Question: 1. Finding predicted values: predict sales for a week with a price of $3.50. = 141,865 24,369.5(3.50) = 56,572 The value we just found is

1. Finding predicted values: predict sales for a week with a price of $3.50. = 141,865 24,369.5(3.50) = 56,572 The value we just found is a __________________ ______________________. We can also report a range of values. JMP instructions for a specific predicted value: Analyze > Fit Y by X Click on sales and then Select columns, Y, Response Click on price and then Select column, X, Factor OK Click on red triangle and select Fit Line Click on red triangle by Linear Fit Click on Save Predicteds Click on Rows (at top, in main menu); Add Rows How many rows to add: 1 Where? at end Dots will appear in row 40 (at the bottom of the data set); in the second column (price), type in 3.50 56572.32222 will appear next to 3.50 in the third column, under predicted sales There are two different confidence intervals we will consider. One is for all items of a particular value, and the other is for individual items. The predicted value, , is the same for both, but there is more variation for individuals than for means of groups, so the two intervals are different. JMP instructions for predicted values of a mean response: Analyze > Fit Y by X Click on sales and then Select Columns, Y, Response Click on price and then Select Column, X, Factor OK Click on red triangle and select Fit Line Click on red triangle by Linear Fit Click on Mean Confidence Limit Formula (you will get 2 new columns, lower 95% mean sales and upper 95% mean sales) Click on Confid Shaded Fit or Confid Curves Indiv (or both) JMP OUTPUT (the first five rows only) The formula for a confidence interval of a predicted mean response is: t n-2 Find a 95% prediction interval for a mean value, given that the average price is $3.66. 56,572 2.026 56,572 3,327 (53,245, 59,899) (Off a bit from JMPs estimate due to rounding error) From JMP data table: Interpretation: We are 95% confident that average _____________________ will be between 53,252 and 59,893 pounds for all weeks in which the price is______________________. The formula for a prediction interval of an individual response is: t n-2 JMP instructions for predicted values of an individual response: Analyze > Fit Y by X Click on sales and then Select Columns, Y, Response Click on price and then Select Column, X, Factor OK Click on red triangle and select Fit Line Click on red triangle by Linear Fit Click on Individual Confidence Limit Formula (you will get 2 new columns, lower 95% indiv sales and upper 95% indiv sales) Click on Confid Shaded Indiv or Confid Curve Fit (or both) Interpretation: we are 95% confident that pizza sales for a single week when price is $3.50/unit is between __________________ and _______________________ pounds. Compare: Which is narrower, the prediction interval for an individual, or the confidence interval for the mean of a group? Why? There is less _________________________ about the mean of a group than about an ____________________________________ response. Things that influence the width of the interval: a. n: larger n makes interval ________________________________ b. se: smaller standard deviation of residuals makes interval ___________________ c. _______________: x close to its mean makes the interval ___________________ 2. Hypothesis tests Test for a significant linear association between X and Y at the 5% level of significance. State the null and alternative hypotheses: H0: H1: Reject H0 if p-value < ; DNR H0 if p-value Since 0.0001 < .05, _________________________ H0. State your conclusion in plain English. 3. Find a 95% confidence interval for 1. JMP instructions for confidence interval of slope: Analyze > Fit Y by X Click on sales and then Select Columns, Y, Response Click on price and then Select Column, X, Factor OK Click on red triangle and select Fit Line Right click on estimate (under parameter estimates section); select columns; lower 95%; upper 95% Interpretation: We are 95% confident that the _____________________________ is in this range. Things that influence consistency of regression slope: (see page 534) Do we want these values to be small or large? a. Spread (variation) around the line. ___________________________________ b. Spread of the Xs. _________________________________________________ c. Sample size. _____________________________________________________ 4. Assumptions and conditions of regression: (see section 16.2 in your text) The assumptions of regression are: 1. _______________________________________________________ 2. ________________________________________________________ 3. ________________________________________________________ 4. ________________________________________________________ To test these assumptions, we will check several conditions. To check conditions: 1. Scatterplot. Checks linearity assumption 2. Residual plot (graph with residuals on vertical axis and predicted values on horizontal axis). Checks equal variance assumption 3. Residual by row plot for time series. Checks independence assumption 4. Shapiro-Wilk test for normality. Checks normality assumption What is a residual? ______________________________________________________ If actual sales for a price of $3.50 turned out to be 60,000 pounds, what would the residual be? ___________________________________________________________ How many residuals can be calculated for this model? _________________________ ********Class Exercise: Residual Plots******** Residual plots are graphs with residuals on the vertical axis and predicted values on the horizontal axis. What to look for: residuals (left-overs) should be random. Residual Plot #1 This is a ___________________ plot. There are no discernable patterns. The variance is equal (constant), there are no non-linear patterns, and no outliers. Residual Plot #2 This graph displays a _________________ shape, which means the variance is not equal. Residual Plot #3 This graph displays a curve, which means the association between X and Y is not _________________________. Residual Plot #4 An _______________________, or unusual observation, is evident in this graph

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