posted by Bev .
The manager of a vacation resort has claimed that, on average, a guest spends at least $3000 at the resort during a one week stay, including meals and entertainment. A member of the accounting staff does not believe the amount is that high. They authorize you to settle their dispute. You take a Simple Random Sample (n = 18) of guests that had stayed a the resort over the last several weeks. Tracking the spending of these guests at the resort you determine that the guests spend a mean of $2925 with a standard deviation of $93. Test the manager's claim that the guests spend a mean of $3000 or more versus the accountant's claim that the mean is actually less than $3000. You may assume a = .05 and the population is symmetric and mounded.
a.p0 (Ho Ha
b. p0 What is the critical value of the test statistic?
c. p0 State the decision rule
d. p0 Sketch the region(s) of acceptance and rejection in the graph
e. p0 Show the calculated test statistic and then state the statistical decision.
f. p0 What would you tell the manager and accountant about their dispute?
Business Statistics -
Ho: µ ≥ 3000 -->this is the null hypothesis.
Ha: µ < 3000 -->this is the alternate or alternative hypothesis.
You can use a one-sample t-test formula for this problem since the sample size is less than 30.
Using a t-table at 0.05 level of significance for a one-tailed test (alternate or alternative hypothesis shows a specific direction) at 17 degrees of freedom (df = n - 1 = 18 - 1 = 17), check for your critical value. This will determine your rejection region.
Does the test statistic from the t-test formula exceed the critical value from the t-table? If it does, reject the null. If the test statistic does not exceed the critical value from the t-table, do not reject the null. You can draw your conclusions from there.
I hope this will help get you started.