Question 1181389: At the Foremost State Bank the average savings account balance in 2012 was $1100. A random sample of 38 savings account balanes for 2013 yielded a mean of $1800 with a standard deviation of $2800. At the α = 0.10 significance level test the claim that the mean savings account balance in 2013 is different from the mean savings account balance in 2012.
(a) Identify the correct alternative hypothesis.
μ _____ $1100
(b) The test statistic value is __________. (Round your answer to two decimal places.)
(c) Using the critical value approach (traditional method), the critical value is ____________ . (Round your answer to three decimal places.)
(d) Based on your answers above, do you
Reject the H^0
Fail to reject the H^0
(e) Explain your decision about the claim.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's a breakdown of the solution:
**(a) Alternative Hypothesis:**
The claim is that the mean savings account balance in 2013 is *different* from the mean savings account balance in 2012. "Different" means *not equal to*. Therefore, the correct alternative hypothesis is:
μ **≠** $1100
**(b) Test Statistic:**
We are given:
* Sample mean (x̄) = $1800
* Population mean (μ) = $1100
* Sample size (n) = 38
* Sample standard deviation (s) = $2800
Since the sample size is greater than 30, we use a t-test. The test statistic is:
t = (x̄ - μ) / (s / √n)
t = (1800 - 1100) / (2800 / √38)
t = 700 / (2800 / 6.1644)
t = 700 / 454.16
t ≈ 1.54
The test statistic value is approximately **1.54**.
**(c) Critical Value:**
We're using a two-tailed test (because H₁ is ≠) with α = 0.10. Degrees of freedom (df) = n - 1 = 38 - 1 = 37.
For a two-tailed test with α = 0.10 and df = 37, you'd consult a t-table or calculator. You're looking for the t-value that cuts off 0.05 in *each* tail (since it's two-tailed).
The critical values are approximately **±1.687**.
**(d) Decision:**
Our test statistic (1.54) falls *between* -1.687 and +1.687. It's *not* in the rejection region (the tails). Therefore, we *fail to reject* the null hypothesis.
**Fail to reject the H₀**
**(e) Explanation:**
Because our calculated test statistic (1.54) falls within the non-rejection region (between the critical values), we do not have sufficient evidence at the α = 0.10 significance level to reject the null hypothesis. In simpler terms, even though the sample mean is higher ($1800 vs $1100), there's too much variability in the sample (indicated by the large standard deviation) to confidently say that the true population mean for 2013 is *different* from $1100. The difference we see in the sample could reasonably be due to random chance.
|
|
|
