Question

To test how long coronavirus can stay viable (infectious) on plastic surfaces, 10 sample plastic surfaces were tested for the duration that the virus could be viable on them. The sample mean and standard deviation were found to be 70 and 6 hours, respectively. We can assume that the population has a normal distribution. (Disclaimer: These numbers are not based on actual research.)

a) Based on this test, what is the 99% confidence interval for the mean viability duration of coronavirus on cardboard? BONUS question: What do you learn from your result, practically? (2 bonus marks)

b) According to the sample that was collected, coronavirus loses its viability on plastic after an average of 70 hours. Assuming that the actual standard deviation of the viability duration on plastic is 3.5 hours, how many samples must be taken to be 99% confident that the error of estimating this time is less than 2 hours?

c) Considering the information about the sample, can we be convinced that the actual viability duration has a standard deviation of 3.5 hours?

Answer 1

R-codes:

n=10
xbar=70
s=6
alpha=0.01
ta2=qt(1-alpha/2,df=n-1)
ta2
xbar-ta2*s/sqrt(n)
xbar+ta2*s/sqrt(n)
qnorm(0.005)
(2.575829*3.5/2)^2
chical=(10-1)*s^2/(3.5^2)
chical
qchisq(alpha/2,df=10-1)
qchisq(1-alpha/2,df=10-1)

R-commands and outputs:

> n=10
> xbar=70
> s=6
> alpha=0.01
> ta2=qt(1-alpha/2,df=n-1)
> ta2
[1] 3.249836

# a)
> xbar-ta2*s/sqrt(n)
[1] 63.83387
> xbar+ta2*s/sqrt(n)
[1] 76.16613

# Cross-check
> b=76.16613
> a=63.83387
> b1=(b-70)/(6/sqrt(10))
> pt(b1,9)
[1] 0.995
> a1=(a-70)/(6/sqrt(10))
> pt(a1,9)
[1] 0.004999997
> pt(b1,9)-pt(a1,9)
[1] 0.99

# b)
> qnorm(0.005)
[1] -2.575829
> n=(2.575829*3.5/2)^2
> n
[1] 20.31937
> round(n)
[1] 20

# c)
> chical=(10-1)*s^2/(3.5^2)
> chical
[1] 26.44898
> qchisq(alpha/2,df=10-1)
[1] 1.734933
> qchisq(1-alpha/2,df=10-1)
[1] 23.58935
# chical > chitab, Reject H0.