Question

Bank ATMs must be stocked with enough cash to satisfy customers making withdrawals over an entire weekend. But if too much cash is unnecessarily kept in the ATMs, the bank is forgoing the opportunity of investing the money and earning interest. Suppose that at a particular branch the population mean amount of money withdrawn from ATMs per customer transaction over the weekend is $150, with a population standard deviation of $20. The branch manager wonders whether the average withdrawal amount has changed, so she takes a random sample of 25 customer transactions and finds that the sample mean withdrawal is $155. Using a significance level of .03, do these results indicate that the population average withdrawal is no longer $150? Do a complete and appropriate hypothesis test.

Step 1 (Hypotheses)

H0: σ/x-bar/π/n/s/p/μ =/≠/≤/>/≥/< ______ HA: μ/x-bar/π/p/σ/s/n =/≠/≤/>/≥/< ______

Step 2 (Decision rule)

Using only the appropriate statistical table in your textbook, the critical value for rejecting H₀ is +/-/± . (report your answer to 3 decimal places, using conventional rounding rules)

Step 3 (Test statistic)

Using the sample data, the calculated value of the test statistic is +/-/± . (report your answer to 2 decimal places, using conventional rounding rules)

Step 4 (Evaluate the null hypothesis)

Should the null hypothesis be rejected? yes/no

Step 5 (Practical conclusion)

Is the population average withdrawal still $150? yes/no

Using only the appropriate statistical table in your textbook, what is the most accurate statement you can make about the numerical value of the p-value of this hypothesis test?      

Answer: (provide a one-sentence statement about the p-value)

Answer 1

H₀: $\mu$ = 150

H₁: $\mu \neq$ 150

At alpha = 0.03, the critical values are +/- z_0.015 = +/- 2.17

Reject H₀, if z < -2.17 or z > 2.17

The test statistic z = ( $\bar x - \mu$ )/( $\sigma/\sqrt n$ )

                              = (155 - 150)/(20/ $\sqrt 25$ )

                              = 1.25

Since the test statistic value is not greater than the upper critical value (1.25 < 2.17), so we should not reject the null hypothesis.

Yes, the population average withdrawal still is $150.

P-value = 2 * P(Z > 1.25)

             = 2 * (1 - P(Z < 1.25))

             = 2 * (1 - 0.8944)

             = 0.2112

P-value is the exact probability of committing a type I error.