Question 7 A restaurant owner is considering opening an ethnic cuisine restaurant and wonders how much people in the target community spend on eating out relative to their annual household incomes. She commissions a telephone survey and builds a regression model for individual respondents, where household income is in $1000's. Based on this model she obtains a Prediction Interval for eating out per week, with 99% confidence, for a family with $120,000 annual income: ($100 to $170). Select the proper interpretation for this prediction Interval. 0 There is a 99% chance that the mean amount spent by family with an annual household income of $120,000 is between $100 and $170. 99% of surveyed households with an annual income of $120,000 spend between $100 and $170 eating out each week. We are 99% confident that a family with an annual income of $120,000 will spend between $100 and $170 each week. 99% of all households spend between $100 and $170 to eat out each week. Question 8 Observation no. 1 2 9 10 Mean Standard Deviation Number of salespeople Sales (in $1,000s) (y) working (x) 10 5.64 There are 10 observations in our data. Some rows are hidden. Regression Line bo 8.47 b1 0.90 Se 2.05 SE(b) 0.11 17.5 5.41 Find Margin of Error ME for a 95% Confidence interval for estimating mean sales (in $1,000s) when 10 salespeople are working. 1.012 1.348 O 1.495 0 2.099 Question 9 For obtaining a 99% Confidence Interval for the difference of the means of two normal populations with equal variance, two random samples, one from each population, both of the same size 11, were drawn. For the first sample, sample mean = 23, and sample standard deviation = 12.22. For the second sample, sample mean = 33, and sample standard deviation = 11.38. The critical t-value t* used for obtaining the confidence interval is: 2.845 0 3.169 0 3.331 0 3.153 Question 10 Suppose a population with random variable Y follows a normal distribution. We want to conduct a Left-sided Test of Hypothesis HO: =142, HA: < 142. We take a random sample of 25 observations and find that the sample standard deviations = 35. We will reject the null hypothesis if our calculated y value is sufficiently smaller than a critical y value. Call this critical value y*. Find this critical value y*, if the level of significance a is 0.1. C 136 132.78 0 133.03 136.11