# HW2 - 2 a Call lm(formula=minutes~copiers,data=copier.data...

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2. a. Call: lm(formula = minutes ~ copiers, data = copier.data) Residuals:      Min       1Q   Median       3Q      Max  -22.7723  -3.7371   0.3334   6.3334  15.4039  Coefficients:             Estimate Std. Error t value Pr(>|t|)     (Intercept)  -0.5802     2.8039  -0.207    0.837     copiers      15.0352     0.4831  31.123   <2e-16 *** --- Signif. codes:  0  *** ’  0.001  ** ’  0.01  * ’  0.05  . ’  0.1  ‘ ’  1 Residual standard error: 8.914 on 43 degrees of freedom Multiple R-squared:  0.9575, Adjusted R-squared:  0.9565  F-statistic: 968.7 on 1 and 43 DF,  p-value: < 2.2e-16 Ho:                            H a : t statistics  is given as 31.12233, which is way more greater than the critical value 1.681071. therefore, we have to reject the mull hypothesis and in favor of the alternative hypothesis. P value is the probability of obtaining an effect at least as extreme as the one in your sample data, assuming the truth of the null hypothesis. In this case P value is extremely small, less than the 0.001 bound. Therefore, we have to reject the null hypothesis at 0.05 significance level that says there is no relationship between x and y. b. t value for the 95% confidence interval is calculated as  qt(c(.025, .975), df=43)  -2.016692  2.016692 estimated value of beta one gives the increase in mean service time when one more copier is served. as calculated in fitting the regression mode. standard error for beta one is 0.4831, the confidence interval can be calculated as (15.0352-0.4831*2.016692 ,15.0352+0.4831*2.016692)=(14.06094, 16.00946) this 95% confidence interval means that in repeated sampling to estimate change in the mean service time when one more copier is served, the resulting increase in service time will be cap- tured by the above interval 95% of the time. c. i. the 95% confidence interval given in b. is (14.06094, 16.00946), which means that the in- creased mean service time will be in this interval 95% of the repeated sampling. the manu- facture states that the mean increase in service time will not be more than 14 mins, which is in contradiction with the number given in the confidence interval. the confidence interval states that the mean increase in service time should be between (14, 16) 95% of the time. ii. Ho:                            H a : t(0.95, 43) = 1.681071
qt(c(.95), df=43)  1.681071 t statistics: (15.0352-14)/0.4832=2.1428 >1.681071 therefore, we have to reject the null hypothesis at 5% significance level and conclude that the mean increase in service time will be more that 14 mins. the conclusions from the 95% confidence interval and the 5% significance test are consistent. d. bo does not give relevant information because time can not be a negative number. also -0.5802, the start time , is not in the observed sample data.
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