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9. An automobile dealer believes that the average cost of accessories in new automobiles is $3,000 over the base sticker price. He selects 50 new automobiles at random and finds that the average cost of the accessories is $3,256. The standard deviation of the sample is $2,300. Test his belief at -0.0s. Use the classical method

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Answer:

There is no enough evidence to claim that the average cost of accesories is different from $3,000.

Explanation:

The significance level for this test is α=0.05.

The classical method is based on regions of rejection of acceptance, according to the sample parameter. In this case, the standard deviation of the population is unknown.

The null and alternative hypothesis are:


H_0: \mu=3000\\\\ H_a: \mu\\eq 3000

This is a two-tailed test, with significance level of 0.05.

The t-value for this sample is:


t=(x-\mu)/(s/√(N)) =(3256-3000)/(2300/√(50))=(256)/(325)=0.787

The degrees of freedom are:


df=n-1=50-1=49

For df=49 and α=0.05 (two-tailed test), the critical values are
|t|>2.009, so the value t=0.787 is within the acceptance region.

The null hypothesis can not be rejected.

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