Question

A client of yours wants to find out the best microbial environment for C. elegans. In previous meetings, the client told you that C. elegans feed on bacteria but may also be killed by certain bacteria. Therefore, it is important to figure out what bacteria are beneficial to C. elegans. In particular, the client was interested in studying the association between the density of Gluconobacter and the density of C. elegans.

Answer 1

Let the model that is being estimated is

celegans = Bo + Bigluconobactor + error

Using the following

Coefficients: Estimate Std. Error t value pr(>lt) (Intercept) 11.92231 9.924 <2e-16 *** Gluconobacter 0.02722 0.07001 0.389 0.698

We know that the estimated value of the slope is

1 = 0.02722

The standard error of this estimate is

Sê, = 0.07001

98% confidence interval indicates a significance level

a=1 - 98/100 = 0.02

The right tail critical value of t is

PT > ta/2) = a/2 = 0.02/2 = 0.01

the number of observations, n=100 and the degrees of freedom are n-2=100-2=98

Using the standard t tables, for df=98 and area under the right tail=0.01 (or the combined area=0.02) we get

ta/2 = 2.365

The 98% confidence interval is

81 £ta/2 X 83. 20.02722 +2.365 x 0.07001 31-0.1384, 0.1928)

ans: The 98% confidence interval for $\begin{align*} &\hat{\beta}_1 \end{align*}$ is [-0.1384, 0.1928]

w/6-01

This indicates that we are 98% confident that for each
1 x 10-4g/ml
increase in the initial density of Gluconobactor, the predicted  final density of C. elegans increases on an average by a value in the interval  [-0.1384, 0.1928]
w/6-01

(Basically we are saying with 98% confidence that for each unit
(1/6-07 XL)
increase in the initial density of Gluconobactor, the predicted  final density of C. elegans will decrease at most by  
0.1384 x 10-4g/ml
or increase by at most
0.1928 x 10-4g/ml
)

4) We can calculate the F statistic by multiple methods

Method 1: Using the t statistic of the slope

The t statistic of the slope estimate is

t = 0,389

The F statistic is

F=+ = 0.3892 = 0.1513

Method II; Using the R-Square estimate

The R-square Estimate is

Residual standard error: 3.352 on 98 degrees of freedom Multiple R-squared: 0.00154, Adjusted R-squared: -0.008649 F-statistic: on 1 and 98 DF, p-value: 0.6983

RP = 0.00154

The F statistic is

RP (1 - R2)/(n - 2) 0.00154 = 0.1512 (1 -0.00154)/(100 - 2)

ans: The F statistic for testing the null hypothesis is 0.1512

Ho: B1 = 0 v.s. Ha: B1 +0

5) The p-value of the test on $\begin{align*}\beta_1 \end{align*}$ is

Coefficients: Estimate Std. Error t value Pr>It) (Intercept) 11.92231 1 9 .924 <2e-16 *** Gluconobacter 0.02722 0.07001 0.389 0.698

ans: The p-value of the test on $\begin{align*}\beta_1 \end{align*}$ is 0.698

The null hypothesis being tested here is

Ho : B1 = 0

That is, the null hypothesis is that there is no linear relationship between the initial density of Glucanobacter and the final density of C. elegans

alternatively, the null hypothesis is that the initial density of Glucanobacter cannot explain the final density of C. elegans.