Question 1181389
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.