Question

There are 350 Families lives in the small town of America. A poll of 50 families revealed the mean annual church contribution is $550 with the standard deviation of $ 75.Construct the 95% confidence interval for the mean annual contribution.

Answer 1

The 95% confidence interval of the mean annual contribution,
`\bar{x}` is

$529.21 <
`\bar{x}` < $570.79

Explanation:

The parameters given are;

The sample size, n = 50

The mean annual church contribution,
`\bar{x}` = $550

The standard deviation, σ = $75

The confidence level = 95%

The formula for finding the confidence interval having a known mean is given as follows;

`CI=\bar{x}\pm z_(\alpha/2) * (\sigma)/(√(n))`

Where, the z value at 95% confidence level = 1.96, we therefore have;

`CI=550 \pm 1.96 * (75)/(√(50))`

Which gives the 95% confidence interval of the mean annual contribution,
`\bar{x}`, as follows;

$529.21 <
`\bar{x}` < $570.79

Answer 2

550 ± 20.789

= [529.211 – 570.789]

Explanation:

Before constructing the confidence interval, we would start by calculating the confidence interval

The formula for confidence interval =

μ ± Z × σ/√n

Where

μ = mean

Z = Z score of the confidence interval

σ = Standard Deviation

n = number of the samples

From the question

μ = $550

Z = we are given a 95% confidence interval. The Z score = 1.96

σ = $75

n = 50

Hence,

Confidence interval = μ ± Z × σ/√n

Confidence interval = 550 ± 1.96 × 75/√50

= 550 ± 1.96(10.606601718)

= 550 ±20.789

= [529.211 – 570.789]

Therefore, the 95% confidence interval for the mean annual contribution is

= 550 ± 20.789

= [529.211 – 570.789]