict the score on a math test would be for a student who practices a musical instrument for five hours a week?<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n

If you are trying to predict an outcome but the correlation coefficient is not significant, we can simply use the mean of the $y$-data as our predicted value, since the $x$-data and the $y$-data are not well correlated in this case.<\/div>\n

 <\/p>\n<\/div>\n","protected":false},"author":1,"menu_order":5,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-112","chapter","type-chapter","status-publish","hentry"],"part":103,"_links":{"self":[{"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/chapters\/112","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/wp\/v2\/users\/1"}],"version-history":[{"count":4,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/chapters\/112\/revisions"}],"predecessor-version":[{"id":404,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/chapters\/112\/revisions\/404"}],"part":[{"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/parts\/103"}],"metadata":[{"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/chapters\/112\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/wp\/v2\/media?parent=112"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/chapter-type?post=112"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/wp\/v2\/contributor?post=112"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/wp\/v2\/license?post=112"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}