To find a 98% confidence interval for the mean weight of all US males based on the sample data provided, follow these steps:

1. **Calculate the sample mean ($\bar{x}$):**
$$
\bar{x} = \frac{149 + 178 + 202 + 206 + 213 + 220 + 252}{7}
$$
$$
\bar{x} = \frac{1420}{7} = 202.9
$$

2. **Calculate the sample standard deviation (s):**
$$
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}
$$
First, find each deviation from the mean, square it, and sum these values:
$$
(149 - 202.9)^2 = 2880.41
$$
$$
(178 - 202.9)^2 = 620.41
$$
$$
(202 - 202.9)^2 = 0.81
$$
$$
(206 - 202.9)^2 = 9.61
$$
$$
(213 - 202.9)^2 = 102.01
$$
$$
(220 - 202.9)^2 = 290.41
$$
$$
(252 - 202.9)^2 = 2400.81
$$
Sum of squared deviations:
$$
2880.41 + 620.41 + 0.81 + 9.61 + 102.01 + 290.41 + 2400.81 = 6304.47
$$
Now, divide by $n-1$ (degrees of freedom):
$$
s^2 = \frac{6304.47}{6} = 1050.745
$$
$$
s = \sqrt{1050.745} \approx 32.4
$$

3. **Determine the t-score for a 98% confidence interval with $n-1$ degrees of freedom:**
For $n-1 = 6$ degrees of freedom, the t-score for a 98% confidence interval is approximately 2.447 (from t-distribution tables).

4. **Calculate the margin of error (ME):**
$$
ME = t \times \frac{s}{\sqrt{n}}
$$
$$
ME = 2.447 \times \frac{32.4}{\sqrt{7}}
$$
$$
ME = 2.447 \times 12.24 \approx 29.9
$$

5. **Construct the confidence interval:**
$$
\text{Lower limit} = \bar{x} - ME = 202.9 - 29.9 = 173.0
$$
$$
\text{Upper limit} = \bar{x} + ME = 202.9 + 29.9 = 232.8
$$

Therefore, the 98% confidence interval for the mean weight of all US males is (173.0, 232.8).

Question

To find a 98% confidence interval for the mean weight of all US males based on the sample data provided, follow these steps:

1. **Calculate the sample mean ($\bar{x}$):**
   $$
   \bar{x} = \frac{149 + 178 + 202 + 206 + 213 + 220 + 252}{7}
   $$
   $$
   \bar{x} = \frac{1420}{7} = 202.9
   $$

2. **Calculate the sample standard deviation (s):**
   $$
   s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}
   $$
   First, find each deviation from the mean, square it, and sum these values:
   $$
   (149 - 202.9)^2 = 2880.41
   $$
   $$
   (178 - 202.9)^2 = 620.41
   $$
   $$
   (202 - 202.9)^2 = 0.81
   $$
   $$
   (206 - 202.9)^2 = 9.61
   $$
   $$
   (213 - 202.9)^2 = 102.01
   $$
   $$
   (220 - 202.9)^2 = 290.41
   $$
   $$
   (252 - 202.9)^2 = 2400.81
   $$
   Sum of squared deviations:
   $$
   2880.41 + 620.41 + 0.81 + 9.61 + 102.01 + 290.41 + 2400.81 = 6304.47
   $$
   Now, divide by $n-1$ (degrees of freedom):
   $$
   s^2 = \frac{6304.47}{6} = 1050.745
   $$
   $$
   s = \sqrt{1050.745} \approx 32.4
   $$

3. **Determine the t-score for a 98% confidence interval with $n-1$ degrees of freedom:**
   For $n-1 = 6$ degrees of freedom, the t-score for a 98% confidence interval is approximately 2.447 (from t-distribution tables).

4. **Calculate the margin of error (ME):**
   $$
   ME = t \times \frac{s}{\sqrt{n}}
   $$
   $$
   ME = 2.447 \times \frac{32.4}{\sqrt{7}}
   $$
   $$
   ME = 2.447 \times 12.24 \approx 29.9
   $$

5. **Construct the confidence interval:**
   $$
   \text{Lower limit} = \bar{x} - ME = 202.9 - 29.9 = 173.0
   $$
   $$
   \text{Upper limit} = \bar{x} + ME = 202.9 + 29.9 = 232.8
   $$

Therefore, the 98% confidence interval for the mean weight of all US males is (173.0, 232.8).

Knowee AI · Accepted Answer

To find a 98% confidence interval for the mean weight of all US males based on the sample data provided, follow these steps:

1. **Calculate the sample mean ($\bar{x}$):**
   $$
   \bar{x} = \frac{149 + 178 + 202 + 206 + 213 + 220 + 252}{7}
   $$
   $$
   \bar{x} = \frac{1420}{7} = 202.9
   $$

2. **Calculate the sample standard deviation (s):**
   $$
   s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}
   $$
   First, find each deviation from the mean, square it, and sum these values:
   $$
   (149 - 202.9)^2 = 2880.41
   $$
   $$
   (178 - 202.9)^2 = 620.41
   $$
   $$
   (202 - 202.9)^2 = 0.81
   $$
   $$
   (206 - 202.9)^2 = 9.61
   $$
   $$
   (213 - 202.9)^2 = 102.01
   $$
   $$
   (220 - 202.9)^2 = 290.41
   $$
   $$
   (252 - 202.9)^2 = 2400.81
   $$
   Sum of squared deviations:
   $$
   2880.41 + 620.41 + 0.81 + 9.61 + 102.01 + 290.41 + 2400.81 = 6304.47
   $$
   Now, divide by $n-1$ (degrees of freedom):
   $$
   s^2 = \frac{6304.47}{6} = 1050.745
   $$
   $$
   s = \sqrt{1050.745} \approx 32.4
   $$

3. **Determine the t-score for a 98% confidence interval with $n-1$ degrees of freedom:**
   For $n-1 = 6$ degrees of freedom, the t-score for a 98% confidence interval is approximately 2.447 (from t-distribution tables).

4. **Calculate the margin of error (ME):**
   $$
   ME = t \times \frac{s}{\sqrt{n}}
   $$
   $$
   ME = 2.447 \times \frac{32.4}{\sqrt{7}}
   $$
   $$
   ME = 2.447 \times 12.24 \approx 29.9
   $$

5. **Construct the confidence interval:**
   $$
   \text{Lower limit} = \bar{x} - ME = 202.9 - 29.9 = 173.0
   $$
   $$
   \text{Upper limit} = \bar{x} + ME = 202.9 + 29.9 = 232.8
   $$

Therefore, the 98% confidence interval for the mean weight of all US males is (173.0, 232.8).

Weights. I sample 7 random males in the US and record their weights in pounds. Some of their data are given below.149 178 202 206 213 220 252Find a 98% confidence interval for the mean weight of all US males. (Round to one decimal place)(,)

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