Answer to Question #181492 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 with a standard deviation of 3.32. 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. 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.