Answer to Question #181367 in Statistics and Probability for Cream

Question

The Coronavirus Disease (COVID-19) is an infectious disease caused by a new strain of 

coronavirus. The World Health Organization (WHO) claims that the incubation period of the virus 

in the infected person has a mean of 5.1 days. The doctors in the Philippines conducted a 

research and they found out that incubation period of the virus in human body is 6.03 days with a 

standard deviation of 3.32. The samples were 46 COVID patients. Is there enough evidence to 

conclude that the incubation period of the virus is 5.1 days as stated, at 𝛼 = 0. 01?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=5.1$

$H_1:\mu\not=5.1$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

Based on the information provided, the significance level is $\alpha=0.01,$ and the critical value for a two-tailed test and degrees of freedom $df=n-1=46-1=45$ is $t_c=2.690.$

The rejection region for this two-tailed test is $R=\{t:|t|>2.690\}.$  

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{6.03-5.1}{3.32/\sqrt{46}}=1.900$$

Since it is observed that $|t|=1.900<2.690=t_c,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is different than 5.1, at the 0.01 significance level.

Using the P-value approach:

The p-value for two-tailed test, $df =45, t=1.900$ is $p=0.063852,$ and since $p=0.063852>0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is different than 5.1, at the 0.01 significance level.