Final answer:
The problem is a hypothesis testing scenario using a t-test to determine if the observed average ATM withdrawal of $171 significantly differs from the expected $160, considering a standard deviation of $34 and sample size of 26. The p-value will guide the decision to accept or reject the null hypothesis.
Step-by-step explanation:
The given scenario involves assessing whether the observed average withdrawal of $171 from a random sample of 26 customer transactions significantly differs from the expected average withdrawal of $160. To analyze this, we can use a hypothesis test, specifically a t-test since we are dealing with a small sample size (< 30) and the population standard deviation is unknown.
The null hypothesis (H0) would assume that the true mean withdrawal is $160. The alternative hypothesis (Ha) posits that the true mean is not $160 (two tailed) or specifically greater than $160 if we suspect higher withdrawals (one-tailed). To execute this test, we calculate the t-statistic using the sample mean ($171), the hypothesized mean ($160), the standard deviation ($34), and the sample size (26).
Using the t-distribution table or a software, we would then compare our calculated t-statistic with the critical t-value to determine if we should reject H0. Our p-value indicates the probability of observing a sample mean as extreme as $171, given that the null hypothesis is true. If the p-value is less than our significance level (α), we reject the null hypothesis and conclude that there is enough evidence that the average withdrawal differs from $160.