Answer:
Step-by-step explanation:
Given that:
The population mean μ = 900
The sample size n = 53
The sample mean = 841
The standard deviation = 190
The level of significance = 0.01
To test the claim that the mean balance of the accounts has decreased during the period.
The null and the alternative hypothesis can be computed as:
[tex]H_o : \mu \ge 900 \\ \\ H_1 : \mu < 900[/tex]
This is a left-tailed test since the alternative is less than 900
The test statistics can be computed as:
[tex]Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }[/tex]
[tex]Z = \dfrac{841 -900}{\dfrac{190}{\sqrt{53}} }[/tex]
[tex]Z = \dfrac{-59}{\dfrac{190}{7.28}}[/tex]
Z = -2.26
The p-value = P(Z < -2.26)
From the z-tables
The p-value = 0.01191
The p-value ≅ 0.012
Decision rule: To reject the null hypothesis if the significance level is > p-value.
We failed to reject the null hypothesis.
Conclusion: Thus, there is insufficient evidence to conclude that such accounts mean balance has decreased during this period.
