In a simple random sample of 135 households, the sample mean number of personal computers was 1.22. Assume the population standard deviation is o=0.67.

Question

(a) Construct a 95% confidence interval for the mean number of personal computers. Round the answer to at least two decimal places.
A 95% confidence interval for the mean number of personal computers is [blank] < u < [blank]

GPT-4o mini · Answer

To construct a 95% confidence interval for the mean number of personal computers, we can use the following formula:

\[ \text{Confidence Interval} = \bar{x} \pm z_{\alpha/2} \left(\frac{\sigma}{\sqrt{n}}\right) \]

Where:

$\bar{x}$ is the sample mean,
$z_{\alpha/2}$ is the z-value that corresponds to the desired confidence level (for 95%, $z_{\alpha/2} \approx 1.96$),
$\sigma$ is the population standard deviation,
$n$ is the sample size.

Given:

$\bar{x} = 1.22$
$\sigma = 0.67$
$n = 135$

Step 1: Calculate the standard error (SE)

\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{0.67}{\sqrt{135}} \approx \frac{0.67}{11.618} \approx 0.0576 \]

Step 2: Find the z-value for 95% confidence level

For a 95% confidence level, $z_{\alpha/2} \approx 1.96$.

Step 3: Calculate the margin of error (ME)

\[ ME = z_{\alpha/2} \cdot SE = 1.96 \cdot 0.0576 \approx 0.1134 \]

Step 4: Create the confidence interval

\[ \text{Confidence Interval} = \bar{x} \pm ME \] \[ \text{Lower limit} = 1.22 - 0.1134 \approx 1.1066 \] \[ \text{Upper limit} = 1.22 + 0.1134 \approx 1.3334 \]

Step 5: Round to at least two decimal places

Lower limit: $1.11$
Upper limit: $1.33$

Thus, the 95% confidence interval for the mean number of personal computers is:

\[ [1.11 < \mu < 1.33] \]

Answer

(b) If the sample size were 95 rather than 135, would the margin of error be larger or smaller than the result in part (a)? Explain.

The margin of error would be [smaller or larger] , since [an increase or decrease?] in the sample size will [increase/decrease?] the standard error.

GPT-4o mini · Answer

The margin of error would be **larger**, since a **decrease** in the sample size will **increase** the standard error. **Explanation:** The standard error (SE) is calculated using the formula: $$ SE = \frac{\sigma}{\sqrt{n}} $$ Where $ \sigma $ is the population standard deviation and $ n $ is the sample size. When the sample size decreases (from 135 to 95 in this case), the denominator of the standard error formula (which is the square root of the sample size) becomes smaller. This results in a larger value for the standard error. Since the margin of error (ME) is calculated as: $$ ME = z_{\alpha/2} \cdot SE $$ An increase in the standard error will lead to an increase in the margin of error. Thus: - The margin of error would be **larger**, since a **decrease** in the sample size increases the standard error.

Answer

(c) If the confidence levels were 80% rather than 95%, would the margin of error be larger or smaller than the result in part (a)? Explain.

GPT-4o mini · Answer

If the confidence level were 80% rather than 95%, the margin of error would be smaller.

Explanation:

The margin of error (ME) is calculated using the formula:

\[ ME = z_{\alpha/2} \cdot SE \]

Where:

$ z_{\alpha/2} $ is the z-value corresponding to the desired confidence level.
$ SE $ is the standard error, which remains the same regardless of the confidence level, as it depends only on the sample size and population standard deviation.

For a 95% confidence level, the z-value (approximately 1.96) is larger than for an 80% confidence level (approximately 1.28).

Since the margin of error incorporates the z-value, if we decrease the confidence level to 80%, the corresponding z-value decreases, which directly leads to a smaller margin of error.

Thus, with an 80% confidence level:

The margin of error would be smaller since reducing the confidence level decreases the z-value used in the margin of error calculation.

Answer

(d) Based on the confidence interval constructed in part (a), is it likely that the mean number of personal computers is less than 0.94?

GPT-4o mini · Answer

To determine whether it is likely that the mean number of personal computers is less than 0.94 based on the 95% confidence interval constructed in part (a), we need to consider the confidence interval values.

From part (a), the 95% confidence interval for the mean number of personal computers was calculated as:

\[ [1.11 < \mu < 1.33] \]

Since the entire confidence interval (1.11 to 1.33) is above 0.94, it suggests that we are 95% confident that the true population mean number of personal computers is between 1.11 and 1.33.

Therefore, based on the confidence interval:

It is not likely that the mean number of personal computers is less than 0.94, because the lower bound of the confidence interval (1.11) is greater than 0.94. In other words, this indicates that values below 0.94 are not supported by our sample data.