Question

a nationwide job recruiting firm wants to compare the annual incomes for childcare workers in pennsylvania and illinois. due to recent trends in the childcare industry, the firm suspects that the mean annual income of childcare workers in the state of pennsylvania is greater than the mean annual income of childcare workers in illinois. to see if this is true, the firm selected a random sample of 15 childcare workers from pennsylvania and an independent random sample of 15 childcare workers from illinois and asked them to report their mean annual income. the data obtained were as follows. the population standard deviations for the annual incomes of childcare workers in pennsylvania and in illinois are estimated as 6400 and 6100, respectively. it is also known that both populations are approximately normally distributed. at the 0.01 level of significance, is there sufficient evidence to support the claim that the mean annual income, μ1, of childcare workers in pennsylvania is greater than the mean annual income, μ2, of childcare workers in illinois? perform a one - tailed test. then complete the parts below. carry your intermediate computations to at least three decimal places. (if necessary, consult a list of formulas.) (a) state the null hypothesis h0 and the alternative hypothesis h1. (b) determine the type of test statistic to use. (c) find the value of the test statistic. (round to three or more decimal places.) (d) find the critical value at the 0.01 level of significance. (round to three or more decimal places.) (e) can we support the claim that the mean annual income of childcare workers in pennsylvania is greater than the mean annual income of childcare workers in illinois? yes no

Explanation:

Step1: State hypotheses

The null hypothesis $H_0$ is that the mean annual income of childcare - workers in Pennsylvania is less than or equal to that in Illinois, i.e., $H_0:\mu_1\leq\mu_2$. The alternative hypothesis $H_1$ is that the mean annual income of childcare - workers in Pennsylvania is greater than that in Illinois, i.e., $H_1:\mu_1 > \mu_2$.

Step2: Determine test - statistic type

Since the population standard deviations $\sigma_1 = 6400$ and $\sigma_2=6100$ are known and the populations are approximately normally distributed, we use a two - sample z - test.

Step3: Calculate the sample means

Let $n_1 = n_2=15$.
For Pennsylvania:
$\bar{x}_1=\frac{43157 + 38402+\cdots+45809}{15}=\frac{704964}{15}=46997.6$
For Illinois:
$\bar{x}_2=\frac{37137 + 41885+\cdots+44904}{15}=\frac{637058}{15}=42470.533$

Step4: Calculate the z - test statistic

The formula for the two - sample z - test statistic is $z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}}$. Under $H_0$, $\mu_1-\mu_2 = 0$.
$z=\frac{(46997.6 - 42470.533)-0}{\sqrt{\frac{6400^{2}}{15}+\frac{6100^{2}}{15}}}$
$=\frac{4527.067}{\sqrt{\frac{40960000}{15}+\frac{37210000}{15}}}$
$=\frac{4527.067}{\sqrt{\frac{40960000 + 37210000}{15}}}$
$=\frac{4527.067}{\sqrt{\frac{78170000}{15}}}$
$=\frac{4527.067}{\sqrt{5211333.333}}$
$=\frac{4527.067}{2282.834}\approx1.983$

Step5: Find the critical value

For a one - tailed test with $\alpha = 0.01$, the critical value $z_{\alpha}$ from the standard normal distribution table is $z_{0.01}=2.326$.

Step6: Make a decision

Since the calculated z - value $z = 1.983

Answer:

(a) $H_0:\mu_1\leq\mu_2$, $H_1:\mu_1 > \mu_2$
(b) z - test
(c) $1.983$
(d) $2.326$
(e) No