Statistical Analysis of ATM Withdrawals and Investment Models

HOA 3 Name: Sakshi Jain SHORT ANSWER 1) The main aim is to test whether there is any significant that the mean withdrawals from a college campus ATM is less than 100 at 5% level of significance. The hypotheses for the test are given below: Null hypothesis: Alternative hypothesis: This test is corresponding to left tailed test and the population standard deviation value is unknown, one sample t- test can be used. From the output, the test statistics value is obtained as -0.68 and the P-value for the test is 0.2518. Decision about the null hypothesis: If the p-value is less than the level of significant, then the null hypothesis should be rejected, otherwise fail to reject the null hypothesis. Here, the p -value is greater than the level of significance value. That is, 0.2518>0.05. Hence the null hypothesis should not be rejected. Mary can conclude that there is no sufficient evidence to conclude that there is a significant that the mean withdrawals from a college campus ATM is less than 100 at 5% level of significance. 2) To Test: H o : U Annarbor = Livonia H 1 : U Annabor < Livonia i) Assuming two population variances are equal Test Statistics is t = -4.26 Dt = 15 P-value = 0.003 Since P-value < 0.05 Therefore, Reject H o at < 0.05 Hence mean age of individual who purchases iPhone 3G at AnnArbor is less than Less than Livonia

Assuming two population variances are equal Test statistics is t = -3.85 Df = 8 P-value = 0.0024 Since P-value < 0.05 This implies that mean age of individual who purchases iPhone 3G at AnnArbor is less than Less than Livoni 3) C) Lars is underpaid, and his salary is an outlier. 4) The p-value of F test is 0.0000. Since p-value is very small so it shows that model is significant. The r-square is 71.1%. It shows that 71.1% of variation is accounted for by the model. It shows that model is a good fit. The standard error of estimate is 3.038. It is small which shows that model is a good fit. The fitted regression line is CityMPG = 47.0484 - 0.0080 * Weight The intercept is 47.084. It shows that when weight of the car is 0 pounds City MPG is 47.0484. It is not possible to have a car with zero pounds weights so intercept is meaningless in this case. The slope is -0.0080. It shows that for each unit pound increase in weight, mileage is decreased by 0.0080 units. The predicted value for weight = 3000 is CityMPG = 47.0484 - 0.0080 * 3000 = 23.0484 The predicted value for weight = 3000 is CityMPG = 47.0484 - 0.0080 * 4000 = 15.0484 For each unit pound decrease in weight, mileage is increased by 0.0080 units. So, for 1000 pounds decrease in weight mileage will be increased by 1000 * 0.0080 = 8 units 5) (1) The first graph is the linear model graph where if the Quarter period increases, the Value of investment increases. The second graph is the exponential model graph where if the Quarter period increases, the Value of investment increases.

The third graph is the quadratic model graph where if the Quarter period increases, the Value of investment increases. (2) Using the first trend, we have: The regression equation is: y = 67.578x + 792.43 Put x = 20 y = 67.578*20 + 792.43 y = 2143.99 Using the second trend, we have: The regression equation is: y = 882.87e0.048x Put x = 20 y = 882.87e0.048*20 y = 2305.788 Using the third trend, we have: The regression equation is: y = 84.697x2 - 12.281x + 1032 Put x = 20 y = 84.697*202 - 12.281*20 + 1032 y = 34665.18 (3) I would choose the quadratic model since it has the highest R-squared value.

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