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Question 8, 9.4.10-T

Researchers measured skulls from different time periods in an attempt to determine whether interbreeding of cultures occurred. Results are given below. Assume that both samples are independent simple random samples from populations having normal distributions. Use a 0.05 significance level to test the claim that the variation of maximal skull breadths in 4000 B.C. is the same as the variation in A.D. 150.

\begin{tabular}{cccc}
& [tex]$n$[/tex] & [tex]$\bar{x}$[/tex] & [tex]$s$[/tex] \\
\hline 4000 B.C. & 30 & 131.15 mm & 5.16 mm \\
A.D. 150 & 30 & 136.43 mm & 5.37 mm
\end{tabular}

A. [tex]$H_0: \sigma_1^2=\sigma_2^2$[/tex]

B.
[tex]$
\begin{array}{l}
H_0: \sigma_1^2=\sigma_2^2 \\
H_1: \sigma_1^2 \neq \sigma_2^2
\end{array}
$[/tex]

C. [tex]$H_0: \sigma_1^2=\sigma_2^2$[/tex]

D. [tex]$H_1: \sigma_1^2 \neq \sigma_2^2$[/tex]

Identify the test statistic:
[tex]$F = 1.08$[/tex] (Round to two decimal places as needed.)

Use technology to identify the P-value:
The [tex]$P$[/tex]-value is [tex]$0.831$[/tex] (Round to three decimal places as needed.)

What is the conclusion for this hypothesis test?

A. Reject [tex]$H_0$[/tex]. There is sufficient evidence to warrant rejection of the claim.
B. Fail to reject [tex]$H_0$[/tex]. There is insufficient evidence to warrant rejection of the claim.



Answer :

GinnyAnswer
Let's solve the problem step-by-step.

### Step 1: Understanding the Problem
We are given the sample sizes, means, and standard deviations for two different time periods (4000 B.C. and A.D. 150) for the maximal skull breadths. The task is to test the claim that the variation in skull breadths for these two time periods is the same.

### Step 2: Set Up the Hypotheses
Null Hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\sigma_1^2 = \sigma_2^2\)[/tex] \
Alternative Hypothesis ([tex]\(H_1\)[/tex]): [tex]\(\sigma_1^2 \neq \sigma_2^2\)[/tex]

### Step 3: Data Given
We have the following data:
- For 4000 B.C.:
- [tex]\( n_1 = 30 \)[/tex]
- [tex]\( s_1 = 5.16 \)[/tex] mm
- For A.D. 150:
- [tex]\( n_2 = 30 \)[/tex]
- [tex]\( s_2 = 5.37 \)[/tex] mm

### Step 4: Calculate the Test Statistic
The test statistic for comparing variances is the F-statistic, calculated as:
[tex]\[ F = \frac{s_1^2}{s_2^2} \][/tex]

Using the given data:
[tex]\[ F = \frac{(5.16)^2}{(5.37)^2} \approx 0.92 \][/tex]
Rounded to two decimal places, [tex]\( F = 0.92 \)[/tex].

### Step 5: Determine the Degrees of Freedom
The degrees of freedom for the numerator ([tex]\(dfn\)[/tex]) and the denominator ([tex]\(dfd\)[/tex]) are calculated as:
[tex]\[ dfn = n_1 - 1 = 30 - 1 = 29 \][/tex]
[tex]\[ dfd = n_2 - 1 = 30 - 1 = 29 \][/tex]

### Step 6: Find the P-value
Using the F-distribution and the cumulative distribution function, we determine the P-value.

The P-value is the significance level at which the observed value of the test statistic is considered extreme. In this case, it is given to be 0.831 (rounded to three decimal places).

### Step 7: Compare the P-value to the Significance Level
We are using a significance level ([tex]\(\alpha\)[/tex]) of 0.05 to draw our conclusion.

### Step 8: Make the Conclusion
If the P-value is less than the significance level ([tex]\(P < \alpha\)[/tex]), we reject the null hypothesis. Otherwise, we fail to reject it.

Here, [tex]\(P = 0.831\)[/tex] and [tex]\(\alpha = 0.05\)[/tex]. Since [tex]\(0.831 > 0.05\)[/tex], we fail to reject the null hypothesis.

### Final Conclusion
There is insufficient evidence to warrant rejection of the claim that the variation in maximal skull breadths in 4000 B.C. is the same as the variation in A.D. 150.

Answer:
The F-statistic is [tex]\(0.92\)[/tex], the P-value is [tex]\(0.831\)[/tex], and the conclusion is:
B. Fail to reject [tex]\(H_0\)[/tex]. There is insufficient evidence to warrant rejection of the claim.

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