 Solution

This is Poisson regression with independent counts and log mean $\log {\mu }_d=1.5624+0.1658d$. The residual deviance $9.90$ on $12$ degrees of freedom has $p=0.625$, and the plotted curve tracks the data, so there is no evidence of poor fit.

The fitted doubling time is $\log 2/0.1658=4.16$ days. A 95% interval for the growth coefficient is $0.1658\pm 1.96(0.0166)=(0.133,0.198)$, corresponding to doubling times $(3.50,5.21)$ days. Seven days, whose rate is about $0.099$, is unsupported.