Question

A recent report states the variance of the fuel consumption of a certain car is equal to 16.81 miles per gallon (mpg). A researcher claims the actual variance is lower. A sample of 21 cars produced a variance of 6.25 mpg. Test the claim at alpha = 0.05, use the P-value method. For question #29, tell what type of test his is (right, left, or two-tailed). For question #30, select from choices given with regard to identifying the alternate hypothesis.

#31, fill in the blank with your Critical Value(s) for Chi-square.

#32, fill in the space provided with your calculated value of Chi-square.

#33, use your answer from #32 to select between which two areas your P-value lies.

#34, based on your answer for #33, answer True or False as to whether the null hypothesis will be rejected

Answer 1

This is a two tailed test as the alternative hypothesis contains $\neq$ sign.

The null hypothesis will be rejected - TRUE

Let o denotes the population standard deviation. Hypotheses : To test null hypothesis H. :02 =282.58 against alternative hypothesis H1 :02 € 282.58 Let s denotes the sample standard deviation and n denotes the sample size. The test statistic can be written as : x? (n-1)s 282.58 which under H, follows a x2 distribution with n-1 df. Decision rule / Rejection region : We reject Ho at 0.05 level of significance if P-value < 0.05 or if xəbs > x6.025,n–1 or if Xəbs < xã.975,1–1 1 1 0.06 0.04 - 0.02 60 0.00 - 0 Now, chisa The value of the test statistic Xobs (21 – 1) *6.252 282.5761 2.764742 2.765 associated degrees of freedom = 20 The smaller critical value is =xô.975,20 =9.590777 The bigger critical value is =xố.025,20 =34.169607 P-value = 2 * P(x2-1 < xəs) 2 + P(x) < 2.76) = 2 + 0.000002 = 0.000004 Conclusion : Since p-value < 0.05 and xəbs > Xéritical = 34.169607, so we reject H, at 0.05 level of significance Hence, we can conclude that the the population variance is significantly different from 282.58.