Question
The National Health Statistics Reports described a study in which a sample of 350 one-year-old baby boys were weighed. Their mean weight was 25.8 pounds with standard deviation 5.3 pounds. A pediatrician claims that the mean weight of one-year-old boys differs from 25 pounds. Do the data provide convincing evidence that the pediatrician's claim is true? Use the a=0.01 level of significance and the P-value method.
State the appropriate null and alternate hypothesis.
Hv0 = [?]
Hv1 = [?]
This test is a [right? left? two? tailed] test
State the appropriate null and alternate hypothesis.
Hv0 = [?]
Hv1 = [?]
This test is a [right? left? two? tailed] test
Answers
Answer
(b) Compute the value of the test statistic. Round the answer to two decimal places.
Answer
One of your digits is wrong. check your work
Answer
The answer is 2.82.
Now compute the P-value , rounded to at least 4 decimal places
Now compute the P-value , rounded to at least 4 decimal places
Answer
No, use 2.82
Answered by
GPT-4o mini
To address the pediatrician's claim, we can formulate our hypotheses as follows:
**Null Hypothesis (H0):** The mean weight of one-year-old boys is equal to 25 pounds.
\[
H_0: \mu = 25
\]
**Alternative Hypothesis (H1):** The mean weight of one-year-old boys differs from 25 pounds.
\[
H_1: \mu \neq 25
\]
Since the alternative hypothesis is stating that the mean weight is either greater than or less than 25 pounds, this is a **two-tailed** test.
**Summary:**
- Null Hypothesis (H0): \(\mu = 25\)
- Alternative Hypothesis (H1): \(\mu \neq 25\)
- This test is a **two-tailed** test.
Next, we can proceed with the test using the significance level \(\alpha = 0.01\) and the P-value method. However, please let me know if you want me to continue with the calculation of the test statistic and P-value!
**Null Hypothesis (H0):** The mean weight of one-year-old boys is equal to 25 pounds.
\[
H_0: \mu = 25
\]
**Alternative Hypothesis (H1):** The mean weight of one-year-old boys differs from 25 pounds.
\[
H_1: \mu \neq 25
\]
Since the alternative hypothesis is stating that the mean weight is either greater than or less than 25 pounds, this is a **two-tailed** test.
**Summary:**
- Null Hypothesis (H0): \(\mu = 25\)
- Alternative Hypothesis (H1): \(\mu \neq 25\)
- This test is a **two-tailed** test.
Next, we can proceed with the test using the significance level \(\alpha = 0.01\) and the P-value method. However, please let me know if you want me to continue with the calculation of the test statistic and P-value!
Answered by
GPT-4o mini
To compute the test statistic for this hypothesis test, we can use the formula for the z-test for the mean. The formula is:
\[
z = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}
\]
where:
- \(\bar{x}\) = sample mean = 25.8 pounds
- \(\mu_0\) = population mean under the null hypothesis = 25 pounds
- \(s\) = sample standard deviation = 5.3 pounds
- \(n\) = sample size = 350
Now, let's plug in the values:
1. Calculate the standard error (SE):
\[
SE = \frac{s}{\sqrt{n}} = \frac{5.3}{\sqrt{350}} \approx \frac{5.3}{18.708} \approx 0.283
\]
2. Now compute the z-test statistic:
\[
z = \frac{25.8 - 25}{0.283} = \frac{0.8}{0.283} \approx 2.83
\]
So, the value of the test statistic is approximately:
\[
\boxed{2.83}
\]
\[
z = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}
\]
where:
- \(\bar{x}\) = sample mean = 25.8 pounds
- \(\mu_0\) = population mean under the null hypothesis = 25 pounds
- \(s\) = sample standard deviation = 5.3 pounds
- \(n\) = sample size = 350
Now, let's plug in the values:
1. Calculate the standard error (SE):
\[
SE = \frac{s}{\sqrt{n}} = \frac{5.3}{\sqrt{350}} \approx \frac{5.3}{18.708} \approx 0.283
\]
2. Now compute the z-test statistic:
\[
z = \frac{25.8 - 25}{0.283} = \frac{0.8}{0.283} \approx 2.83
\]
So, the value of the test statistic is approximately:
\[
\boxed{2.83}
\]
Answered by
GPT-4o mini
Let’s go through the calculations step-by-step again to ensure accuracy.
### Given Data:
- Sample mean (\(\bar{x}\)) = 25.8 pounds
- Null hypothesis mean (\(\mu_0\)) = 25 pounds
- Sample standard deviation (\(s\)) = 5.3 pounds
- Sample size (\(n\)) = 350
### Steps:
1. **Calculate the Standard Error (SE)**:
\[
SE = \frac{s}{\sqrt{n}} = \frac{5.3}{\sqrt{350}}
\]
First, compute \(\sqrt{350}\):
\[
\sqrt{350} \approx 18.71
\]
Now, compute the SE:
\[
SE \approx \frac{5.3}{18.71} \approx 0.283
\]
2. **Calculate the z-test statistic**:
The formula for the z-test statistic is:
\[
z = \frac{\bar{x} - \mu_0}{SE} = \frac{25.8 - 25}{SE}
\]
Substituting in the values:
\[
z = \frac{25.8 - 25}{0.283} = \frac{0.8}{0.283} \approx 2.83
\]
To double-check the division:
\[
\frac{0.8}{0.283} \approx 2.826
\]
Rounding this to two decimal places gives:
\[
z \approx 2.83
\]
Upon reviewing, the calculations seem correct, and the value of the test statistic is accurately computed as:
\[
\boxed{2.83}
\]
Please let me know if there is a specific part of the calculation you would like me to clarify or reassess!
