A sample of 64 account balances from a credit company showed an average daily balance of $1,040. The standard deviation of the population is known to be $200. We are interested in determining if the mean of all account balances (i. e., population mean) is significantly different from $1,000.

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Business

Crystal Hodges

27 April, 23:23

A sample of 64 account balances from a credit company showed an average daily balance of $1,040. The standard deviation of the population is known to be $200. We are interested in determining if the mean of all account balances (i. e., population mean) is significantly different from $1,000.

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Phillip Bailey · Answer 1

With 95% of confidence, there is no statistical evidence to confirm that the mean of all account balances is different from $1,000.

Explanation:

Null hypothesis (H0) : μ=1000

Alternative hypothesis (H1) : μ≠1000

We must use the z distribution because we know the population standard deviation.

z-statistic formula:

z = (xbar-m) / (σ / (sqrt (n)))

xbar: sample mean

m: hypothesized value

σ: population standard deviation

n: number of observations

z = (1,040-1,000) / (200/sqrt (64))

z-statistic = 1.6

Because the problem do not specify the significance level, we will use 5%

The critical value from the z distribution with, 64-1 degrees of freedom and 2.5% significance level (because is a two-tail test), is 1,96.

Because the z-statistic is less than the critical value, there is no statistical evidence to confirm that the mean of all account balances is different from $1,000.