Question

for the given data, (a) find the test statistic, (b) find the standardized test statistic, (c) decide whether the standardized test statistic is in the rejection region, and (d) decide whether you should reject or fail to reject the null hypothesis. the samples are random and independent. claim: $\mu_1 < \mu_2$, $\alpha = 0.01$. sample statistics: $\bar{x}_1 = 1235$, $n_1 = 35$, $\bar{x}_2 = 1195$, and $n_2 = 55$. population parameters: $\sigma_1 = 75$ and $\sigma_2 = 100$. (a) the test statistic for $\mu_1 - \mu_2$ is 40. (b) the standardized test statistic for $\mu_1 - \mu_2$ is 2.16. (round to two decimal places as needed.) (c) is the standardized test statistic in the rejection region? no yes (d) should you reject or fail to reject the null hypothesis? $h_0: \mu_1 \geq \mu_2$; $h_a: \mu_1 < \mu_2$. fail to reject $h_0$. at the 1% significance level, there is evidence to support the claim.

Explanation:

Step1: Identify the test type

This is a two - sample z - test for the difference between two population means since the population standard deviations \(\sigma_1\) and \(\sigma_2\) are known. The formula for the standardized test statistic (z - statistic) for the difference between two means \(\mu_1\) and \(\mu_2\) is:

$$z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}}$$

For the null hypothesis \(H_0:\mu_1\geq\mu_2\) and alternative hypothesis \(H_a:\mu_1 < \mu_2\), the hypothesized difference \((\mu_1-\mu_2) = 0\) (under the null hypothesis).

Step2: Calculate the numerator

The numerator of the z - statistic formula is \((\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)\). We know that \(\bar{x}_1 = 1235\), \(\bar{x}_2=1195\) and \((\mu_1 - \mu_2)=0\) (from the null hypothesis). So, \((\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)=(1235 - 1195)-0=40\).

Step3: Calculate the denominator

The denominator is \(\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}\). We know that \(\sigma_1 = 75\), \(n_1 = 35\), \(\sigma_2=100\) and \(n_2 = 55\).
First, calculate \(\frac{\sigma_1^{2}}{n_1}=\frac{75^{2}}{35}=\frac{5625}{35}\approx160.7143\)
Second, calculate \(\frac{\sigma_2^{2}}{n_2}=\frac{100^{2}}{55}=\frac{10000}{55}\approx181.8182\)
Then, \(\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}=160.7143 + 181.8182=342.5325\)
And \(\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}=\sqrt{342.5325}\approx18.51\)

Step4: Calculate the z - statistic

Using the formula \(z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}}\), substitute the values of numerator and denominator:
\(z=\frac{40}{18.51}\approx2.16\) (rounded to two decimal places)

Step5: Determine the rejection region

For a left - tailed test with \(\alpha = 0.01\), the critical value \(z_{\alpha}\) is the z - score such that \(P(Z < z_{\alpha})=\alpha\). Looking up in the standard normal table, \(z_{0.01}=- 2.33\) (since it is a left - tailed test, the critical value is negative). The rejection region is \(z < - 2.33\). Our calculated z - statistic is \(z = 2.16\), which is greater than \(- 2.33\), so it is not in the rejection region.

Step6: Decision about the null hypothesis

Since the standardized test statistic is not in the rejection region, we fail to reject the null hypothesis \(H_0\). At the \(\alpha=0.01\) significance level, there is not enough evidence to support the claim that \(\mu_1<\mu_2\).

Answer:

(a) The test statistic for \(\mu_1-\mu_2\) is \(\boldsymbol{40}\).
(b) The standardized test statistic for \(\mu_1 - \mu_2\) is \(\boldsymbol{2.16}\).
(c) No
(d) Fail to reject \(H_0\). At the 1% significance level, there is not enough evidence to support the claim.