Answer:
The P-value is 0.0234.
Step-by-step explanation:
We are given that a statistics practitioner calculated the mean and the standard deviation from a sample of 400. They are x = 98 and s = 20.
Let [tex]\mu[/tex] = population mean.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 100 {means that the population mean is equal to 100}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 100 {means that the population mean is more than 100}
The test statistics that will be used here is One-sample t-test statistics because we don't know about population standard deviation;
T.S. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean = 98
s = sample standard deviation = 20
n = sample size = 400
So, the test statistics = [tex]\frac{98-100}{\frac{20}{\sqrt{400} } }[/tex] ~ [tex]t_3_9_9[/tex]
= -2
The value of t-test statistics is -2.
Now, the P-value of the test statistics is given by;
P([tex]t_3_9_9[/tex] < -2) = 0.0234 {using the t-table}