AN20231216-925

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Institute of Health Profession Education & Research, KMU Peshawar *

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Accounting

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Nov 24, 2024

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Last Name 1 Student’s Name Professor’s Name Course Number Date FINAL PROBLEMS 1. a. To calculate the expected mean annual payment ( μ ), we can use the following formula: μ = p 1 ∗ x 1 + p 2 ∗ x 2 + p 3 ∗ x 3 where ; p i = probability of event i x i = payment associated with event i p 1 = 0.02 x 1 = $ 15000 p 2 = 0.05 x 2 = $ 8000 p 3 = 0.1 x 3 = $ 5000 μ = ( 0.02 ∗ 15000 ) + ( 0.05 ∗ 8000 ) +( 0.1 ∗ 5000 ) μ = $ 1200 So, the expected mean annual payment to a client ( μ ) is $1,200. b. σ = √ p 1 ( x 1 − μ ) 2 + p 2 ( x 2 − μ ) 2 + p 3 ( x 3 − μ ) 2 ¿ √ 0.02 ( 15000 − 1200 ) 2 + 0.05 ( 8000 − 1200 ) 2 + 0.1 ( 5000 − 1200 ) 2 σ = 2750.42

Last Name 2 So, the expected standard deviation of the annual payments to clients ( σ ) is approximately $2750.42 2. n= 500000 μ = 950 σ = 4220 P ( X < z ) = ∅ ( x − μ σ √ n ) P ( X < 960 ) = ∅ ( 960 − 950 4220 √ 500000 ) ¿ ∅ ( 1.676 ) Using the normal distribution table, P ( X < 960 ) = 0.9531 The probability that the plan will be profitable in the upcoming year is 0.9531. 3. a. To make predictions about the mean annual income ( μ ) of all NY teachers based on a sample, we can use the confidence interval formula: Confidenceinterval = x ±z ( s √ n ) Since we have very large n, we can use z In this case, you want to calculate a 99% confidence interval. The z-score for a 99% confidence level is approximately 2.582. So, CI = 60000 ± 2.582 ∗ ( 7500 √ 800 )

Last Name 3 CI = 60000 ± 684.66 Now, calculate the upper and lower bounds of the confidence interval: Upper bound: CI = 60000 + 684.66 = 60684.66 Lower bound: CI= 60000 − 684.66 = 59315.34 Prediction about μ : The prediction is that the mean annual income ( μ ) of all NY teachers is within the range of approximately $59,315.34 to $60,684.66. b. Margin of error= z ( s √ n ) = 2.582 ∗ ( 7500 √ 800 ) Margin of error= 684.66 4. a. Margin of error= t ( s √ n ) sample size= n=40 sample mean = x = ¿ 60,000 sample standard deviation = s= 7500 df= 40- 1=39 t (0.01/2) , 39 = 2.708 = 2.708 ∗ ( 7500 √ 40 ) = 3211.29 b. Confidenceinterval = x ±t ( s √ n ) = 60000 ± 3211.29 = (56788.71, 63211.29)

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Last Name 4 If repeated samples of equal size are taken then in 99% of the time the population mean annual income will fall within $56888.71 and $63211.29 5. a. The null hypothesis ( H 0) is a statement of no effect or no difference. In this case, it would be that drinking ZXZ does not affect the average time to run the marathon. The alternative hypothesis ( Ha ) is a statement of an effect or a difference. In this case, it would be that drinking ZXZ helps runners to run the marathon faster. So, the hypotheses are: H 0: μ =289 H 1: μ <289 b. t = x − μ s √ n Given the information:  Sample mean (ˉ X ˉ) = 281 minutes  Population mean under the null hypothesis ( μ ) = 289 minutes  Sample standard deviation ( s ) is 26 minutes  Sample size ( n ) = 30 t = 281 − 289 26 √ 30 t= -1.69 df= n - 1 = 30 - 1= 29 P-value using excel =TDIST(-1.69,29,1)=0.0509

Last Name 5 Since the P-value(0.0509) is greater than alpha(0.05) we fail to reject the null hypothesis So we do not have sufficient evidence to support the claim that ZXZ helps runners to run the marathon faster. 6. a. The null hypothesis ( H 0) is a statement of no effect or no difference. In this case, it would be that drinking ZXZ does not affect the average time to run the marathon. The alternative hypothesis ( Ha ) is a statement of an effect or a difference. In this case, it would be that drinking ZXZ helps runners to run the marathon faster. So, the hypotheses are: H 0: μ =289 H 1: μ <289 b. t = x − μ s √ n Given the information:  Sample mean (ˉ X ˉ) = 281 minutes  Population mean under the null hypothesis ( μ ) = 289 minutes  Sample standard deviation ( s ) is 26 minutes  Sample size ( n ) = 30 t = 281 − 289 26 √ 30 t= -1.69

Last Name 6 df= n - 1 = 30 - 1= 29 P-value using excel =TDIST(-1.69,29,1)=0.0509 Since the P-value(0.0509) is greater than alpha(0.05) we fail to reject the null hypothesis So we do not have sufficient evidence to support the claim that ZXZ helps runners to run the marathon faster.

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