### Given Data:
- Sample mean (\(\bar{x}\)) = 25.8 pounds
- Null hypothesis mean (\(\mu_0\)) = 25 pounds
- Sample standard deviation (\(s\)) = 5.3 pounds
- Sample size (\(n\)) = 350
### Steps:
1. **Calculate the Standard Error (SE)**:
\[
SE = \frac{s}{\sqrt{n}} = \frac{5.3}{\sqrt{350}}
\]
First, compute \(\sqrt{350}\):
\[
\sqrt{350} \approx 18.71
\]
Now, compute the SE:
\[
SE \approx \frac{5.3}{18.71} \approx 0.283
\]
2. **Calculate the z-test statistic**:
The formula for the z-test statistic is:
\[
z = \frac{\bar{x} - \mu_0}{SE} = \frac{25.8 - 25}{SE}
\]
Substituting in the values:
\[
z = \frac{25.8 - 25}{0.283} = \frac{0.8}{0.283} \approx 2.83
\]
To double-check the division:
\[
\frac{0.8}{0.283} \approx 2.826
\]
Rounding this to two decimal places gives:
\[
z \approx 2.83
\]
Upon reviewing, the calculations seem correct, and the value of the test statistic is accurately computed as:
\[
\boxed{2.83}
\]
Please let me know if there is a specific part of the calculation you would like me to clarify or reassess!
Answered by
GPT-4o mini
Let's calculate the \( P \)-value for the test statistic \( z = 2.83 \) (even though you mentioned it to be \( 2.82 \), we'll proceed with \( 2.83 \) for now, and you can adjust if necessary).
Since this is a two-tailed test, we need to calculate the area in both tails of the normal distribution.
### Step 1: Calculate the \( P \)-value
1. **Find the area to the right of \( z = 2.83 \)**:
Using a standard normal distribution table or calculator, the area to the right of \( z = 2.83 \) corresponds to the \( P(Z > 2.83) \).
Typically, the standard normal table gives the area to the left, so we will look up \( z = 2.83 \):
From standard normal tables or calculators, we find:
\[
P(Z < 2.83) \approx 0.9977
\]
Therefore, the area to the right is:
\[
P(Z > 2.83) = 1 - P(Z < 2.83) \approx 1 - 0.9977 = 0.0023
\]
2. **Since it’s a two-tailed test**, multiply this area by 2:
\[
P\text{-value} = 2 \times P(Z > 2.83) = 2 \times 0.0023 \approx 0.0046
\]
### Conclusion
The \( P \)-value (rounded to at least 4 decimal places) is:
\[
\boxed{0.0046}
\]
Since this is a two-tailed test, we need to calculate the area in both tails of the normal distribution.
### Step 1: Calculate the \( P \)-value
1. **Find the area to the right of \( z = 2.83 \)**:
Using a standard normal distribution table or calculator, the area to the right of \( z = 2.83 \) corresponds to the \( P(Z > 2.83) \).
Typically, the standard normal table gives the area to the left, so we will look up \( z = 2.83 \):
From standard normal tables or calculators, we find:
\[
P(Z < 2.83) \approx 0.9977
\]
Therefore, the area to the right is:
\[
P(Z > 2.83) = 1 - P(Z < 2.83) \approx 1 - 0.9977 = 0.0023
\]
2. **Since it’s a two-tailed test**, multiply this area by 2:
\[
P\text{-value} = 2 \times P(Z > 2.83) = 2 \times 0.0023 \approx 0.0046
\]
### Conclusion
The \( P \)-value (rounded to at least 4 decimal places) is:
\[
\boxed{0.0046}
\]
Answered by
GPT-4o mini
Apologies for the oversight. Let's compute the P-value using the z-test statistic \( z = 2.82 \).
### Step 1: Calculate the P-value
1. **Find the area to the right of \( z = 2.82 \)**:
Using a standard normal distribution table or calculator, we look up the value for \( z = 2.82 \).
From standard normal tables or calculators:
\[
P(Z < 2.82) \approx 0.9974
\]
Therefore, the area to the right is:
\[
P(Z > 2.82) = 1 - P(Z < 2.82) \approx 1 - 0.9974 = 0.0026
\]
2. **Since it’s a two-tailed test**, we multiply this area by 2:
\[
\text{P-value} = 2 \times P(Z > 2.82) = 2 \times 0.0026 \approx 0.0052
\]
### Conclusion
The P-value (rounded to at least 4 decimal places) is:
\[
\boxed{0.0052}
\]
If you have any further questions or need additional calculations, feel free to ask!
### Step 1: Calculate the P-value
1. **Find the area to the right of \( z = 2.82 \)**:
Using a standard normal distribution table or calculator, we look up the value for \( z = 2.82 \).
From standard normal tables or calculators:
\[
P(Z < 2.82) \approx 0.9974
\]
Therefore, the area to the right is:
\[
P(Z > 2.82) = 1 - P(Z < 2.82) \approx 1 - 0.9974 = 0.0026
\]
2. **Since it’s a two-tailed test**, we multiply this area by 2:
\[
\text{P-value} = 2 \times P(Z > 2.82) = 2 \times 0.0026 \approx 0.0052
\]
### Conclusion
The P-value (rounded to at least 4 decimal places) is:
\[
\boxed{0.0052}
\]
If you have any further questions or need additional calculations, feel free to ask!