91Ó°ÊÓ

a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume all assumptions are met. A medical researcher wishes to see if hospital patients in a large hospital have the same blood type distribution as those in the general population. The distribution for the general population is as follows: type \(\mathrm{A}, 20 \%\); type \(\mathrm{B}, 28 \%\); type \(\mathrm{O}, 36 \%\); and type \(\mathrm{AB}=16 \% .\) He selects a random sample of 50 patients and finds the following: 12 have type A blood, 8 have type \(\mathrm{B}, 24\) have type \(\mathrm{O},\) and 6 have type \(\mathrm{AB}\) blood. At \(\alpha=0.10,\) can it be concluded that the distribution is the same as that of the general population?

Step by step solution

01

State the Hypotheses

The null hypothesis \( H_0 \) is that the distribution of blood types in the hospital is the same as in the general population: \( P(A) = 0.20, \ P(B) = 0.28, \ P(O) = 0.36, \ P(AB) = 0.16 \). The alternative hypothesis \( H_1 \) is that at least one proportion is different from the general population distribution.

02

Identify the Claim

The medical researcher claims that the distribution of blood types in the hospital matches that of the general population. Thus, the claim is aligned with the null hypothesis \( H_0 \).

03

Find the Critical Value

We use the chi-square test to determine the critical value. The degrees of freedom \( df \) is calculated as \( k - 1 = 4 - 1 = 3 \), where \( k \) is the number of categories. Using a chi-square distribution table and \( \alpha = 0.10 \), the critical value for \( df = 3 \) is approximately 6.251 (this value may vary slightly depending on the table used).

04

Compute the Test Value

Calculate the expected frequencies: \( E_A = 10, \, E_B = 14, \, E_O = 18, \, E_{AB} = 8 \) using \( n \times P \) for each blood type. Compute the test statistic using \( \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \). For the observed values, \( O_A = 12, O_B = 8, O_O = 24, O_{AB} = 6 \):\[\chi^2 = \frac{(12-10)^2}{10} + \frac{(8-14)^2}{14} + \frac{(24-18)^2}{18} + \frac{(6-8)^2}{8} \approx 4.29\]

05

Make the Decision

Compare the test statistic to the critical value. Since \( 4.29 < 6.251 \), we fail to reject the null hypothesis at the 0.10 significance level.

06

Summarize the Results

There is insufficient evidence to conclude that the blood type distribution in the hospital is different from that of the general population at the \( \alpha = 0.10 \) level.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chi-Square Test

The Chi-square test is a statistical method used to examine if there is a significant difference between observed and expected frequencies in categorical data. In our scenario about blood type distribution among hospital patients, the chi-square test is used to determine if the distribution of blood types in the hospital differs from that of the general population. This test compares the observed frequencies from the hospital sample with the expected frequencies if the hospital's blood type distribution were the same as the general population's. With this approach, we can determine if any differences are due to random chance or if they are statistically significant.

The chi-square test requires assumptions to be met, such as having independent observations and a sufficiently large sample size. In this problem, all these assumptions were stated to be met.

Critical Value

The critical value in hypothesis testing serves as a threshold. It helps decide whether to reject the null hypothesis. To find it, we use a chi-square distribution table. For our test, given a significance level of \(\alpha = 0.10\) and degrees of freedom \(df = k - 1\), where \(k\) is the number of blood type categories (4 in this case), the degrees of freedom are 3.

Using the chi-square distribution table, we find the critical value for \(df = 3\) at \(\alpha = 0.10\), which is approximately 6.251. This critical value essentially acts as a cutoff point; if our test statistic is greater than 6.251, we would reject the null hypothesis.

Null Hypothesis

The null hypothesis \(H_0\) is the starting assumption of no effect or no difference. In the context of this blood type distribution study, \(H_0\) assumes that the hospital blood type distribution is the same as the general population distribution. This means the proportions of blood types A, B, O, and AB in the hospital should match 20%, 28%, 36%, and 16%, respectively.

The purpose of conducting the hypothesis test is to challenge this assumption. If the statistical evidence is strong enough, we may reject \(H_0\), suggesting that the actual distribution is different. However, in this study, the result indicated insufficient evidence to reject \(H_0\). Thus, we maintain that the blood type distribution is likely the same as that of the general population.

Observed and Expected Frequencies

Observed frequencies are the actual counts of occurrences observed in the data. For the hospital blood types, these were: 12 patients with type A, 8 with type B, 24 with type O, and 6 with type AB.

On the other hand, expected frequencies are the counts we would anticipate based on certain assumptions or previous data. For example, the expected frequencies if the hospital's blood type distribution is similar to the general population would be calculated as follows:

• Type A: \( E_A = 50 \times 0.20 = 10 \)
• Type B: \( E_B = 50 \times 0.28 = 14 \)
• Type O: \( E_O = 50 \times 0.36 = 18 \)
• Type AB: \( E_{AB} = 50 \times 0.16 = 8 \)

The chi-square test uses both observed and expected frequencies to compute the test statistic, which assesses how well the observed data fit the expected distribution.