Question

Part A.

Score on 25 point test is normally distributed with mean 22 and standard deviation 5.        

You took a sample of 9 students. The mean for this group is 19.111111       

Test the hypothesis that the performance of this group is different than the regular group. Use α=.05.

1. What is the alternative hypothesis?
2. What is the rejection region?
3. Calculated z
4. What is your decision? Yes or no?
5. You believe that the group from an honors and would perform better than the regular class.    If you were to test this hypothesis, would your decision change? You are still using α=.05.

Part B.  

Score on 100 points examination are is normally distributed with mean 90 and standard deviation 5.  An honors class was started last year.   It is believed that the average score for the honors class would be significantly higher than the score of the regular class. Test the hypothesis that the average score of the honors class is greater than the regular class. Assume that the standard deviation for the honors class is the same as that of the regular class. The average for this class of 25 students is 92.  Use α=.05.

1. What is the alternative hypothesis?
2. Rejection region is?
3. Calculated value of z is
4. What is your decision? Yes or no

Part C

x is normally distributed with mean 10 and standard deviation 1. What is the value of the following probability:

P(x>11)?

Answer 1

Part A)

Let $\mu$ denotes the mean score for the selected group.

Hypotheses : To test null hypothesis Ho u =22 against alternative hypothesis H u + 22 Let i denotes the sample mean, o denotes the population standard deviation and n denotes the sample size. Here, m = 19.1111, 62 = 25.0000,0 = 5.00, n = 9 The test statistic can be written as: _ V3 (z – 22) which under Ho follows a standard normal distribution. Decision rule / Rejection region : We reject Ho at 0.05 level of significance if P-value < 0.05 or if |zobs > 20.025 = 1.959964 04- 0.3- 20.2- 30.1- 4 -2 2 4 0 Z Now, Value of test statistic: The value of the test statistic is Pobs =- 19(19.1111 – 22) 5.00 = -1.733333 -1.73 The critical value == 1.959964 +1.96 P-value = P(Z > Zobs) = 2 * P(Z < -1.733333) = 2 * 0.041518 = 0.083036 ~ 0.0830 Conclusion: Since p-value > 0.05 and Zobs > critical = 1.96, so We fail to reject H, at 0.05 level of significance Hence, we can conclude that the population mean is not different from 22.

Part B)

Let $\mu$ denotes the average score of the honors class

Hypotheses : To test null hypothesis Hou < 90 against alternative hypothesis H : > 90 Let denotes the sample mean, o denotes the sample standard deviation and n denotes the sample size. Here, m = 92.000, o2 = 25.0000,0 = 5.00, n = 25 The test statistic can be written as: _vnic – 90.0) which under H, follows a standard normal distribution. Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if |zobs | > 20.025 = 1.645 0.4 - 0.3- density -2 - 2 4 NO. Now, Value of test statistic: The value of the test statistic is Pobs = 25 (92.000 – 90) L = 2.000000 ~ 2.00 5.00 The critical value == 1.644854 1.645 P-value = P(Z > Zobs) = 2 * P(Z > 2.000000) = 2*(1 – 0.977250) = 0.022750 ~ 0.0228 Conclusion: Since p-value < 0.05 and Zobs > Zcritical = 1.645, so we reject Ho at 0.05 level of significance Hence, we can conclude that the the population mean is greater than 90.

Part C)

Here, X - Normal(10.000, 1.002) Now, P(X > 11.00) = 1- P(X < 11.00) = 1 - PIX – 10.000 11.00 – 10.000 1.00 - 1.00 = P(X > 11.00) = 1- P(Z < 11.00 – 10.000 1.00 = P(X > 11.00) = 1 - P(Z < 1.00) = P(X > 11.00) = 1 – 0.841345 = 0.158655 ~ 0.1587