An increase in the rate of consumer savings is frequently tied to a lack of confidence in the economy and is said to be an indicator of a recessional tendency in the economy. A random samping of = 240 savings accounts in a local community showed a man increase in savings account values of 75% over the past 12 months, with a standard deviation of 5.9%. Estimate the mean percent increase savings account values over the past 12 months for depositors in the community Find the 95% margin of error in ) for your estimate. (Round your answer to the decimal places)

Question

An increase in the rate of consumer savings is frequently tied to a lack of confidence in the economy and is said to be an indicator of a recessional tendency in the economy. A random samping of = 240 savings accounts in a local community showed a man increase in savings account values of 75% over the past 12 months, with a standard deviation of 5.9%. Estimate the mean percent increase savings account values over the past 12 months for depositors in the community Find the 95% margin of error in ) for your estimate. (Round your answer to the decimal places)

Answer 1

To estimate the mean percent increase in savings account values over the past 12 months for depositors in the community, we can use the formula for the margin of error:

Margin of Error = Z * (σ / sqrt(n))

Where:
Z = the z-score corresponding to the desired level of confidence (in this case, 95% confidence)
σ = the standard deviation of the population (5.9%)
n = the sample size (240)

To find the z-score for a 95% confidence level, we can use a z-table or a statistical calculator. The value is approximately 1.96.

Using these values in the formula, we have:

Margin of Error = 1.96 * (5.9% / sqrt(240))

Calculating this, we get:

Margin of Error = 1.96 * (0.059 / sqrt(240))
= 1.96 * (0.059 / 15.4919)
= 0.00755

Rounded to two decimal places, the 95% margin of error is 0.01 or 1%.

Answer 2

To estimate the mean percent increase in savings account values over the past 12 months for depositors in the community, we can use a 95% confidence interval.

The formula for the margin of error in estimating the mean is:

Margin of Error = Z * (Standard Deviation / √n)

Where:
Z = Z-score for the desired confidence level (in this case, 95% confidence level)
Standard Deviation = 5.9% (provided)
n = sample size = 240 (provided)

First, we need to find the Z-score for a 95% confidence level. The Z-score can be obtained from a Z-table or a calculator. For a 95% confidence level, the Z-score is approximately 1.96.

Now, we can calculate the margin of error using the formula:

Margin of Error = 1.96 * (5.9% / √240)

Calculating this:

Margin of Error = 1.96 * (0.059 / √240) = 0.051 or 5.1% (rounded to one decimal place)

Therefore, the 95% margin of error for the estimate of the mean percent increase in savings account values over the past 12 months for depositors in the community is 5.1%.