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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 ‘ ’ 1Residual standard error: 8.914 on 43 degrees of freedomMultiple R-squared: 0.9575,Adjusted R-squared: 0.9565 F-statistic: 968.7 on 1 and 43 DF, p-value: < 2.2e-16Ho: Ha:t statistics is given as 31.12233, which is way more greater than the critical value 1.681071. therefore, wehave 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 sampledata, 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 rejectthe 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 2.016692estimated value of beta one gives the increase in mean service time when one more copier isserved.as calculated in fitting the regression mode. standard error for beta one is 0.4831, the confidenceinterval 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 meanservice 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 isin contradiction with the number given in the confidence interval. the confidence intervalstates that the mean increase in service time should be between (14, 16) 95% of the time.ii.Ho: Ha:t(0.95, 43) = 1.681071
qt(c(.95), df=43) 1.681071t statistics: (15.0352-14)/0.4832=2.1428 >1.681071therefore, we have to reject the null hypothesis at 5% significance level and conclude that themean 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.