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 \\( \square \\). (round to two decimal places as needed.)

Explanation:

Step1: Recall the formula for standardized test statistic (z - test) for two - sample means

The formula for the standardized test statistic \(z\) when testing the difference between two population means \(\mu_1\) and \(\mu_2\) (when population standard deviations \(\sigma_1\) and \(\sigma_2\) are known) 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 left - tailed test with the claim \(\mu_1<\mu_2\), the null hypothesis \(H_0:\mu_1\geq\mu_2\) and the alternative hypothesis \(H_a:\mu_1 < \mu_2\). Under the null hypothesis, \(\mu_1-\mu_2 = 0\).

Step2: Substitute the given values into the formula

We know that \(\bar{x}_1 = 1235\), \(\bar{x}_2=1195\), \(\sigma_1 = 75\), \(\sigma_2 = 100\), \(n_1 = 35\), \(n_2=55\) and \(\mu_1-\mu_2 = 0\) (under \(H_0\)).
First, calculate the numerator: \((\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)=(1235 - 1195)-0=40\)
Then, calculate the denominator:
\(\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}=\sqrt{\frac{75^{2}}{35}+\frac{100^{2}}{55}}\)
Calculate \(\frac{75^{2}}{35}=\frac{5625}{35}\approx160.7143\)
Calculate \(\frac{100^{2}}{55}=\frac{10000}{55}\approx181.8182\)
Sum these two values: \(160.7143 + 181.8182=342.5325\)
Take the square root: \(\sqrt{342.5325}\approx18.51\)

Step3: Calculate the z - statistic

Now, divide the numerator by the denominator:
\(z=\frac{40}{18.51}\approx2.16\) (Wait, no, wait. Wait, the claim is \(\mu_1<\mu_2\), so \(\bar{x}_1-\bar{x}_2=1235 - 1195 = 40\), but since we are testing \(\mu_1<\mu_2\), the numerator is \((\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)\). But in the left - tailed test, let's re - check. Wait, the formula is correct. Wait, \(\bar{x}_1-\bar{x}_2=40\), \(\mu_1-\mu_2 = 0\) (under \(H_0\)). But wait, if the claim is \(\mu_1<\mu_2\), then \(\bar{x}_1-\bar{x}_2\) is positive here (1235>1195), which seems a bit counter - intuitive, but let's proceed with the calculation.
Wait, \(\sqrt{\frac{75^{2}}{35}+\frac{100^{2}}{55}}=\sqrt{\frac{5625}{35}+\frac{10000}{55}}\)
\(\frac{5625}{35}\approx160.714\), \(\frac{10000}{55}\approx181.818\)
Sum: \(160.714 + 181.818=342.532\)
Square root: \(\sqrt{342.532}\approx18.51\)
Then \(z=\frac{40}{18.51}\approx2.16\)? Wait, no, wait. Wait, the test is left - tailed, but the difference \(\bar{x}_1-\bar{x}_2\) is positive. But let's check the formula again. The formula for the two - sample z - test for means (when \(\sigma_1\) and \(\sigma_2\) are known) is:
\[z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}}\]
Since the null hypothesis is \(H_0:\mu_1-\mu_2\geq0\) (because the claim is \(\mu_1<\mu_2\), so \(H_0\) is the complement), so \(\mu_1 - \mu_2 = 0\) (the equality part of the null hypothesis). So the numerator is \((\bar{x}_1-\bar{x}_2)-0=\bar{x}_1-\bar{x}_2 = 40\)
The denominator is \(\sqrt{\frac{75^{2}}{35}+\frac{100^{2}}{55}}=\sqrt{\frac{5625}{35}+\frac{10000}{55}}\approx\sqrt{160.714 + 181.818}=\sqrt{342.532}\approx18.51\)
Then \(z=\frac{40}{18.51}\approx2.16\)? Wait, but if the claim is \(\mu_1<\mu_2\), we expect \(\bar{x}_1-\bar{x}_2\) to be negative, but here \(\bar{x}_1 = 1235\) and \(\bar{x}_2 = 1195\), so \(\bar{x}_1-\bar{x}_2=40\) is positive. But let's just follow the formula.

Wait, maybe I made a mistake in the sign. Wait, the claim is \(\mu_1<\mu_2\), so \(\mu_1-\mu_2<0\). The 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 \(H_0\), \(\mu_1-\mu_2 = 0\) (for the left - tailed test, \(H_0:\mu_1\geq\mu_2\), so the critical value is negative). But the sample difference is positive. Let's recalculate the denominator:

\(\frac{\sigma_1^{2}}{n_1}=\frac{75^2}{35}=\frac{5625}{35}\approx160.7143\)

\(\frac{\sigma_2^{2}}{n_2}=\frac{100^2}{55}=\frac{10000}{55}\approx181.8182\)

\(\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}=160.7143 + 181.8182 = 342.5325\)

\(\sqrt{342.5325}\approx18.51\)

Numerator: \((\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)=(1235 - 1195)-0 = 40\)

So \(z=\frac{40}{18.51}\approx2.16\)? Wait, but that seems wrong because the claim is \(\mu_1<\mu_2\), and the sample mean of the first group is higher than the second. But maybe the calculation is correct. Let's check with another approach.

Wait, the formula for the z - score in two - sample z - test (when population standard deviations are known) is:

\[z=\frac{(\bar{x}_1-\bar{x}_2)-D_0}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}\]

where \(D_0=\mu_1-\mu_2\) under the null hypothesis. For \(H_0:\mu_1\geq\mu_2\), \(D_0 = 0\) (since we are testing if \(\mu_1-\mu_2\geq0\) against \(\mu_1-\mu_2<0\)).

So \(\bar{x}_1-\bar{x}_2=1235 - 1195 = 40\), \(D_0 = 0\)

\(\sqrt{\frac{75^2}{35}+\frac{100^2}{55}}=\sqrt{\frac{5625}{35}+\frac{10000}{55}}\approx\sqrt{160.71 + 181.82}=\sqrt{342.53}\approx18.51\)

Then \(z=\frac{40 - 0}{18.51}\approx2.16\)

Answer:

(a) The test statistic for \(\mu_1-\mu_2\) is \(40\) (as given).
(b) The standardized test statistic for \(\mu_1-\mu_2\) is \(\approx2.16\)