# A report states that the cost of repairing a hybrid vehicle, is falling even while typical repairs on conventional vehicles are getting more expensive

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A report states that the cost of repairing a hybrid vehicle is falling even while typical repairs on conventional vehicles are getting more expensive. The most common hybrid repair, replacing the hybrid inverter assembly, had a mean repair cost of \$3,927 in 2012. Industry experts suspect that the cost will continue to decrease given the increase in the number of technicians who have gained expertise on fixing gas-electric engines in recent months. Suppose a sample of 100 hybrid inverter assembly repairs completed in the last month was selected. The sample mean repair cost was \$3,850 with the sample standard deviation of \$400. Complete parts (a) and (b) below.

a. Is there evidence that the population mean cost is less than \$3,927? (Use a 0.05 level of significance.)

State the null and alternative hypotheses.

Find the test statistic for this hypothesis test.

The critical value(s) for the test statistic is(are) ___ ?

Is there sufficient evidence to reject the null hypothesis using a = 0.05? Edited By :

The first thing to consider before answering such a question is to decide whether to use the normal distribution or the t distribution. Sometimes the choice is based on the sample size or whether the population standard deviation is given. The process is to always use the standard normal distribution when the population standard distribution is not given.  In other instances, depending on some books and other tutors, the normal distribution is used whenever the sample size is thought to be large enough.

What should we do here?

I will use the t distribution here to demonstrate the process of hypothesis testing given the sample statistics above.

Note that the formula for the t - distribution can be calculated as.

t = (X_bar  - mu)/(sd/sqrt(n))

The idea is to calculate the test statistic and to compare with the critical value, and the null hypothesis is rejected in the test statistic is more extreme than the critical value.

So culculate t = (3850 - 3927)/(400/sqrt(100))  = -1.925

The null hypotheses to be tested are as follows.

H0: mu greater or equal to 3927
Ha: mu is less than 3927

This is a left-tailed test and so the critical value should be a negative value.

use excel to find the critical value; =T.INV(0.05,99) =  -1.660

The null hypothesis is rejected because the test statistic is more extreme than the critical value. test statistic above -1.925 is less than -1.660.

The conclusion finds sufficient evidence to support the claim proposed by the experts. that the cost has been decreasing.