Hello,

This is a more theoretical question as i'm quite new to this... I was wondering if i'm using this: ln wage = b1 + ... + b3 experience + b4 experience ^2

If i'm using a squared X- variabele to know the effect I should use derivates etc but i'm wondering what my hypothesistests will be for the variables experience and experience ^2. Do I need 2 seperate hypotheses or just one and use in the alternative what i'm expecting experiences effect to be on wage over time?

Thanks a lot, this is much appreciated!!

Consider the following three models

fit2 <- lm(mpg ~ hp + drat + I(drat^2) , data = mtcars)

summary(fit2)
#>
#> Call:
#> lm(formula = mpg ~ hp + drat + I(drat^2), data = mtcars)
#>
#> Residuals:
#>     Min      1Q  Median      3Q     Max
#> -5.1847 -2.4796 -0.3974  1.2020  7.5564
#>
#> Coefficients:
#>              Estimate Std. Error t value Pr(>|t|)
#> (Intercept)  2.314340  23.589436   0.098    0.923
#> hp          -0.052384   0.009573  -5.472 7.64e-06 ***
#> drat         9.499096  13.095200   0.725    0.474
#> I(drat^2)   -0.658769   1.789196  -0.368    0.715
#> ---
#> Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#>
#> Residual standard error: 3.219 on 28 degrees of freedom
#> Multiple R-squared:  0.7424, Adjusted R-squared:  0.7148
#> F-statistic:  26.9 on 3 and 28 DF,  p-value: 2.138e-08

fit <- lm(mpg ~ hp + drat, data = mtcars)

summary(fit)
#>
#> Call:
#> lm(formula = mpg ~ hp + drat, data = mtcars)
#>
#> Residuals:
#>     Min      1Q  Median      3Q     Max
#> -5.0369 -2.3487 -0.6034  1.1897  7.7500
#>
#> Coefficients:
#>              Estimate Std. Error t value Pr(>|t|)
#> (Intercept) 10.789861   5.077752   2.125 0.042238 *
#> hp          -0.051787   0.009293  -5.573 5.17e-06 ***
#> drat         4.698158   1.191633   3.943 0.000467 ***
#> ---
#> Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#>
#> Residual standard error: 3.17 on 29 degrees of freedom
#> Multiple R-squared:  0.7412, Adjusted R-squared:  0.7233
#> F-statistic: 41.52 on 2 and 29 DF,  p-value: 3.081e-09

fit_l <- lm(log(mpg) ~ hp + drat + I(drat^2), data = mtcars)

summary(fit_l)
#>
#> Call:
#> lm(formula = log(mpg) ~ hp + drat + I(drat^2), data = mtcars)
#>
#> Residuals:
#>       Min        1Q    Median        3Q       Max
#> -0.290377 -0.097397  0.001702  0.080792  0.305609
#>
#> Coefficients:
#>               Estimate Std. Error t value Pr(>|t|)
#> (Intercept)  1.5194249  1.1299116   1.345    0.190
#> hp          -0.0027235  0.0004585  -5.940 2.15e-06 ***
#> drat         0.8044142  0.6272477   1.282    0.210
#> I(drat^2)   -0.0798903  0.0857008  -0.932    0.359
#> ---
#> Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#>
#> Residual standard error: 0.1542 on 28 degrees of freedom
#> Multiple R-squared:  0.7579, Adjusted R-squared:  0.732
#> F-statistic: 29.22 on 3 and 28 DF,  p-value: 9.06e-09


The null hypothesis, H_0, is that the independent variables hp and drat have no association with mpg  at a given level of confidence equal to 1 - \alpha where \alpha (such as 0.05) is the probability of observing a test statistic of at least as extreme purely by chance. For these models, we can reject H_1 for the hp variable, but not drat or drat^2.

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