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Minitab - Assignment Example

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We evaluated the traffic offenders imprisonment in Australia, we subjected the analysis one-simple t test and using a hypothesized mean value of 16.7 at a significance α = .05. Assuming an equal mean variance ; the sample mean = 17.8 (SD = 6.0) is statistically significantly…
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Minitab
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T – test with hypothesized mean Problem one Ho = Traffic offenders get an average imprisonment of 16.7 months. Ha = Traffic offenders get longer sentences more than 16.7 monthsTable 1One-Sample T Test of μ = 16.7 vs ≠ 16.7 N Mean StDev SE Mean 95% CI T P100 17.800 6.000 0.600 (16.609, 18.991) 1.83 0.070We evaluated the traffic offenders imprisonment in Australia, we subjected the analysis one-simple t test and using a hypothesized mean value of 16.7 at a significance α = .05. Assuming an equal mean variance ; the sample mean = 17.

8 (SD = 6.0) is statistically significantly different from 16.7, t= 1.83, p = .07. At 95% CI [16.6, 19 ] the sample mean lies within the parameters. Thus we reject null hypothesis that the average jail sentence of traffic offenders is 16.7.We can conclude that the traffic offenders in Australia get longer sentences of more than 16.7 months.Problem 2Ho = The American’s still drive the same average distance as 2005Ha = The American’s do not drive the same average distance as 2005a)Figure 1The graph shows a normal distribution of the data with most average distance driven is between 6 – 12.b)The sample N = 500 is a large size thus we have to detect the gross non-normality, thus we evaluate the normal distribution curve with mean and variance for the normal distribution for distance sample data.c)Table 2One-Sample T: DISTANCE Test of μ = 12.

4 vs ≠ 12.4Variable N Mean StDev SE Mean 95% CI T PDISTANCE 500 11.802 6.002 0.268 (11.275, 12.329) -2.23 0.026One mean t-test was conducted to evaluate whether the Americans drive the same mean distance as in 2005. With a sample N = 500 , with a sample mean 11.8, t =-2.23, p = .03 at a significance level α = .05 with a variation of SD 6.00; thus the test is statistically significant at a hypothesized mean value of 12.4. Thus we can assume that with equal mean variance test the evaluation is significant, and at 95% CI [ 11.3, 12.3], the distance mean fall within the parameter of the CI.

We accept the null hypothesis that American’s still drive the mean distance like 2005.We conclude that the distance driven by drivers in America don’t vary with time rather its constant.

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