Question

find the standardized test statistic, t, to test the claim that $\mu_1 = \mu_2$. two samples populations that are normal. the sample statistics are given below. assume that $n_1 = 25, n_2 = 30, \bar{x}_1 = 27, \bar{x}_2 = 25, s_1 = 1.5, s_2 = 1.9$ \\(\bigcirc\\) a. 3.287 \\(\bigcirc\\) b. 1.986 \\(\bigcirc\\) c. 2.892 \\(\bigcirc\\) d. 4.361

Explanation:

Step1: Recall the formula for the t - statistic for two - sample means

The formula for the t - statistic when testing the claim that \(\mu_1=\mu_2\) (assuming unequal variances, we can also use the pooled - variance formula, but since the problem doesn't specify, we can use the formula for the two - sample t - test: \(t=\frac{(\bar{x}_1 - \bar{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}}\). Since the claim is \(\mu_1 = \mu_2\), then \(\mu_1-\mu_2 = 0\), so the formula simplifies to \(t=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}}\)

Step2: Substitute the given values into the formula

We are given that \(n_1 = 25\), \(n_2=30\), \(\bar{x}_1 = 27\), \(\bar{x}_2 = 25\), \(s_1 = 1.5\), \(s_2=1.9\)

First, calculate the numerator: \(\bar{x}_1-\bar{x}_2=27 - 25=2\)

Then, calculate the denominator:
\(\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}=\sqrt{\frac{1.5^{2}}{25}+\frac{1.9^{2}}{30}}\)
\(=\sqrt{\frac{2.25}{25}+\frac{3.61}{30}}\)
\(=\sqrt{0.09+\frac{3.61}{30}}\)
\(=\sqrt{0.09 + 0.1203\overline{3}}\)
\(=\sqrt{0.2103\overline{3}}\approx\sqrt{0.2103}\approx0.4586\)

Step3: Calculate the t - statistic

\(t=\frac{2}{0.4586}\approx4.361\) Wait, no, wait. Wait, I made a mistake in the denominator calculation. Let's recalculate the denominator:

\(\frac{s_1^{2}}{n_1}=\frac{1.5^{2}}{25}=\frac{2.25}{25} = 0.09\)

\(\frac{s_2^{2}}{n_2}=\frac{1.9^{2}}{30}=\frac{3.61}{30}\approx0.1203\)

Then \(\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}=0.09 + 0.1203=0.2103\)

\(\sqrt{0.2103}\approx0.4586\)

Then \(t=\frac{2}{0.4586}\approx4.361\)? Wait, but let's check again. Wait, maybe I used the wrong formula. Wait, the formula for the two - sample t - test when the population variances are unknown (which is the case here) is correct. Wait, but let's recalculate:

Wait, \(\bar{x}_1-\bar{x}_2 = 2\)

\(\frac{s_1^{2}}{n_1}=\frac{2.25}{25}=0.09\)

\(\frac{s_2^{2}}{n_2}=\frac{3.61}{30}\approx0.1203\)

Sum: \(0.09 + 0.1203=0.2103\)

Square root of sum: \(\sqrt{0.2103}\approx0.4586\)

Then \(t=\frac{2}{0.4586}\approx4.361\)

Wait, but let's check the answer options. Option D is 4.361. But wait, maybe I made a mistake in the formula. Wait, another way: the formula for the two - sample t - test (pooled variance) is \(t=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{s_p^{2}(\frac{1}{n_1}+\frac{1}{n_2})}}\), where \(s_p^{2}=\frac{(n_1 - 1)s_1^{2}+(n_2 - 1)s_2^{2}}{n_1 + n_2-2}\)

Let's try the pooled - variance formula.

First, calculate \(s_p^{2}\):

\(n_1-1 = 24\), \(n_2 - 1=29\)

\(s_p^{2}=\frac{24\times1.5^{2}+29\times1.9^{2}}{25 + 30-2}=\frac{24\times2.25+29\times3.61}{53}\)

\(24\times2.25 = 54\), \(29\times3.61=104.69\)

\(s_p^{2}=\frac{54 + 104.69}{53}=\frac{158.69}{53}\approx2.994\)

Then \(\sqrt{s_p^{2}(\frac{1}{n_1}+\frac{1}{n_2})}=\sqrt{2.994\times(\frac{1}{25}+\frac{1}{30})}\)

\(\frac{1}{25}+\frac{1}{30}=\frac{6 + 5}{150}=\frac{11}{150}\approx0.0733\)

\(2.994\times0.0733\approx0.2195\)

\(\sqrt{0.2195}\approx0.4685\)

Then \(t=\frac{2-0}{0.4685}\approx4.269\). Hmm, there is a discrepancy. Wait, maybe the problem assumes unequal variances. Let's go back to the first formula (unequal variances, Welch - Satterthwaite approximation, but for the purpose of calculation, we can use the formula \(t=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}}\))

\(\frac{s_1^{2}}{n_1}=\frac{2.25}{25}=0.09\), \(\frac{s_2^{2}}{n_2}=\frac{3.61}{30}\approx0.1203\)

Sum: \(0.09 + 0.1203 = 0.2103\)

Square root: \(\sqrt{0.2103}\approx0.4586\)

\(t=\frac{2}{0.4586}\approx4.361\), which is option D.

Wait, but let's check the calculation again. Wait, \(\bar{x}_1-\bar{x}_2=2\), \(\frac{s_1^{2}}{n_1}=\frac{2.25}{25}=0.09\), \(\frac{s_2^{2}}{n_2}=\frac{3.61}{30}\approx0.1203\). Then \(\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}=0.09 + 0.1203 = 0.2103\). The square root of 0.2103 is approximately \(\sqrt{0.2103}\approx0.4586\). Then \(t=\frac{2}{0.4586}\approx4.361\), which matches option D.

Answer:

D. 4.361