{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Concordia-Comm215 Mock Final Solutions

# Concordia-Comm215 Mock Final Solutions - COMM 215 MOCK...

This preview shows pages 1–3. Sign up to view the full content.

Question 1 Using these results: 75 56 98 73 89 92 82 52 78 65 final calculus mark 52 34 75 28 47 57 64 21 43 39 achievement test a) Calculate the coefficient of correlation for this data. Solution: data: a8 x = 460, a8 x 2 = 23634 a8 y = 760 a8 y 2 = 59816 a8 xy = 36854 r = n a8 xy a8 x a8 y n ( a8 x 2 ) ( a8 x ) 2 n ( a8 y 2 ) ( a8 y ) 2 r = 10 ( 36854 ) − ( 460 )( 760 ) 10 ( 23634 ) ( 460 ) 2 10 ( 59816 ) ( 760 ) 2 = 0.8398 l 0.84 b) Find the equation of the line of best fit for this data. Solution: b 0 = 40.78 b 1 = 0.766 y = 40.78 + 0.766 x c) Predict the final calculus grade for a student who scored 50 on the Achievement Test. Solution: y = 40.78 + 0.766 (50) = 79.08 or 79%. Question 2 a) For a random sample of 16 summer days in Montreal, the mean level of CO was found to be 4.9 ppm. With a standard deviation of 1.2. Based on this, construct a 95% confidence interval estimate of the true mean level of CO in summer in Montreal. Solution: Since n = 16, this a small sample t -distribution situation. We know that , s = 1.2, and x = 4.9 s x = 1.2 16 = 0.3 degrees of freedom = 15, so t a/2 , 15 = 2.131, so our confidence limits are 4.9 ± 2.131(0.3) = 4.9 ± 0.639 the confidence interval for the mean level of CO is [ 4.26, 5.54 ] parts per million. b) A preliminary study indicated that a new drug was effective in 75% of its trials. Accordingly, how large a sample would be required to estimate the true effective rate of the drug to within 3% with 95% confidence. Solution: We’re finding sample size given a proportion, the size of the error and the level of confidence. The formula for sample size under these conditions is: n = p ( 1 p ) z a14 /2 E 2 95% confidence means z = 1.96, we know p = 0.75 and E = 0.03 , so sample size is 801 people. n = 0.75 ( 0.25 ) 1.96 0.03 2 = 800.3333 COMM 215 MOCK FINAL EXAM SOLUTIONS exam solutions page 1 Solutions will be posted at www.prep101.com/solutions

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Question 3 A report said the average height of CEGEP basketball players is at least 180 cm. At a recent tournament a random sample of 25 players yielded the following data 196 182 176 177 187 190 184 180 185 178 183 180 189 183 181 197 195 180 179 177 180 178 200 181 176 height in cm of 25 Cégep basketball players Do these results support the report at the 0.05 Level of Significance? Solution: Since n = 25, we’ll use the t -distribution for this upper tail test statistic. a8 x = 4594, x = 185.76, s = 6.98, t 0.05, 24 = 1.711 Conclusion: the test supports the claim that the average height of CEGEP basketball players is at least 180 cm.. Decision: 4.12 is WAY BIGGER than 1.711, we must heave Ho. Test Statistic: t = x a25 s / n = 185.76 180 6.98/ 25 = 4.12 Criteria: if t > 1.711, heave Ho Level of significance: = 5 % , df = 24 a14 Ha: a25 > 180 Hypotheses: Ho: a25 [ 180 Value of interest: µ , mean height of players Small Sample Hypothesis Test on a Mean Question 4 In a study of air-bag effectiveness, it was found that in 821 crashes of midsize cars equipped with air-bags, 46 of the crashes resulted in the hospitalization of the drivers. Use a 0.01 level of significance to test the claim that the hospitalization rate for crashes of midsize cars equipped with air-bags is less than 7.8%.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 8

Concordia-Comm215 Mock Final Solutions - COMM 215 MOCK...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online