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Answer :
Answer:
There is sufficient evidence to conclude that the mean weight of males is less than 150 pounds at the 0.05 significance level.
Explanation:
To make a decision using the rejection region method for a claim that the mean value of a male's weight is less than 150 pounds with a significance level of 0.05, we follow these steps:
1. State 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 less than 150 pounds ([tex]\( \mu < 150 \)[/tex]).
2. Determine the critical value(s) for the test. Since it's a one-tailed test (less than), we find the critical value from the standard normal distribution corresponding to the given significance level [tex]\( \alpha = 0.05 \)[/tex] . Using a z-table or statistical software, [tex]\( z_{\alpha} \approx -1.645 \).[/tex]
3. Calculate the test statistic, which is the z-score for the sample mean:
[tex]\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \][/tex]
[tex]\[ z = \frac{145 - 150}{\frac{16}{\sqrt{41}}} \][/tex]
[tex]\[ z \approx \frac{-5}{\frac{16}{\sqrt{41}}} \][/tex]
[tex]\[ z \approx \frac{-5}{2.5} \][/tex]
[tex]\[ z = -2 \][/tex]
4. Make a decision:
- If the test statistic falls in the rejection region (i.e., if [tex]\( z < z_{\alpha} \)[/tex]), reject the null hypothesis.
- Otherwise, fail to reject the null hypothesis.
Since [tex]\( z = -2 \)[/tex] falls in the rejection region ([tex]\( z < -1.645 \)[/tex]), we reject the null hypothesis.
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Rewritten by : Barada
based on this analysis, we do not have sufficient evidence to reject the claim that the mean weight of males is 150 pounds at a significance level of 0.04.
To make a decision about a claim regarding the mean value of male's weight using the rejection region method, we typically perform a hypothesis test. Let's break down the steps for this scenario:
1. **Formulate Hypotheses:**
- Null Hypothesis[tex](\( H_0 \))[/tex]: The mean weight of males is 150 pounds.
- Alternative Hypothesis [tex](\( H_1 \))[/tex]: The mean weight of males is not 150 pounds (two-tailed test).
2. **Select Significance Level[tex](\( \alpha \))[/tex]:**
The significance level [tex](\( \alpha \))[/tex] is given as 0.04 in this case.
3. **Calculate Test Statistic:**
The test statistic for a one-sample z-test for the mean (\( \mu \)) can be calculated using:
[tex]\[ z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} \][/tex]
where [tex]\( \bar{x} \)[/tex] is the sample mean, [tex]\( \mu_0 \)[/tex] is the hypothesized population mean, \( \sigma \) is the population standard deviation, and \( n \) is the sample size.
Given:
- Sample mean (\( \bar{x} \)): 145 pounds
- Population standard deviation (\( \sigma \)): 16 pounds
- Sample size (\( n \)): 36
Plugging in the values:
[tex]\[ z = \frac{145 - 150}{\frac{16}{\sqrt{36}}} = \frac{-5}{\frac{16}{6}} = \frac{-30}{16} = -1.875 \][/tex]
4. **Determine Rejection Region:**
For a two-tailed test at [tex]\( \alpha = 0.04 \)[/tex], the critical z-values can be found using a z-table or a statistical calculator. Since this is a two-tailed test, we look for the z-value that corresponds to an area of [tex]\( \frac{\alpha}{2} = 0.02 \)[/tex] in each tail.
Using a z-table or calculator, we find that the critical z-values are approximately [tex]\( \pm 2.054 \) for \( \alpha = 0.04 \).[/tex]
5. **Make a Decision:**
- Since ( -1.875 ) is not in the rejection region (-2.054, 2.054) , we do not reject the null hypothesis.
[tex]- \( -1.875 \) is less than \( -1.751 \), so \( -1.875 < -1.751 \).\\ - \( -1.875 \) is less than \( 2.054 \), so \( -1.875 < 2.054 \).\\ - \( -1.875 \) is not less than or equal to \( 0.04 \), so \( -1.875 > 0.04 \).[/tex]
Therefore, based on this analysis, we do not have sufficient evidence to reject the claim that the mean weight of males is 150 pounds at a significance level of 0.04.
complete question:-
How do you make a decision for a claim that the mean value of male's weight is 150 pounds for a significant level of 0.04 using a rejection region method? A sample with a mean of 145 pounds, size of 36, and a standard deviation of 16 pounds is selected. Is
-1.875 ≤ -2.054? Is -1.875 < -1.751? Is -1.875 > 2.054? Is -1.875 ≤ 0.04?
