Question 1192251
(a) State the null hypothesis H0 and the alternative hypothesis H1.

H0: μ1 = μ2 (The mean annual income of childcare workers in Arizona is equal to the mean annual income of childcare workers in Nevada)

H1: μ1 < μ2 (The mean annual income of childcare workers in Arizona is less than the mean annual income of childcare workers in Nevada)

(b) Determine the type of test statistic to use.

Since the population standard deviations are known and the populations are approximately normally distributed, we will use a z-test for the difference between two population means.

(c) Find the value of the test statistic. (Round to three or more decimal places.)

To calculate the z-test statistic, we can use the following formula:

```
z = (x̄1 - x̄2) / √((σ1²/n1) + (σ2²/n2))
```

Where:

* x̄1 and x̄2 are the sample means for Arizona and Nevada, respectively
* σ1 and σ2 are the population standard deviations for Arizona and Nevada, respectively
* n1 and n2 are the sample sizes for Arizona and Nevada, respectively

Plugging in the given values:

```
z = (39495.7 - 43425.9) / √((6300²/10) + (6500²/10))
z = -3930.2 / √((39690000/10) + (42250000/10))
z = -3930.2 / √(8194000)
z ≈ -1.371
```

(d) Find the critical value at the 0.05 level of significance. (Round to three or more decimal places.)

Since this is a one-tailed test with a significance level of 0.05, we need to find the z-value that corresponds to the 5th percentile in the standard normal distribution. Using a z-table or calculator, we find that the critical value is approximately -1.645.

(e) Can we support the claim that the mean annual income of childcare workers in Arizona is less than the mean annual income of childcare workers in Nevada?

No. The test statistic (-1.371) is greater than the critical value (-1.645). Therefore, we fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the mean annual income of childcare workers in Arizona is less than the mean annual income of childcare workers in Nevada.