Question

find the standardized test statistic to test the claim that $\mu_1 < \mu_2$. two samples are random 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 z - test statistic for two - sample means

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

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

Since we are testing the claim that \(\mu_1<\mu_2\), the null hypothesis is \(H_0:\mu_1=\mu_2\) (so \(\mu_1-\mu_2 = 0\)). So 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:
First, calculate \(\frac{\sigma_1^{2}}{n_1}=\frac{(2.9)^{2}}{35}=\frac{8.41}{35}\approx0.2403\)
Second, calculate \(\frac{\sigma_2^{2}}{n_2}=\frac{(2.8)^{2}}{42}=\frac{7.84}{42}\approx0.1867\)
Then, \(\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}=\sqrt{0.2403 + 0.1867}=\sqrt{0.427}\approx0.6535\)

Step3: Calculate the z - statistic

Now, substitute the numerator and denominator into the z - formula:

$$z=\frac{-2.55}{0.6535}\approx - 3.90$$

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

Wait, let's recalculate \(\frac{\sigma_1^{2}}{n_1}=\frac{2.9^{2}}{35}=\frac{8.41}{35}\approx0.240286\)
\(\frac{\sigma_2^{2}}{n_2}=\frac{2.8^{2}}{42}=\frac{7.84}{42}\approx0.186667\)
Sum of the two fractions: \(0.240286+0.186667 = 0.426953\)
Square root of the sum: \(\sqrt{0.426953}\approx0.6534\)
Now, numerator: \(27.09 - 29.64=-2.55\)
\(z=\frac{-2.55}{0.6534}\approx - 3.90\) (Wait, but let's check again. Wait, maybe I made a mistake in the numerator? Wait, 27.09 - 29.64 is - 2.55? Wait, 29.64 - 27.09 is 2.55, so 27.09 - 29.64 is - 2.55. Then denominator:

Wait, maybe I miscalculated the denominator. Let's do it more precisely:

\(\sigma_1^2=2.9\times2.9 = 8.41\), \(n_1 = 35\), so \(\frac{\sigma_1^2}{n_1}=\frac{8.41}{35}\approx0.2402857\)

\(\sigma_2^2 = 2.8\times2.8=7.84\), \(n_2 = 42\), so \(\frac{\sigma_2^2}{n_2}=\frac{7.84}{42}\approx0.1866667\)

Sum: \(0.2402857+0.1866667 = 0.4269524\)

Square root: \(\sqrt{0.4269524}\approx0.6534\)

Then \(z=\frac{27.09 - 29.64}{0.6534}=\frac{-2.55}{0.6534}\approx - 3.90\)

Wait, but let's check the options. Option C is - 3.90. But wait, maybe I made a mistake in the numerator? Wait, 27.09 - 29.64 is - 2.55? Let's check 29.64 - 27.09 = 2.55, so yes, 27.09 - 29.64=-2.55.

Wait, but let's recalculate the denominator again:

\(\frac{\sigma_1^2}{n_1}=\frac{2.9^2}{35}=\frac{8.41}{35}\approx0.240286\)

\(\frac{\sigma_2^2}{n_2}=\frac{2.8^2}{42}=\frac{7.84}{42}\approx0.186667\)

Sum: 0.240286 + 0.186667=0.426953

Square root: \(\sqrt{0.426953}\approx0.6534\)

Then \(z=\frac{-2.55}{0.6534}\approx - 3.90\)

Answer:

C. - 3.90