Question #97198
An automobile manufacturing company, while introducing a new model of car during 2007, claimed that the model is fuel efficient as compared to the old model which average 10.5km ler litre with standard deviation of 3.5km. the first 50 cars which came out of the factory were tested for fuel efficiency under normal conditions and found to average 11.3km per litre. Would you conclude that the new model is fuel efficient?
1
Expert's answer
2019-10-28T12:33:10-0400

μ=10.5,σ=3.5,n=50,xˉ=11.3Formulate the hypothesisH0:μ=10.5H1:μ>10.5This is a one tailed testα=0.05Since σ is known we are using the z test10.05=0.95, from ztables we have 1.645The z critical value(s) is 1.645z=xˉμσn=11.310.53.550=1.616Since the test statistics isless than the critical value, that is 1.616<1.645We fail to reject the the null hypothesis and conclude that the new model is not fuel efficientas compared to the old model with average of 10.5km per litre at 5% level ofsignificance.\mu=10.5, \sigma=3.5, n=50, \bar{x} =11.3\\ Formulate\ the \ hypothesis\\ H_0:\mu=10.5\\ H_1:\mu>10.5\\ This \ is \ a\ one\ tailed\ test\\ \alpha=0.05\\ Since\ \sigma \ is \ known\ we\ are\ using\ the\ z\ test\\ 1-0.05=0.95,\ from\ z-tables\ we \ have\ 1.645\\ The\ z\ critical\ value(s)\ is\ 1.645\\ z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{11.3-10.5}{\frac{3.5}{\sqrt{50}}}=1.616\\ Since\ the\ test\ statistics\ is \,less \ than\ the\ critical\ value,\ that\ is\ 1.616\lt1.645\\ We\ fail\ to\ reject\ the\ the\ null\ hypothesis\ and\ conclude\ that\ the\ new\ model\ is\ not\ fuel\ efficient\\ as\ compared\ to\ the\ old\ model\ with\ average\ of\ 10.5km\ per\ litre\ at\ 5\%\ level\ of significance.


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