Question

55.

(same story as previous question)

A sample of 64 account balances from a credit company showed an average daily balance of $1145.

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 $1100. (In other words, we are testing the Null Hypothesis that the mean account balance for the population is $1100)

Which of the following is true

	The Null Hypothesis can be rejected at the 2.5% significance level
	The Null Hypothesis cannot be rejected at the 10% significance level
	The Null Hypothesis cannot be rejected at the 20% significance level
	The Null Hypothesis can be rejected at the 1% significance level
	The Null Hypothesis cannot be rejected at the 5% significance level

44.

A sample of 64 account balances from a credit company showed an average daily balance of $1145.

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 $1100.

Compute the test statistic.

(note: enter answer as a number with two decimal places - such as 2.77 or 0.95)

Answer 1

55)

Sample size = n = 64

Sample mean = = 1145

Population standard deviation = = 200

Claim: The mean of all account balances is significantly different from $1100.

The null and alternative hypothesis is

Here population standard deviation is known so we have to use z-test statistic.

Test statistic is

P-value = 2*P(Z > 1.8) = 0.0719

If p-value < level of significance we reject null hypothesis.

At 10% and 20% we reject null hypothesis.

alpha	Reject
0.025	not
0.1	yes
0.2	yes
0.01	not
0.05	not

Correct option is

The Null Hypothesis cannot be rejected at the 5% significance level

44)

Sample size = n = 64

Sample mean = = 1145

Population standard deviation = = 200

Here population standard deviation is known so we have to use z-test statistic.

Test statistic is