Question

Problem 5 (7 marks) A survey of 25 retail stores revealed that the average price of a DVD was $375 with a standard deviation of $20. a) What is the 95% confidence interval to estimate the true cost of the DVD? (4 marks) b) What sample size would be needed to estimate the true average price of a DVD with an error of +55 and a 99% confidence? (3 marks) Problem 6 (10 marks) Based on the BBM TV ratings, the Sudbury local television claims its 11:00 PM newscast reaches at least 41% of the viewing audience in the area. In a survey of 100 viewers, 36 indicated that they watch the late evening news on this Sudbury local television. At a significance level a -0.05, can you infer that the Sudbury local television claim is false? 24 - Problem 7 (20 marks) A sales manager for an advertising agency believes there is a relationship between the number of contacts and the amount of the sales. To verify this believe, the following data was collected Salesperson Number of Contacts Sales (in thousands) 14 2 12 14 20 28 16 30 S 80 6 7 18 90 8 50 85 9 55 10 50 110 3 4 16 23 120 Assume normality of variables. a) Calculate the coefficient of correlation r. Provide an interpretation of the computed value of r. (9 marks) b) Calculate the coefficient de determination r square (R2). Provide an interpretation of the computed value of r square (R2). (3 marks) c) Determine the least squares line. Provide an interpretation of your results. (8 marks) 5

Answer 1

We would be looking at the first question all parts here as:

Q5) a) For n - 1 = 24 degrees of freedom, we have from t distribution tables here:
P( t₂₄ < 2.064) = 0.975,

Therefore, due to symmetry, we have here:
P( -2.064 < t₂₄ < 2.064) = 0.95

The confidence interval here is obtained as:

$\bar X \pm t^*\frac{s}{\sqrt{n}}$

20 375 +2.064* 25

$375 \pm 8.256$

This is the required 95% confidence interval for the true cost of the DVD here.

b) From standard normal tables, we have here:
P(-2.576 < Z < 2.576) = 0.99

The margin of error here is computed as:

$MOE = z^*\frac{\sigma}{\sqrt{n}}$

$n = (z^*\frac{\sigma}{MOE})^2 = (\frac{2.576*20}{5})^2 = 106.17$

Therefore 107 is the required sample size here.