Videos
(a)
Identify the p-value that indicates predictor significance at
(a)
Answer to Problem 32CE
The p-value for ‘% with Bachelor's Degree’ indicates predictor significance at
Explanation of Solution
Calculation:
The given information is that, the dataset of ‘Noodles & Company Sales, Seating, and Demographic data’ contains
Software procedure:
Step by step procedure to obtain p-value for predictor using MegaStat software is given as,
- • Choose MegaStat >
Correlation /Regression>Regression Analysis . - • SelectInput ranges, enter the variable
range for ‘Seats-Inside, Seats-Patio, MedIncome, MedAge, BachDeg%’ as the column of X, Independent variable(s) - • Enter the variable range for ‘Sales/SqFt’ as the column of Y, Dependent variable.
- • Click OK.
Output using MegaStatsoftware is given below:
The p-value for predictor seats-inside is 0.0733.
The p-value for predictor seats-patio is 0.2350.
The p-value for predictor MedIncome is 0.0589.
The p-value for predictor MedAge is 0.9972.
The p-value for predictor BachDeg% is 0.0015.
For seats-inside:
Let
Null hypothesis:
The predictor variable seats-inside is not related to annual sales.
Alternative hypothesis:
The predictor variable seats-inside is related to annual sales.
Rejection rules:
- • If p-value is less than the level of significance then the null hypothesis is rejected. The predictor is significant.
- • If p-value is greater than the level of significance then the null hypothesis is not rejected. The predictor is not significant.
Conclusion:
The p-value for predictor seats-inside is 0.0733.
The level of significance is 0.05.
The p-value is greater than the level of significance.
That is,
The null hypothesis is not rejected.
The predictor variable seats-inside is not related to annual sales.
The predictor seats-inside is not significant.
For seats-Patio:
Let
Null hypothesis:
The predictor variable seats-Patio is not related to annual sales.
Alternative hypothesis:
The predictor variable seats-Patio is related to annual sales.
Conclusion:
The p-value for predictor seats-patio is 0.2350.
The level of significance is 0.05.
The p-value is greater than the level of significance.
That is,
The null hypothesis is not rejected.
The predictor variable seats-patio is not related to annual sales.
The predictor seats-patio is not significant.
For Median Income:
Let
Null hypothesis:
The predictor variable median income is not related to annual sales.
Alternative hypothesis:
The predictor variable median income is related to annual sales.
Conclusion:
The p-value for predictor median income is 0.0589.
The level of significance is 0.05.
The p-value is greater than the level of significance.
That is,
The null hypothesis is not rejected.
The predictor variable median income is not related to annual sales.
The predictor median income is not significant.
For Median Age of population:
Let
Null hypothesis:
The predictor variable median age of population is not related to annual sales.
Alternative hypothesis:
The predictor variable median age of population is related to annual sales.
Conclusion:
The p-value for predictor median age of population is 0.9972.
The level of significance is 0.05.
The p-value is greater than the level of significance.
That is,
The null hypothesis is not rejected.
The predictor variable median age of population is not related to annual sales.
The predictor median age of population is not significant.
For % with Bachelor's Degree:
Let
Null hypothesis:
The predictor variable % with Bachelor's Degree is not related to annual sales.
Alternative hypothesis:
The predictor variable % with Bachelor's Degree is related to annual sales.
Conclusion:
The p-value for predictor ‘% with Bachelor's Degree’ is 0.0015.
The level of significance is 0.05.
The p-value is less than the level of significance.
That is,
The null hypothesis is rejected.
The predictor variable ‘% with Bachelor's Degree’ is related to annual sales.
The predictor ‘% with Bachelor's Degree’of population is significant.
(b)
Explain whether p-values support the conclusions reached from the t tests.
(b)
Answer to Problem 32CE
Yes, p-values support the conclusions reached from the t tests.
Explanation of Solution
Justification: The conclusion reached from the t test is that the predictor variable ‘% with Bachelor's Degree’ is significant predictor and the predictor variables ‘seats-inside, seats-patio, median income, median age of population, ‘% with Bachelor's Degree’’ are not significant predictors. The conclusions for the predictor variables with the p-values at
Hence, p-values support the conclusions reached from the t tests.
(c)
Explain whether the t test or the p-value approach is preferred.
(c)
Answer to Problem 32CE
The p-value approach is preferred because it determines the strength of the significance for the predictor.
Explanation of Solution
Justification: The conclusion using the t test and p-value approach are same, but in most of the tests p-value is preferred because the strength of the significance for the predictor is better determined using the p-value when compared with the t test statistic. Hence, the p-value approach is preferred in tests.
Want to see more full solutions like this?
