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Question
The manager of a vacation resort has claimed that, on average, a guest spends at least $1600 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 sample of 16 guests that had stayed at the resort over the last several weeks. Tracking the spending of these guests at the resort you determine that the guests spent a mean of $1478. Assume that the population standard deviation is $160. Test the accountant's claim that the mean is actually less than $1600 versus the manager's claim that the guests spend a mean of $1600 or more Show all problem steps. You may assume alpha = .05 .
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
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To test the accountant's claim against the manager's claim, we will use a hypothesis test.
Step 1: State the hypotheses.
- Null Hypothesis (H0): The population mean spending is greater than or equal to $1600. (μ ≥ $1600)
- Alternative Hypothesis (H1): The population mean spending is less than $1600. (μ < $1600)
Step 2: Set the significance level.
The significance level, denoted as α, is given as α = 0.05. This means that we are willing to accept a 5% chance of making a Type I error (rejecting the null hypothesis when it is true).
Step 3: Calculate the test statistic.
The test statistic for this problem is the t-statistic since the population standard deviation is unknown and we have a small sample size.
The formula for the t-statistic is:
t = (sample mean - hypothesized mean) / (sample standard deviation / √n)
Where:
- Sample mean = $1478
- Hypothesized mean = $1600
- Sample standard deviation = $160 (given)
- n = 16 (sample size)
Calculating the values and substituting them into the formula, we get:
t = (1478 - 1600) / (160 / √16)
t = -122 / (160 / 4)
t = -122 / 40
t ≈ -3.05
Step 4: Determine the critical value.
Since this is a one-tailed test (less than), we need to find the critical value using the t-distribution table. With a sample size of 16 and α = 0.05, the critical value is -1.753.
Step 5: Make a decision.
If the test statistic is more extreme than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, the test statistic (-3.05) is more extreme than the critical value (-1.753). Therefore, we reject the null hypothesis.
Step 6: State the conclusion.
Based on the sample evidence, we have enough evidence to conclude that the mean spending of guests at the resort is less than $1600. Therefore, we support the accountant's claim and do not support the manager's claim.