Question

25)With a previous contractor, the average time to replace a lamppost in the city was 3.0 days. A city councilwoman thinks the new contractor is not quickly replacing the streetlights. She selects a random sample of 12 streetlight service calls and gets the following replacement times (in days). 6.2, 7.1, 5.4, 5.5, 7.5, 2.6, 4.3, 2.9, 3.7, 0.7, 5.6 ,1.7 Because the sample size is small, you should verify that the replacement time is normally distributed and the sample does not contain outliers. The normal probability diagram and boxplot are provided. Are the conditions to test the hypothesis satisfied? YAN 0 1 2 3 4 5 6 7 8 tiempo de reemplazo dias) replacement time in days b) Is there sufficient evidence to support what the councilor believes at the significance level a = 0.05?

Answer 1

Form the normal probability diagrams since all data point lie in close proximity of line the data is adequately normal

From the boxplot , the mean is little shifted toward the Q3 ( third quartile) the data is moderately negatively skewed and not symmetric.

The condition for one sample t-test are satisfied since data is randomly selected and also approximately normal.

Let $\mu$ be the average time to replace a lamppost in the city.

To test the claim by the previou contractor the average time to replace a lamppost in the city is 3.0 days

the null hypothesis is given by

HO: 3.0

i.e average time to replace a lamppost in the city is 3.0 days

vs the alternative hypothesis is given as

Н. :> 3.0

i,e average time to replace a lamppost in the city is more than 3.0 days

The sample of 12 streetlight service calls is selected

and given as

6.2, 7.1, 5.4, 5.5, 7.5, 2.6, 4.3, 2.9, 3.7, 0.7, 5.6 ,1.7

The sample mean is given as

C n

6.2 + 7.1+5.4 +5.5+ 7.5 +2.6 + 4.3+2.9 +3.7 + 0.7 +5.6 +1.7 12

T= 53.2 12

= 4.43333

The sample standard deviation (s)

2 2 Στί – ) η –1

s? 50.74667 12 - 1

=  4.613333

s = 2.147867

The test statistic is given as

T- tn-1 s/n

tn 1 4.4333 – 3.0 2.147867/V12

tn-1 2.311694

Obtainig the critical value from the t-table

df = n-1

= 11

tn-1,0 = t11,0.05-

37 0.50 1.00 . 0.25 0.50 0.20 0.40 0.15 0.30 0.10 0.20 0.05 0.025 0.10 0.05 0.01 0.02 0.005 0.001 0.0005 0.01 0.002 0.001 0.000 AN cum. probl one-tail two-tails df 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 nnnn 1.000 0.816 0.765 0.741 0.727 0.718 0.711 0.706 0.703 0.700 0.697 0.695 0.694 0.692 0.691 nen 1.376 1.061 0.978 0.941 0.920 0.906 0.896 0.889 0.883 0.879 0.876 0.873 0.870 0.868 0.866 ORG 1.963 1.386 1.250 1.190 1.156 1.134 1.119 1.108 1.100 1.093 1.088 1.083 1.079 1.076 1.074 1 071 3.078 1.886 1.638 1.533 1.476 1.440 1.415 1.397 1.383 1.372 1.363 1.356 1.350 1.345 1.341 6.314 2.920 2.353 2.132 2015 1.943 1.895 1.680 LO 1.812 1.796 12.71 4.303 3.182 2.776 2.571 2.447 2.365 2.306 2.262 2.228 2.201 2.179 2.160 2.145 2.131 31.82 6.965 4.541 3.747 3.365 3.143 2.998 2.896 2.821 2.764 2.718 2.681 2.650 2.624 2.602 63.66 9.925 5.841 4604 4.032 3.707 3.499 3.355 3.250 3.169 3.106 3.055 3.012 2 977 2 947 1021 318.31 22.327 10.215 7.173 5.893 5.208 4.785 4.501 4.297 4.144 4.025 3.930 3.852 3.787 3.733 RR 636.62 31.599 12.924 8.610 6.869 5.959 5.408 5.041 4.781 4.587 4.437 4.318 4.221 4.140 4.073 1.771 1.761 1.753 427 TA COD

t11,0.05 1.795

Reject H0 if

t11 t11,0.05

i.e if t( calculated) > t(critical value)

SInce    t( calculated) =2.311694 >  t(critical value) = 1.795

we reject the null hypothesis

( Also the p-value is given as

for two tailed hypothesis

Pt11 > 2.311694 0.02058786

Since p-value < = 0.05

(a)

we reject the null hypothesis)

There is suffiecient evidence to support councilor belief that the new contractor is taking more than 3.0 days and not quickly replacing the streetlights

(Verifying using the R-output

)

> t.test(x, mu= 3, alternative = "greater") One Sample t-test data: t = 2.3117, df = 11, p-value = 0.02059 alternative hypothesis: true mean is greater than 3 95 percent confidence interval: 3.31982 Inf sample estimates: mean of x 4.433333

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HO: 3.0

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C n

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37 0.50 1.00 . 0.25 0.50 0.20 0.40 0.15 0.30 0.10 0.20 0.05 0.025 0.10 0.05 0.01 0.02 0.005 0.001 0.0005 0.01 0.002 0.001 0.000 AN cum. probl one-tail two-tails df 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 nnnn 1.000 0.816 0.765 0.741 0.727 0.718 0.711 0.706 0.703 0.700 0.697 0.695 0.694 0.692 0.691 nen 1.376 1.061 0.978 0.941 0.920 0.906 0.896 0.889 0.883 0.879 0.876 0.873 0.870 0.868 0.866 ORG 1.963 1.386 1.250 1.190 1.156 1.134 1.119 1.108 1.100 1.093 1.088 1.083 1.079 1.076 1.074 1 071 3.078 1.886 1.638 1.533 1.476 1.440 1.415 1.397 1.383 1.372 1.363 1.356 1.350 1.345 1.341 6.314 2.920 2.353 2.132 2015 1.943 1.895 1.680 LO 1.812 1.796 12.71 4.303 3.182 2.776 2.571 2.447 2.365 2.306 2.262 2.228 2.201 2.179 2.160 2.145 2.131 31.82 6.965 4.541 3.747 3.365 3.143 2.998 2.896 2.821 2.764 2.718 2.681 2.650 2.624 2.602 63.66 9.925 5.841 4604 4.032 3.707 3.499 3.355 3.250 3.169 3.106 3.055 3.012 2 977 2 947 1021 318.31 22.327 10.215 7.173 5.893 5.208 4.785 4.501 4.297 4.144 4.025 3.930 3.852 3.787 3.733 RR 636.62 31.599 12.924 8.610 6.869 5.959 5.408 5.041 4.781 4.587 4.437 4.318 4.221 4.140 4.073 1.771 1.761 1.753 427 TA COD

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