Respuesta :
Answer:
a) The P-value is 0.
At a significance level of 0.05, there is enough evidence to support the claim that the online platform was accessed significantly more so than before the stay at home order.
b) Effect size d = 0.77
Medium.
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that the online platform was accessed significantly more so than before the stay at home order.
Then, the null and alternative hypothesis are:
[tex]H_0: \mu=1000\\\\H_a:\mu> 1000[/tex]
The significance level is 0.05.
The sample has a size n=108000.
The sample mean is M=1378.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=489.
The estimated standard error of the mean is computed using the formula:
s_M=\dfrac{s}{\sqrt{n}}=\dfrac{489}{\sqrt{108000}}=1.488
Then, we can calculate the t-statistic as:
t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{1378-1000}{1.488}=\dfrac{378}{1.488}=254.036
The degrees of freedom for this sample size are:
[tex]df=n-1=108000-1=107999[/tex]
This test is a right-tailed test, with 107999 degrees of freedom and t=254.036, so the P-value for this test is calculated as (using a t-table):
[tex]\text{P-value}=P(t>254.036)=0[/tex]
As the P-value (0) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
At a significance level of 0.05, there is enough evidence to support the claim that the online platform was accessed significantly more so than before the stay at home order.
The effect size can be estimated with the Cohen's d.
This can be calculated as:
[tex]d=\dfrac{M-\mu}{\sigma}=\dfrac{1378-1000}{489}=\dfrac{378}{489}=0.77[/tex]
The values of Cohen's d between 0.2 and 0.8 are considered "Medium", so in this case, the effect size d=0.77 is medium.
