A hotel chain is interested in evaluating processes. Guests can reserve a room by using either a telephone system or an online system that is accessed through the hotel's web site. Independent random samples of 80 guests who reserved a room by phone and 60 guests who reserved a room online were selected. Of those who reserved by phone, 57 reported that they were satisfied with the reservation process. Of those who reserved online, 50 reported that they were satisfied. Based on the data, is it reasonable to conclude that the proportion who are satisfied is greater for those who reserve a room online? Test the appropriate hypotheses using a 5% significance level.
Have cells indicating satisfied/dissatisfied for each group. Calculate column, row and grand totals.
Use the Chi-square (X^2) method.
X^2 = ∑ (O-E)^2/E, where O = observed frequency and E = expected frequency.
∑ = sum of all the cells.
E = (column total * row total)/grand total
df = n - 1, where n = number of cells
Look up value in X^2 table in the back of your textbook.
To determine if the proportion of guests satisfied when reserving a room online is greater than those satisfied when reserving by phone, we can conduct a hypothesis test.
Let's define our hypotheses:
Null hypothesis (H₀): The proportion of guests satisfied when reserving a room online is equal to or less than the proportion of guests satisfied when reserving by phone.
Alternative hypothesis (H₁): The proportion of guests satisfied when reserving a room online is greater than the proportion of guests satisfied when reserving by phone.
To test these hypotheses, we can use a one-sided z-test for the difference in proportions.
Step 1: Specify the hypotheses
H₀: p1 ≤ p2
H₁: p1 > p2
where p1 represents the proportion satisfied when reserving online and p2 represents the proportion satisfied when reserving by phone.
Step 2: Set the significance level (α)
In this case, we are using a 5% significance level.
Step 3: Compute the test statistic
We will calculate the test statistic, which is the z-score for the difference in proportions:
z = (p1 - p2) / sqrt (p * (1-p) * (1/n1 + 1/n2))
where p is the combined proportion, n1 is the sample size for phone reservations, and n2 is the sample size for online reservations.
p = (x1 + x2) / (n1 + n2)
where x1 is the number of satisfied phone reservations, and x2 is the number of satisfied online reservations.
Step 4: Determine the critical value
Since we are conducting a one-sided test, we need to find the critical value for a 95% confidence level. In this case, the critical value is 1.645.
Step 5: Compare the test statistic with the critical value
If the test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Step 6: Make a conclusion
Based on the comparison of the test statistic and critical value, we can make a conclusion about whether the proportion of guests satisfied when reserving a room online is greater.
Please provide the number of satisfied phone reservations and the number of satisfied online reservations, and we can calculate the test statistic to make the final conclusion.
To test the hypothesis whether the proportion of guests who are satisfied is greater for those who reserve a room online, we can use the hypothesis testing for two independent proportions.
Let's define the two populations:
Population 1: Guests who reserved a room by phone.
Population 2: Guests who reserved a room online.
Now we can set up the null and alternative hypotheses:
Null Hypothesis: The proportion of guests who are satisfied is the same for those who reserve a room by phone and those who reserve online. In mathematical terms:
H0: p1 = p2
Alternative Hypothesis: The proportion of guests who are satisfied is greater for those who reserve a room online. In mathematical terms:
Ha: p1 < p2
Here, p1 represents the proportion of guests who are satisfied when reserving by phone, and p2 represents the proportion of guests who are satisfied when reserving online.
Next, we can calculate the test statistic Z using the following formula:
Z = (p̂1 - p̂2) / sqrt((p̂(1-p̂)(1/n1 + 1/n2)))
Where:
p̂1 = Number of guests satisfied in phone reservation / Number of guests who reserved by phone
p̂2 = Number of guests satisfied in online reservation / Number of guests who reserved online
n1 = Sample size for phone reservation
n2 = Sample size for online reservation
Now, let's calculate the values required to compute the test statistic:
p̂1 = 57/80 = 0.7125
p̂2 = 50/60 = 0.8333
n1 = 80
n2 = 60
Substituting these values into the formula, we get:
Z = (0.7125 - 0.8333) / sqrt((0.7125 * (1-0.7125) / 80) + (0.8333 * (1-0.8333) / 60))
Calculating this expression yields: Z ≈ -1.92.
To determine the critical value for a one-sided test at a 5% significance level, we look up the z-score in the standard normal distribution table. At a 5% significance level, the critical value is approximately -1.645.
Since our test statistic (-1.92) is less than the critical value (-1.645), we have enough evidence to reject the null hypothesis. This means that it is reasonable to conclude that the proportion who are satisfied is greater for those who reserve a room online.
Therefore, based on the data and the hypothesis test, we can conclude that the online reservation system has a higher proportion of satisfied guests compared to the phone reservation system.