Answer:
The P-value for this test is P=0.2415.
Step-by-step explanation:
We have to perform an hypothesis testing on the mean of alla account balances.
The claim is that the mean of all account balances is significantly greater than $1,150.
Then, the null and alternative hypothesis are:
[tex]H_0: \mu=1150\\\\H_a: \mu>1150[/tex]
The sample size is n=20, with a sample mean is 110 and standard deviation is 125.
We can calculate the t-statistic as:
[tex]t=\dfrac{\bar x-\mu}{s/\sqrt{n}}=\dfrac{1170-1150}{125/\sqrt{20}}=\dfrac{20}{27.95}=0.7156[/tex]
The degrees of freedom fot this test are:
[tex]df=n-1=20-1=19[/tex]
For this one-tailed test and 19 degrees of freedom, the P-value is:
[tex]P-value=P(t>0.7156)=0.2415[/tex]
