Using predict()?

I'm trying to solve a homework problem using R. The problem gives a list
of cricket chirps per second and corresponding temperature, and asks to
give the equation for the linear model and then predict the temperature
to produce 18 chirps per second. So far, I have:

# Homework 11.2.1 and 11.3.3
chirps <- scan()

1: 20
2: 16
3: 19.8
4: 18.4
5: 17.1
6: 15.5
7: 14.7
8: 17.1
9: 15.4
10: 16.2
11: 15
12: 17.2
13: 16
14: 17
15: 14.4
16: 
Read 15 items

temp <- scan()

1: 88.6
2: 71.6
3: 93.3
4: 84.3
5: 80.6
6: 75.2
7: 69.7
8: 82
9: 69.4
10: 83.3
11: 79.6
12: 82.5
13: 80.6
14: 83.5
15: 76.3
16: 
Read 15 items

chirps

 [1] 20.0 16.0 19.8 18.4 17.1 15.5 14.7 17.1 15.4 16.2 15.0 17.2 16.0
17.0 14.4

temp

 [1] 88.6 71.6 93.3 84.3 80.6 75.2 69.7 82.0 69.4 83.3 79.6 82.5 80.6
83.5 76.3

chirps.res <- lm(chirps ~ temp)
summary(chirps.res)

Call:
lm(formula = chirps ~ temp)

Residuals:
     Min       1Q   Median       3Q      Max 
-1.56146 -0.58088  0.02972  0.58807  1.53047 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) -0.31433    3.10963  -0.101 0.921028    
temp         0.21201    0.03873   5.474 0.000107 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 

Residual standard error: 0.9715 on 13 degrees of freedom
Multiple R-Squared: 0.6975,     Adjusted R-squared: 0.6742 
F-statistic: 29.97 on 1 and 13 DF,  p-value: 0.0001067

# From the linear model summary output above, the equation for the

least squares line is:

#    y = -0.3143 + 0.2120*x or chirps = -0.3143 + 0.2120*temp

I can then determine the answer to the prediction, using algebra and R:

pred_temp <- (18+0.3143)/0.2120
pred_temp

[1] 86.3882

However, I'd like to try to use the predict() function. Since 'chirps'
and 'temp' are just vectors of numbers, and not dataframes, these
failed:
predict(chirps.res, newdata=data.frame(chirp=18))
predict(chirps.res, newdata="chirp=18")
predict(chirps.res, newdata=18)

I then tried to turn my two vectors into a dataframe. I would have bet
money that this would have worked, but it didn't:

df <- data.frame(chirps, temp)
 chirps.res <- lm(chirps ~ temp, data=df)
predict(chirps.res, newdata=data.frame(chirps=18))

Can anyone tell me how to use predict() in this circumstance?

Thanks for your help and advice.

-Kevin

Kevin Zembower
Internet Services Group manager
Center for Communication Programs
Bloomberg School of Public Health
Johns Hopkins University
111 Market Place, Suite 310
Baltimore, Maryland  21202
410-659-6139