Chapter 13 Solutions
APPLIED STAT.IN BUS.+ECONOMICS
- Discuss and explain in the picturearrow_forwardBob and Teresa each collect their own samples to test the same hypothesis. Bob’s p-value turns out to be 0.05, and Teresa’s turns out to be 0.01. Why don’t Bob and Teresa get the same p-values? Who has stronger evidence against the null hypothesis: Bob or Teresa?arrow_forwardReview a classmate's Main Post. 1. State if you agree or disagree with the choices made for additional analysis that can be done beyond the frequency table. 2. Choose a measure of central tendency (mean, median, mode) that you would like to compute with the data beyond the frequency table. Complete either a or b below. a. Explain how that analysis can help you understand the data better. b. If you are currently unable to do that analysis, what do you think you could do to make it possible? If you do not think you can do anything, explain why it is not possible.arrow_forward
- 0|0|0|0 - Consider the time series X₁ and Y₁ = (I – B)² (I – B³)Xt. What transformations were performed on Xt to obtain Yt? seasonal difference of order 2 simple difference of order 5 seasonal difference of order 1 seasonal difference of order 5 simple difference of order 2arrow_forwardCalculate the 90% confidence interval for the population mean difference using the data in the attached image. I need to see where I went wrong.arrow_forwardMicrosoft Excel snapshot for random sampling: Also note the formula used for the last column 02 x✓ fx =INDEX(5852:58551, RANK(C2, $C$2:$C$51)) A B 1 No. States 2 1 ALABAMA Rand No. 0.925957526 3 2 ALASKA 0.372999976 4 3 ARIZONA 0.941323044 5 4 ARKANSAS 0.071266381 Random Sample CALIFORNIA NORTH CAROLINA ARKANSAS WASHINGTON G7 Microsoft Excel snapshot for systematic sampling: xfx INDEX(SD52:50551, F7) A B E F G 1 No. States Rand No. Random Sample population 50 2 1 ALABAMA 0.5296685 NEW HAMPSHIRE sample 10 3 2 ALASKA 0.4493186 OKLAHOMA k 5 4 3 ARIZONA 0.707914 KANSAS 5 4 ARKANSAS 0.4831379 NORTH DAKOTA 6 5 CALIFORNIA 0.7277162 INDIANA Random Sample Sample Name 7 6 COLORADO 0.5865002 MISSISSIPPI 8 7:ONNECTICU 0.7640596 ILLINOIS 9 8 DELAWARE 0.5783029 MISSOURI 525 10 15 INDIANA MARYLAND COLORADOarrow_forward
- Suppose the Internal Revenue Service reported that the mean tax refund for the year 2022 was $3401. Assume the standard deviation is $82.5 and that the amounts refunded follow a normal probability distribution. Solve the following three parts? (For the answer to question 14, 15, and 16, start with making a bell curve. Identify on the bell curve where is mean, X, and area(s) to be determined. 1.What percent of the refunds are more than $3,500? 2. What percent of the refunds are more than $3500 but less than $3579? 3. What percent of the refunds are more than $3325 but less than $3579?arrow_forwardA normal distribution has a mean of 50 and a standard deviation of 4. Solve the following three parts? 1. Compute the probability of a value between 44.0 and 55.0. (The question requires finding probability value between 44 and 55. Solve it in 3 steps. In the first step, use the above formula and x = 44, calculate probability value. In the second step repeat the first step with the only difference that x=55. In the third step, subtract the answer of the first part from the answer of the second part.) 2. Compute the probability of a value greater than 55.0. Use the same formula, x=55 and subtract the answer from 1. 3. Compute the probability of a value between 52.0 and 55.0. (The question requires finding probability value between 52 and 55. Solve it in 3 steps. In the first step, use the above formula and x = 52, calculate probability value. In the second step repeat the first step with the only difference that x=55. In the third step, subtract the answer of the first part from the…arrow_forwardIf a uniform distribution is defined over the interval from 6 to 10, then answer the followings: What is the mean of this uniform distribution? Show that the probability of any value between 6 and 10 is equal to 1.0 Find the probability of a value more than 7. Find the probability of a value between 7 and 9. The closing price of Schnur Sporting Goods Inc. common stock is uniformly distributed between $20 and $30 per share. What is the probability that the stock price will be: More than $27? Less than or equal to $24? The April rainfall in Flagstaff, Arizona, follows a uniform distribution between 0.5 and 3.00 inches. What is the mean amount of rainfall for the month? What is the probability of less than an inch of rain for the month? What is the probability of exactly 1.00 inch of rain? What is the probability of more than 1.50 inches of rain for the month? The best way to solve this problem is begin by a step by step creating a chart. Clearly mark the range, identifying the…arrow_forward
- Client 1 Weight before diet (pounds) Weight after diet (pounds) 128 120 2 131 123 3 140 141 4 178 170 5 121 118 6 136 136 7 118 121 8 136 127arrow_forwardClient 1 Weight before diet (pounds) Weight after diet (pounds) 128 120 2 131 123 3 140 141 4 178 170 5 121 118 6 136 136 7 118 121 8 136 127 a) Determine the mean change in patient weight from before to after the diet (after – before). What is the 95% confidence interval of this mean difference?arrow_forwardIn order to find probability, you can use this formula in Microsoft Excel: The best way to understand and solve these problems is by first drawing a bell curve and marking key points such as x, the mean, and the areas of interest. Once marked on the bell curve, figure out what calculations are needed to find the area of interest. =NORM.DIST(x, Mean, Standard Dev., TRUE). When the question mentions “greater than” you may have to subtract your answer from 1. When the question mentions “between (two values)”, you need to do separate calculation for both values and then subtract their results to get the answer. 1. Compute the probability of a value between 44.0 and 55.0. (The question requires finding probability value between 44 and 55. Solve it in 3 steps. In the first step, use the above formula and x = 44, calculate probability value. In the second step repeat the first step with the only difference that x=55. In the third step, subtract the answer of the first part from the…arrow_forward
Holt Mcdougal Larson Pre-algebra: Student Edition...AlgebraISBN:9780547587776Author:HOLT MCDOUGALPublisher:HOLT MCDOUGAL
Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw Hill
Big Ideas Math A Bridge To Success Algebra 1: Stu...AlgebraISBN:9781680331141Author:HOUGHTON MIFFLIN HARCOURTPublisher:Houghton Mifflin Harcourt
Functions and Change: A Modeling Approach to Coll...AlgebraISBN:9781337111348Author:Bruce Crauder, Benny Evans, Alan NoellPublisher:Cengage Learning