Math / Data & Statistics

QuestionThis question: 1 point(\$) possible
Of the 94 participants in a drug trial who were given a new experimental treatment for arthritis, 54 showed improvement. Of the 89 participants given a placebo, 46 showed improvement. Construct a two-way table for these data, and then use a 0.05 significance level to test the claim that improvement is independent of whether the participant was given the drug or a placebo.
Click the icon to view the critical values of $\chi^{2}$ table.
Complete the following two-way table. \begin{tabular}{|l|c|} \hline & improvement \\ \hline Drug & $\square$ \\ \hline Placebo & $\square$ \\ \hline \end{tabular} a. State the null and the al A. The null hypothesis The alternative hyp B. The alternative hyThe null hypothesis b. Assuming independenc
Critical values of chi-square \begin{tabular}{|c|c|c|} \hline \multicolumn{3}{|c|}{Critical Values of $\chi^{2}$ : Reject $H_{0}$ Only If $\chi^{2}>$ Critical Value} \\ \hline \multirow[t]{2}{*}{Table size (rows $\times$ columns)} & \multicolumn{2}{|l|}{Significance level} \\ \hline & 0.05 & 0.01 \\ \hline $2 \times 2$ & 3.841 & 6.635 \\ \hline $2 \times 3$ or $3 \times 2$ & 5.991 & 9.210 \\ \hline $3 \times 3$ & 9.488 & 13.277 \\ \hline $2 \times 4$ or $4 \times 2$ & 7.815 & 11.345 \\ \hline $2 \times 5$ or $5 \times 2$ & 9.488 & 13.277 \\ \hline \end{tabular}
Placebo (Round to the nearest hun c. Find the value of the $\chi^{2}$
Print Done $\chi^{2}=$ $\square$ (Round to the ne d. Use the given significance level to find the $\chi^{2}$ critical value.
The critical value is $\square$ . (Round to the nearest thousandth as needed.) e. Using the given significance level, complete the test of the claim that the two variables are independent. State the conclusion that addresses the original claim. Choose the correct answer below. There does not appear to be a relationship between improvement and treatment (drug or placebo).

Studdy Solution

STEP 1

1. We have two groups: Drug and Placebo.
2. We are testing the independence of improvement from the type of treatment.
3. The significance level is $\alpha = 0.05$ .
4. We will use a chi-square test for independence.

STEP 2

1. Construct the two-way table.
2. State the null and alternative hypotheses.
3. Calculate the expected frequencies.
4. Compute the chi-square statistic.
5. Determine the critical value.
6. Make a decision and state the conclusion.

STEP 3

Construct the two-way table with the given data:
\begin{tabular}{|l|c|c|} \hline & \text{Improvement} & \text{No Improvement} & \text{Total} \\ \hline \text{Drug} & 54 & 94 - 54 = 40 & 94 \\ \hline \text{Placebo} & 46 & 89 - 46 = 43 & 89 \\ \hline \text{Total} & 54 + 46 = 100 & 40 + 43 = 83 & 183 \\ \hline \end{tabular}

STEP 4

State the null and alternative hypotheses:
- Null hypothesis ( $H_0$ ): Improvement is independent of the treatment (drug or placebo). - Alternative hypothesis ( $H_a$ ): Improvement is dependent on the treatment.

STEP 5

Calculate the expected frequencies for each cell in the table using the formula:
$E = \frac{(\text{Row Total}) \times (\text{Column Total})}{\text{Grand Total}}$
For Drug and Improvement:
$E_{DI} = \frac{94 \times 100}{183} \approx 51.37$
For Drug and No Improvement:
$E_{DNI} = \frac{94 \times 83}{183} \approx 42.63$
For Placebo and Improvement:
$E_{PI} = \frac{89 \times 100}{183} \approx 48.63$
For Placebo and No Improvement:
$E_{PNI} = \frac{89 \times 83}{183} \approx 40.37$

