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Answer :
the correct option is the first one: 0.0304.
To calculate the P-value for a claim about the mean weight of males being 150 pounds with a significance level of 0.04 using the P-value method, we can follow these steps:
1. Define the null and alternative hypotheses:
- Null hypothesis[tex](\( H_0 \))[/tex]: The mean weight of males is 150 pounds [tex](\( \mu = 150 \)).[/tex]
- Alternative hypothesis [tex](\( H_1 \))[/tex]: The mean weight of males is not equal to 150 pounds[tex](\( \mu \neq 150 \)).[/tex]
2. Calculate the test statistic using the formula:
[tex]\[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \][/tex]
where [tex]\( \bar{x} \)[/tex] is the sample mean (145 pounds), [tex]\( \mu \)[/tex] is the hypothesized population mean (150 pounds), \( s \) is the standard deviation of the sample (16 pounds), and \( n \) is the sample size (36).
Plugging in the values:
[tex]\[ t = \frac{145 - 150}{\frac{16}{\sqrt{36}}} \] \[ t = \frac{-5}{\frac{16}{6}} \] \[ t = -\frac{30}{16} \] \[ t = -1.875 \][/tex]
3. Determine the degrees of freedom[tex](\( df \))[/tex] for the t-distribution, which is [tex]\( n - 1 = 36 - 1 = 35 \).[/tex]
4. Using a t-distribution table or a statistical calculator, find the P-value associated with the test statistic [tex]\( t = -1.875 \) and \( df = 35 \)[/tex]. The P-value represents the probability of obtaining a test statistic at least as extreme as the one observed, assuming the null hypothesis is true.
After looking up the P-value, we find that it is approximately 0.0304.
So, the correct answer is:
[tex]\[ \text{P-value} \approx 0.0304 \][/tex]
Therefore, the correct option is the first one: 0.0304.
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