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A body mass index (BMI) of more than 25 is considered unhealthful. The data output given is from 50 randomly and independently selected people from a health agency's study. Test the hypothesis that the mean BMI is more than 25 using a significance level of 0.05. Assume that conditions are met.

One-Sample T-Test of [tex]$\mu=25$[/tex] vs [tex]$\ \textgreater \ 25$[/tex]

\begin{tabular}{rrrrcc}
N & Mean & StDev & SE Mean & T & P \\
50 & 27.804 & 7.983 & 1.129 & 2.54 & 0.007 \\
\hline
\end{tabular}

Determine the null and alternative hypotheses. Choose the correct answer below.

A. [tex]$H_0: \mu=25$[/tex] \quad [tex]$H_a: \mu\ \textgreater \ 25$[/tex]
B. [tex]$H_0: \mu=25$[/tex] \quad [tex]$H_a: \mu \leq 25$[/tex]
C. [tex]$H_0: \mu\ \textless \ 25$[/tex] \quad [tex]$H_a: \mu\ \textless \ 25$[/tex]
D. [tex]$H_0: \mu=25$[/tex]
E. [tex]$H_0: \mu \neq 25$[/tex] \quad [tex]$H_a: \mu \neq 25$[/tex]
F. [tex]$H_0: \mu\ \textgreater \ 25$[/tex] \quad [tex]$H_a: \mu \leq 25$[/tex]

The test statistic is 2.54.
(Type an integer or decimal rounded to two decimal places as needed.)

The p-value is 0.007.
(Type an integer or decimal rounded to three decimal places as needed.)

Interpret the results of the test:
Since the p-value is less than the significance level, 0.05, reject [tex]$H_0$[/tex]. There is sufficient evidence to conclude that the mean BMI is more than 25 using a significance level of 0.05.



Answer :

GinnyAnswer
Solution:

We are tasked with testing the hypothesis that the mean BMI is more than 25 at a significance level of 0.05. Let's walk through the steps to find the solution.

Step 1: Formulate the hypotheses
The null hypothesis (H₀) represents the status quo or a statement of no effect or no difference. In this context:

[tex]\( H_0: \mu = 25 \)[/tex]

The alternative hypothesis (Hₐ) is what we want to test against the null hypothesis:

[tex]\( H_a: \mu > 25 \)[/tex]

Correct answer: (A)
- [tex]\( H_0: \mu = 25 \)[/tex]
- [tex]\( H_a: \mu > 25 \)[/tex]

Step 2: Identify the test statistic
From the given data, we have the test statistic (T). According to the output, the test statistic is:

[tex]\( T = 2.54 \)[/tex]

So, the test statistic to two decimal places is:

[tex]\( T = 2.54 \)[/tex]

Step 3: Determine the p-value
The p-value given in the output represents the probability of obtaining test results at least as extreme as the results actually observed, under the assumption that the null hypothesis is correct. The given p-value is:

[tex]\( P = 0.007 \)[/tex]

So, the p-value to three decimal places is:

[tex]\( P = 0.007 \)[/tex]

Step 4: Interpret the p-value in the context of the significance level
Given the significance level [tex]\( \alpha = 0.05 \)[/tex]:

- If the p-value is less than the significance level, we reject the null hypothesis.
- If the p-value is greater than the significance level, we fail to reject the null hypothesis.

Since [tex]\( 0.007 < 0.05 \)[/tex], we reject the null hypothesis.

Step 5: Conclusion
Since the p-value is less than the significance level, there is sufficient evidence to conclude that the mean BMI is more than 25.

Final interpretation:
Since the p-value is less than the significance level, 0.05, we reject [tex]\( H_0 \)[/tex]. There is sufficient evidence to conclude that the mean BMI is more than 25 using a significance level of 0.05.

Here is the completed solution with the answer blanks filled in:
- The null hypothesis: [tex]\( H_0: \mu = 25 \)[/tex]
- The alternative hypothesis: [tex]\( H_a: \mu > 25 \)[/tex]
- The test statistic: 2.54
- The p-value: 0.007
- Since the p-value is less than the significance level, 0.05, we reject [tex]\( H_0 \)[/tex]. There is sufficient evidence to conclude that the mean BMI is more than 25 using a significance level of 0.05.

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