STEP 6

Compute the chi-square statistic using the formula:
$\chi^2 = \sum \frac{(O - E)^2}{E}$
Where $O$ is the observed frequency and $E$ is the expected frequency.
$\chi^2 = \frac{(54 - 51.37)^2}{51.37} + \frac{(40 - 42.63)^2}{42.63} + \frac{(46 - 48.63)^2}{48.63} + \frac{(43 - 40.37)^2}{40.37}$
$\chi^2 \approx \frac{2.63^2}{51.37} + \frac{2.63^2}{42.63} + \frac{2.63^2}{48.63} + \frac{2.63^2}{40.37}$
$\chi^2 \approx 0.134 + 0.162 + 0.142 + 0.171 \approx 0.609$

STEP 7

Determine the critical value from the chi-square distribution table for $1$ degree of freedom at $\alpha = 0.05$ :
Critical value = 3.841

STEP 8

Compare the chi-square statistic to the critical value:
Since $0.609 < 3.841$ , we fail to reject the null hypothesis.
Conclusion: There does not appear to be a relationship between improvement and treatment (drug or placebo).

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Studdy Solution

STEP 1

1. We have two groups: Drug and Placebo.
2. We are testing the independence of improvement from the type of treatment.
3. The significance level is $\alpha = 0.05$ .
4. We will use a chi-square test for independence.

STEP 2

1. Construct the two-way table.
2. State the null and alternative hypotheses.
3. Calculate the expected frequencies.
4. Compute the chi-square statistic.
5. Determine the critical value.
6. Make a decision and state the conclusion.

STEP 3

STEP 4

State the null and alternative hypotheses:
- Null hypothesis ( $H_0$ ): Improvement is independent of the treatment (drug or placebo). - Alternative hypothesis ( $H_a$ ): Improvement is dependent on the treatment.

STEP 5

STEP 6

STEP 7

Determine the critical value from the chi-square distribution table for $1$ degree of freedom at $\alpha = 0.05$ :
Critical value = 3.841

STEP 8

Compare the chi-square statistic to the critical value:
Since $0.609 < 3.841$ , we fail to reject the null hypothesis.
Conclusion: There does not appear to be a relationship between improvement and treatment (drug or placebo).

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Studdy Solution

STEP 1

1. We have two groups: Drug and Placebo.2. We are testing the independence of improvement from the type of treatment.3. The significance level is α=0.05 \alpha = 0.05 α=0.05.4. We will use a chi-square test for independence.

STEP 2

1. Construct the two-way table.2. State the null and alternative hypotheses.3. Calculate the expected frequencies.4. Compute the chi-square statistic.5. Determine the critical value.6. Make a decision and state the conclusion.

STEP 3

STEP 4

State the null and alternative hypotheses:- Null hypothesis (H0H_0H0​): Improvement is independent of the treatment (drug or placebo). - Alternative hypothesis (HaH_aHa​): Improvement is dependent on the treatment.

STEP 5

STEP 6

STEP 7

Determine the critical value from the chi-square distribution table for 111 degree of freedom at α=0.05\alpha = 0.05α=0.05:Critical value = 3.841

STEP 8

Compare the chi-square statistic to the critical value:Since 0.609<3.8410.609 < 3.8410.609<3.841, we fail to reject the null hypothesis.Conclusion: There does not appear to be a relationship between improvement and treatment (drug or placebo).

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1. We have two groups: Drug and Placebo.
2. We are testing the independence of improvement from the type of treatment.
3. The significance level is $\alpha = 0.05$ .
4. We will use a chi-square test for independence.

1. Construct the two-way table.
2. State the null and alternative hypotheses.
3. Calculate the expected frequencies.
4. Compute the chi-square statistic.
5. Determine the critical value.
6. Make a decision and state the conclusion.

State the null and alternative hypotheses:
- Null hypothesis ( $H_0$ ): Improvement is independent of the treatment (drug or placebo). - Alternative hypothesis ( $H_a$ ): Improvement is dependent on the treatment.

Determine the critical value from the chi-square distribution table for $1$ degree of freedom at $\alpha = 0.05$ :
Critical value = 3.841

Compare the chi-square statistic to the critical value:
Since $0.609 < 3.841$ , we fail to reject the null hypothesis.
Conclusion: There does not appear to be a relationship between improvement and treatment (drug or placebo).