Question

find the standardized test statistic to test the claim that $\mu_1 < \mu_2$. two sampl population. the sample statistics are given below. $n_1 = 35, n_2 = 42, \bar{x}_1 = 27.09, \bar{x}_2 = 29.64, \sigma_1 = 2.9, \sigma_2 = 2.8$ \bigcirc a. $-2.63$ \bigcirc b. $-3.16$ \bigcirc c. $-3.90$ \bigcirc d. $-1.66$

Explanation:

Step1: Recall the formula for the standardized test statistic (z - test) for two - sample means when population standard deviations are known.

The formula for the z - test statistic for testing the difference between two population means \(\mu_1\) and \(\mu_2\) is:

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

In the claim \(\mu_1<\mu_2\), the null hypothesis is \(H_0:\mu_1=\mu_2\), so \(\mu_1-\mu_2 = 0\). The formula simplifies to:

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

Step2: Substitute the given values into the formula.

We are given \(n_1 = 35\), \(n_2=42\), \(\bar{x}_1 = 27.09\), \(\bar{x}_2=29.64\), \(\sigma_1 = 2.9\), \(\sigma_2 = 2.8\).
First, calculate the numerator: \(\bar{x}_1-\bar{x}_2=27.09 - 29.64=- 2.55\)
Then, calculate the denominator:

$$
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Step3: Calculate the z - statistic.

Now, divide the numerator by the denominator:

$$z=\frac{- 2.55}{0.653}\approx - 3.90$$

(Wait, let's recalculate the denominator more accurately)

Wait, let's recalculate the denominator:

\(\frac{\sigma_1^{2}}{n_1}=\frac{2.9^{2}}{35}=\frac{8.41}{35}\approx0.2403\)

\(\frac{\sigma_2^{2}}{n_2}=\frac{2.8^{2}}{42}=\frac{7.84}{42}\approx0.1867\)

Sum: \(0.2403 + 0.1867=0.427\)

Square root of \(0.427\): \(\sqrt{0.427}\approx0.653\)

Wait, numerator is \(27.09 - 29.64=-2.55\)

Wait, maybe I made a mistake in the first calculation of the numerator. Wait, \(27.09-29.64=- 2.55\)? Wait, 29.64 - 27.09 = 2.55, so 27.09 - 29.64=-2.55. Then \(z=\frac{-2.55}{\sqrt{\frac{2.9^{2}}{35}+\frac{2.8^{2}}{42}}}\)

Wait, let's recalculate the denominator:

\(\frac{2.9^2}{35}=\frac{8.41}{35}\approx0.2403\)

\(\frac{2.8^2}{42}=\frac{7.84}{42}\approx0.1867\)

Sum: \(0.2403+0.1867 = 0.427\)

\(\sqrt{0.427}\approx0.653\)

Then \(z=\frac{- 2.55}{0.653}\approx - 3.90\)? Wait, but let's check the calculation again.

Wait, maybe I miscalculated the numerator. Wait, \(\bar{x}_1 = 27.09\), \(\bar{x}_2 = 29.64\), so \(\bar{x}_1-\bar{x}_2=27.09 - 29.64=-2.55\). Correct.

Denominator:

\(\frac{\sigma_1^2}{n_1}=\frac{2.9^2}{35}=\frac{8.41}{35}\approx0.2403\)

\(\frac{\sigma_2^2}{n_2}=\frac{2.8^2}{42}=\frac{7.84}{42}\approx0.1867\)

Sum: \(0.2403 + 0.1867=0.427\)

Square root of \(0.427\): \(\sqrt{0.427}\approx0.653\)

Then \(z=\frac{-2.55}{0.653}\approx - 3.90\). So the test statistic is approximately - 3.90.

Answer:

C. - 3.90