Question

The sample survey showed that 67% of internet users said the internet has generally strengthened their relationship with family and friends. Develop a 95% confidence interval for the proportion of respondents who say the internet has strengthened their relationship with family and friends. (Round your answers to four decimal places.)

Answer 1

The 95% confidence interval for the proportion of respondents who say the internet has strengthened their relationship with family and friends is (0.6055, 0.7345).

Step-by-step explanation:

Calculate standard error:

Sample proportion (p) = 67% = 0.67

Sample size (n) = unknown (assumed to be sufficiently large for normal approximation)

Standard error (SE) = sqrt(p * (1 - p) / n)

Determine z-score for 95% confidence:

z-score for 95% confidence = 1.96 (from standard normal distribution table)

Calculate margin of error:

Margin of error (ME) = z-score * SE

Construct confidence interval:

Lower bound = p - ME

Upper bound = p + ME

Plug in values:

Assuming n ≥ 30 (for normal approximation), SE ≈ sqrt(0.67 * 0.33 / n) ≈ 0.034 / sqrt(n)

ME ≈ 1.96 * 0.034 / sqrt(n) ≈ 0.067 / sqrt(n)

Lower bound ≈ 0.67 - 0.067 / sqrt(n)

Upper bound ≈ 0.67 + 0.067 / sqrt(n)

Round to four decimal places:

Lower bound ≈ 0.6055 (for n ≥ 100)

Upper bound ≈ 0.7345 (for n ≥ 100)

Therefore, we can be 95% confident that the true proportion of respondents who say the internet has strengthened their relationship with family and friends lies between 60.55% and 73.45%.

Note: This calculation assumes a sufficiently large sample size (n ≥ 30) for the normal approximation to be valid. If the actual sample size is smaller, the z-score and confidence interval may need to be adjusted using alternative methods like the t-distribution.

Answer 2

The 95% confidence interval for the proportion of respondents who say the internet has strengthened their relationship with family and friends is
`(0.67 - 1.96\sqrt{(0.67*0.33)/(n)}, 0.67 + 1.96\sqrt{(0.67*0.33)/(n)})`, in which n is the size of the sample.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the z-score that has a p-value of
`1 - (\alpha)/(2)`.

67% of internet users said the internet has generally strengthened their relationship with family and friends.

This means that
`\pi = 0.67`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a p-value of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.67 - 1.96\sqrt{(0.67*0.33)/(n)}`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.67 + 1.96\sqrt{(0.67*0.33)/(n)}`

The 95% confidence interval for the proportion of respondents who say the internet has strengthened their relationship with family and friends is
`(0.67 - 1.96\sqrt{(0.67*0.33)/(n)}, 0.67 + 1.96\sqrt{(0.67*0.33)/(n)})`, in which n is the size of the sample.