In exercise 12, the following data on x = average daily hotel room rate and y = amount spent on entertainment (The Wall Street Journal, August 18, 2011) lead to the estimated regression equation ŷ = 17.49 + 1.0334x. For these data SSE = 1541.4.
City: Boston Denver Nashville New Orleans Phoenix San Diego San Francisco San Jose Tamp
Room Rate ($) 148 96 91 110 90 102 136 90 82
Entertainment: 161 105 101 142 100 120 167 140 98
a. Predict the amount spent on entertainment for a particular city that has a daily room rate of $89 (to 2 decimals).
Answer: 109.46
b. Develop a 95% confidence interval for the mean amount spent on entertainment for all cities that have a daily room rate of $89 (to 2 decimals
$____ to $_____.
c. The average room rate in Chicago is $128. Develop a 95% prediction interval for the amount spent on entertainment in Chicago (to 2 decimals).
$___ to $____.
I already solved part A. I just need help with B and C.
For part B I believe the equation to use is t alpha/2(n-2)*s SQRT (1/n + (x - xbar)^2 / SSxx).
For part C I think the equation is t alpha/2(n-2)*s SQRT (1 + 1/n + (x - xbar)^2 / SSxx).
The issue I am having is I don't know how what to plug into each of these equations. I feel like I am missing some information. Please help, thank you.
To solve part B, you are correct in using the equation for the confidence interval. Let's break down the formula and see what information we need to plug in.
The formula for the confidence interval is:
CI = x̄ ± tα/2 * (s × √(1/n + (x - x̄)² / SSxx))
Where:
- CI is the confidence interval
- x̄ is the sample mean
- tα/2 is the critical value from the t-distribution for a specific confidence level
- s is the sample standard deviation
- n is the sample size
- x is the value for which you want to estimate the mean
- SSxx is the sum of squares of the independent variable (x)
For this problem, you want to find a confidence interval for the mean amount spent on entertainment for all cities that have a daily room rate of $89.
Here's what you need to do:
1. Calculate x̄ (sample mean) and s (sample standard deviation) for the given data. In this case, x̄ and s represent the average amount spent on entertainment and the standard deviation of that amount for cities with a daily room rate of $89.
2. Determine the sample size, n. This is the number of observations for cities with a daily room rate of $89.
3. Look up the critical value, tα/2, from the t-distribution table for a 95% confidence level and the degrees of freedom (n-2), where n is the sample size.
4. Plug in the values into the formula to calculate the confidence interval.
For part C, you need to calculate the prediction interval. The formula is similar, but there are a couple of differences. Let's break it down:
PI = ŷ ± tα/2 * (s × √(1 + 1/n + (x - x̄)² / SSxx))
Where:
- PI is the prediction interval
- ŷ is the predicted value
- tα/2 is the critical value from the t-distribution for a specific confidence level
- s is the sample standard deviation
- n is the sample size
- x is the value for which you want to estimate the amount spent on entertainment (in this case, the average room rate in Chicago)
- x̄ is the sample mean of the independent variable (x)
- SSxx is the sum of squares of the independent variable (x)
To solve part C, you need to follow a similar process as in part B but with the appropriate values for x (average room rate in Chicago), x̄, and SSxx.
I hope this helps and clarifies the steps you need to take for parts B and C! Let me know if you have any further questions.
For part B, to develop a confidence interval for the mean amount spent on entertainment for all cities that have a daily room rate of $89, you will need to calculate the standard error and use the t-distribution.
First, let's calculate the standard error:
s = sqrt(SSE / (n-2))
s = sqrt(1541.4 / (9-2))
s ≈ 11.80 (rounded to 2 decimal places)
Next, calculate the standard error of the mean:
SE = s / sqrt(n)
SE = 11.80 / sqrt(9)
SE ≈ 3.93 (rounded to 2 decimal places)
Since we have a t-distribution, we need to find the critical value for a 95% confidence interval with (n-2) degrees of freedom. With (9-2) = 7 degrees of freedom, the critical value is approximately 2.365 (you can look this up on a t-distribution table or use software).
Now, calculate the margin of error:
ME = critical value * SE
ME ≈ 2.365 * 3.93
ME ≈ 9.29 (rounded to 2 decimal places)
Finally, calculate the confidence interval:
CI = mean ± ME
CI ≈ 109.46 ± 9.29
CI ≈ $100.17 to $118.75
Therefore, the 95% confidence interval for the mean amount spent on entertainment for all cities that have a daily room rate of $89 is approximately $100.17 to $118.75.
For part C, to develop a prediction interval for the amount spent on entertainment in Chicago, you will also need to use the t-distribution.
First, let's calculate the standard error following the same steps as in part B:
s = sqrt(SSE / (n-2))
s = sqrt(1541.4 / (9-2))
s ≈ 11.80 (rounded to 2 decimal places)
Next, calculate the standard error of the prediction:
SE_pred = s * sqrt(1 + 1/n + (x - xbar)^2 / SSxx)
SE_pred = 11.80 * sqrt(1 + 1/9 + (128 - mean(x))^2 / SSxx)
To calculate SSxx, we need to know the sum of squares of x (SSx) and the number of observations (n). However, this information is not provided in the question.
Without the sum of squares of x (SSx), we cannot calculate the standard error of the prediction or the prediction interval for Chicago. Therefore, we are missing crucial information for part C. If you have the SSx value or any additional information, please provide it, and I will be happy to assist you further.