Question

2. An automobile manufacturer claims that their leading compact car averages 36 mpg in the city. The population standard deviation is 8 mpg for city driving. Suppose a city police department purchases 64 cars from this auto manufacturer. If these cars were driven exclusively under city conditions and averaged 34 mpg, can one argue that the manufacturer's claim is too high at 1%. State the null and alternative hypothesis b. Determine and draw the hypothesis test model (Label everything) Determine the critical value(s) and test statistic d. Determine the conclusion a. C. 3. Presently, the mean life expectancy of a rare strain of bacteria is 12 hours. A scientist claims that she has developed a medium and therefore disagrees with this hypothesis that it is not 12 hours. The scientist tests 25 cultures and finds that they have a mean life of 13.25 hours with a sample standard eviation - 2 hour. Do these results show that the medium is effective in increasing the bacteria's life expectancy? (Use <= 5%) State the null and alternative hypothesis b. Determine and draw the hypothesis test model (Label everything) Determine the critical value(s) and test statistic d. Determine the conclusion a. c.

Answer 1

2)As poppulation variance is known we will be using the Z-test for testing mean

So we'll be testing the claim that actual mean < 36 mpg

The provided sample mean is X = 34 and the known population standard deviation is o = 8, and the sample size is n = 64. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u = 36 Ha: u < 36 This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.01, and the critical value for a left- tailed test is ze = -2.33. The rejection region for this left-tailed test is R = {z : z < -2.33} (3) Test Statistics The Z-statistic is computed as follows: X - μo Z = 34 - 36 8/164 -2 oln (4) Decision about the null hypothesis 2 > ze = -2.33, it is then concluded that the null hypothesis is not Since it is observed that z = rejected. Using the P-value approach: The p-value is p= 0.0228, and since p = 0.0228 > 0.01, it is concluded that the null hypothesis is not rejected.


(5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean u is less than 36, at the 0.01 significance level. Confidence Interval The 99% confidence interval is 31.424 < p < 36.576.

Hence we conlcude the claim made by company is true at level 1%

3) As population variance is not kinown here, so we'll be using sample variance. Hence the test to be used here is the t-test for mean
We'll be testing the claim that mean not equal to 12 at level 0.05

The provided sample mean is 7 = 13.25 and the sample standard deviation is s = 2, and the sample size is n = 25. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: A = 12 Ha: u = 12 This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a two- tailed test is te = 2.064. The rejection region for this two-tailed test is R= {t: |t| > 2.064} (3) Test Statistics The t-statistic is computed as follows: X - μo t= 13.25 – 12 2/25 - 3.125 s/n (4) Decision about the null hypothesis Since it is observed that t| = 3.125 >te = 2.064, it is then concluded that the null hypothesis is rejected. 0.0046, and since p = 0.0046 < 0.05, it is Using the P-value approach: The p-value is p = concluded that the null hypothesis is rejected.