A data.frame: $15\backslash$times $5$

pH nisin temp brix growth

$15\mathrm{.}57043190$

$25\mathrm{.}55043131$

$35\mathrm{.}55035151$

$45\mathrm{.}53035131$

$55\mathrm{.}53025110$

$65\mathrm{.}5050190$

$75\mathrm{.}5025151$

$83\mathrm{.}57043150$

$93\mathrm{.}57035110$

$103\mathrm{.}55050130$

$113\mathrm{.}55035190$

$123\mathrm{.}53050110$

$133\mathrm{.}53043150$

$143\mathrm{.}5025190$

$155\mathrm{.}07025150$

Call:

glm$($formula $=$ growth ~ pH $+$ nisin $+$ temp $+$ brix, family $=$ "binomial",

data $=$ apple.data$)$

Deviance Residuals:

Min $1$Q Median $3$Q Max

$-2\mathrm{.}3245-0\mathrm{.}4325-0\mathrm{.}14150\mathrm{.}53081\mathrm{.}5593$

Coefficients:

Estimate Std$.$ Error z value Pr$(>|$z$|)$

$($Intercept$)-7\mathrm{.}683633\mathrm{.}28201-2\mathrm{.}3410\mathrm{.}019225*$

pH $2\mathrm{.}049080\mathrm{.}574813\mathrm{.}5650\mathrm{.}000364***$

nisin $-0\mathrm{.}062730\mathrm{.}01910-3\mathrm{.}2830\mathrm{.}001026**$

temp $0\mathrm{.}125320\mathrm{.}050792\mathrm{.}4670\mathrm{.}013614*$

brix $-0\mathrm{.}380000\mathrm{.}15909-2\mathrm{.}3890\mathrm{.}016915*$

$---$

Signif. codes: $0***0\mathrm{.}001**0\mathrm{.}01*0\mathrm{.}05.0\mathrm{.}11$

$($Dispersion parameter for binomial family taken to be $1)$

Null deviance: $95\mathrm{.}072$ on $72$ degrees of freedom

Residual deviance: $49\mathrm{.}844$ on $68$ degrees of freedom

AIC: $59\mathrm{.}844$

Number of Fisher Scoring iterations: $6$

$1.$ The Effects of Temp $($The correct code for question $1$ is shown below. Please help me answer question $2)$

What if we want to determine how a specific predictor affects the probability $($or odds, in the Logistic Regression case$)$ of growth$=1?$ One idea would be to calculate the odds of growth, given different levels of that predictor, while keeping all other predictors constant. Then we could compare the difference between the odds, to see if a larger predictor resulted in a larger probability.

Using your model, calculate the odds of growth with a temperature at the first quartile and at the third quartile, assuming all other features are held constant. Then calculate the difference between the two and store that value as temp.odds.diff.

To calculate this difference, it may be helpful to first think through this equation. Note that oi is the odds of growth for the it$$ quantile.

d$=$o$1/$o$3=$ exp$($log$($o$1/$o$3))=$ exp$($n$1$n$3)=...$

If we let this difference be d$,$ then this value can be interpreted as "The odds of showing evidence of growth is d$\%$ more$/$less when the temperature is in the first quantile than in the third quantile, when adjusted for other predictors."

mod$1<-$ predict$($glmod$.$apple, newdata $=$ data.frame$($temp $=$ quantile$($apple$.$data$temp, $0\mathrm{.}25),$

pH $=$ mean$($apple$.$data$pH$),$

nisin $=$ mean$($apple$.$data$nisin$),$

brix $=$ mean$($apple$.$data$brix$)))$

mod$3<-$ predict$($glmod$.$apple, newdata $=$ data.frame$($temp $=$ quantile$($apple$.$data$temp, $0\mathrm{.}75),$

pH $=$ mean$($apple$.$data$pH$),$

nisin $=$ mean$($apple$.$data$nisin$),$

brix $=$ mean$($apple$.$data$brix$)))$

odds$\_$first $<-$ exp$($mod$1)$

odds$\_$third $<-$ exp$($mod$3)$

temp.odds.diff $=$ odds$\_$first $/$ odds$\_$third

temp.odds.diff $=0\mathrm{.}366951038477902$

$2.$ But there's more than that.

Remember, we're assuming all of our predictors come from some distribution, meaning there is some inherent randomness in our values and calculations. A point$-$value is only so helpful. If we really want to understand the difference, we should calculate the range of values that the difference could potentially fall within.

Calculate the $95\%$ confidence interval for this difference. Store the lower bound in temp.odds.lower and the upper bound in temp.odds.upper.

Hint: You can get the Standard Error of temp from your model.

temp.odds.lower $=$ NA

temp.odds.upper $=$ NA

# your code here