Math / Data & Statistics

QuestionUse the given statistics to complete parts (a) and (b). Assume that the populations are normally distributed. (a) Test whether $\mu_{1}>\mu_{2}$ at the $\alpha=0.05$ level of significance for the given sample data. (b) Construct a $99 \%$ confidence interval about $\mu_{1}-\mu_{2}$ . \begin{tabular}{ccc} & Population 1 & Population 2 \\ \hline $\mathbf{n}$ & 22 & 15 \\ \hline $\overline{\mathbf{x}}$ & 50.7 & 42.1 \\ \hline $\mathbf{s}$ & 4.6 & 10.6 \end{tabular} (a) Identify the null and alternative hypotheses for this test. A. $H_{0}: \mu_{1}=\mu_{2}$ B. $H_{0}: \mu_{1}>\mu_{2}$ C. $H_{0}: \mu_{1} \neq \mu_{2}$ $H_{1}: \mu_{1}<\mu_{2}$ $H_{1}: \mu_{1}=\mu_{2}$ $H_{1}: \mu_{1}=\mu_{2}$ D. $H_{0}: \mu_{1}<\mu_{2}$ E. $H_{0}: \mu_{1}=\mu_{2}$ F. $H_{0}: \mu_{1}=\mu_{2}$ $H_{1}: \mu_{1}=\mu_{2}$ $H_{1}: \mu_{1} \neq \mu_{2}$ $H_{1}: \mu_{1}>\mu_{2}$

Studdy Solution

STEP 1

What is this asking? We're checking if there's a real difference between the averages of two groups, and if so, how big that difference might be. Watch out! Don't mix up the sample sizes and standard deviations – each group has its own!
Also, remember that a confidence interval is about the difference between the means, not the means themselves.

STEP 2

1. Hypothesis Test
2. Confidence Interval

STEP 3

We're testing if the mean of population 1, $\mu_1$ , is greater than the mean of population 2, $\mu_2$ .

STEP 4

Our null hypothesis $H_0$ is that there's no difference, meaning $\mu_1 = \mu_2$ .
The alternative hypothesis $H_1$ is what we're trying to prove: $\mu_1 > \mu_2$ .
This matches option F.

STEP 5

Since we don't know the true population standard deviations, we'll use a t-test.
The formula for the test statistic is: $t = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$ where $\bar{x}_1$ and $\bar{x}_2$ are the sample means, $s_1$ and $s_2$ are the sample standard deviations, and $n_1$ and $n_2$ are the sample sizes.

STEP 6

Under the null hypothesis, $\mu_1 - \mu_2 = 0$ .
Plugging in our values, we get: $t = \frac{(50.7 - 42.1) - 0}{\sqrt{\frac{4.6^2}{22} + \frac{10.6^2}{15}}} = \frac{8.6}{\sqrt{\frac{21.16}{22} + \frac{112.36}{15}}} = \frac{8.6}{\sqrt{0.9618 + 7.4907}} = \frac{8.6}{\sqrt{8.4525}} \approx \frac{8.6}{2.9073} \approx 2.958$ So our t-statistic is approximately 2.958.

STEP 7

With $\alpha = 0.05$ and degrees of freedom $df = \min(n_1 - 1, n_2 - 1) = \min(21, 14) = 14$ , our critical t-value is approximately 1.761.
Since our calculated t-statistic (2.958) is greater than the critical t-value (1.761), we reject the null hypothesis.
There's evidence to suggest $\mu_1 > \mu_2$ !

STEP 8

For a 99% confidence interval, our $\alpha$ is $1 - 0.99 = 0.01$ , so $\alpha/2 = 0.005$ .
With 14 degrees of freedom, our critical t-value for a two-tailed test is approximately 2.977.

STEP 9

The margin of error is calculated as: $E = t_{\alpha/2} \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$ Plugging in our values: $E = 2.977 \cdot \sqrt{\frac{4.6^2}{22} + \frac{10.6^2}{15}} \approx 2.977 \cdot 2.9073 \approx 8.644$

STEP 10

The confidence interval is: $(\bar{x}_1 - \bar{x}_2) \pm E$ $(50.7 - 42.1) \pm 8.644$ $8.6 \pm 8.644$ This gives us an interval of approximately $(-0.044, 17.244)$ .

STEP 11

(a) We reject the null hypothesis and conclude that $\mu_1 > \mu_2$ at the 0.05 significance level. (b) The 99% confidence interval for $\mu_1 - \mu_2$ is approximately $(-0.044, 17.244)$ .

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Studdy Solution

STEP 1

STEP 2

1. Hypothesis Test
2. Confidence Interval

STEP 3

We're testing if the mean of population 1, $\mu_1$ , is greater than the mean of population 2, $\mu_2$ .

STEP 4

Our null hypothesis $H_0$ is that there's no difference, meaning $\mu_1 = \mu_2$ .
The alternative hypothesis $H_1$ is what we're trying to prove: $\mu_1 > \mu_2$ .
This matches option F.

STEP 5

STEP 6

STEP 7

STEP 8

For a 99% confidence interval, our $\alpha$ is $1 - 0.99 = 0.01$ , so $\alpha/2 = 0.005$ .
With 14 degrees of freedom, our critical t-value for a two-tailed test is approximately 2.977.

STEP 9

The margin of error is calculated as: $E = t_{\alpha/2} \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$ Plugging in our values: $E = 2.977 \cdot \sqrt{\frac{4.6^2}{22} + \frac{10.6^2}{15}} \approx 2.977 \cdot 2.9073 \approx 8.644$

STEP 10

The confidence interval is: $(\bar{x}_1 - \bar{x}_2) \pm E$ $(50.7 - 42.1) \pm 8.644$ $8.6 \pm 8.644$ This gives us an interval of approximately $(-0.044, 17.244)$ .

STEP 11

(a) We reject the null hypothesis and conclude that $\mu_1 > \mu_2$ at the 0.05 significance level. (b) The 99% confidence interval for $\mu_1 - \mu_2$ is approximately $(-0.044, 17.244)$ .

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Studdy Solution

STEP 1

STEP 2

1. Hypothesis Test2. Confidence Interval

STEP 3

We're testing if the mean of population 1, μ1\mu_1μ1​, is *greater than* the mean of population 2, μ2\mu_2μ2​.

STEP 4

Our **null hypothesis** H0H_0H0​ is that there's *no difference*, meaning μ1=μ2\mu_1 = \mu_2μ1​=μ2​.The **alternative hypothesis** H1H_1H1​ is what we're trying to prove: μ1>μ2\mu_1 > \mu_2μ1​>μ2​.This matches option F.

STEP 5

STEP 6

STEP 7

STEP 8

For a 99% confidence interval, our α\alphaα is 1−0.99=0.011 - 0.99 = 0.011−0.99=0.01, so α/2=0.005\alpha/2 = 0.005α/2=0.005.With 14 degrees of freedom, our **critical t-value** for a two-tailed test is approximately **2.977**.

STEP 9

STEP 10

The **confidence interval** is: (xˉ1−xˉ2)±E (\bar{x}_1 - \bar{x}_2) \pm E (xˉ1​−xˉ2​)±E (50.7−42.1)±8.644 (50.7 - 42.1) \pm 8.644 (50.7−42.1)±8.6448.6±8.644 8.6 \pm 8.644 8.6±8.644This gives us an interval of approximately (−0.044,17.244)(-0.044, 17.244)(−0.044,17.244).

STEP 11

(a) We reject the null hypothesis and conclude that μ1>μ2\mu_1 > \mu_2μ1​>μ2​ at the 0.05 significance level. (b) The 99% confidence interval for μ1−μ2\mu_1 - \mu_2μ1​−μ2​ is approximately (−0.044,17.244)(-0.044, 17.244)(−0.044,17.244).

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1. Hypothesis Test
2. Confidence Interval

We're testing if the mean of population 1, $\mu_1$ , is greater than the mean of population 2, $\mu_2$ .

Our null hypothesis $H_0$ is that there's no difference, meaning $\mu_1 = \mu_2$ .
The alternative hypothesis $H_1$ is what we're trying to prove: $\mu_1 > \mu_2$ .
This matches option F.

For a 99% confidence interval, our $\alpha$ is $1 - 0.99 = 0.01$ , so $\alpha/2 = 0.005$ .
With 14 degrees of freedom, our critical t-value for a two-tailed test is approximately 2.977.

The confidence interval is: $(\bar{x}_1 - \bar{x}_2) \pm E$ $(50.7 - 42.1) \pm 8.644$ $8.6 \pm 8.644$ This gives us an interval of approximately $(-0.044, 17.244)$ .

(a) We reject the null hypothesis and conclude that $\mu_1 > \mu_2$ at the 0.05 significance level. (b) The 99% confidence interval for $\mu_1 - \mu_2$ is approximately $(-0.044, 17.244)$ .