Question

Check my w Consider the following hypotheses: No! = 290 BM 290 The population is normally distributed with a population standard deviation of 71. (You may find it useful to reference the appropriate table: z table or t table) a-1. Calculate the value of the test statistic with x = 300 and 85. (Round Intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) Test statistic nces a-2. What is the conclusion at the 5% significance level? O Reject so since the p-value is less than the significance level. O Reject Hosince the p value is greater than the significance level Do not reject #, since the pvalue is less than the significance level O Do not reject wo since the p-value is greater than the significance level.

Answer 1

a)

a-2) Do not reject H₀ since p-value is greater than the significance level.

Let u denotes the population mean. Hypotheses : To test null hypothesis Ho :je =290 against alternative hypothesis H : # 290 Let I denotes the sample mean, o denotes the population standard deviation and n denotes the sample size. Here, @ = 300.0000, 02 = 5041.0000,0 = 71.00, n = 85 The test statistic can be written as: na – 290) which under Ho 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 = 1.959964 0.4- 0.3 density 0.2 0.1 0.0 -2 0 N Now, Value of test statistic: The value of the test statistic is Zobs 85 (300.0000 - 290) 71.00 = 1.298527 1.30 The critical value = 1.959964 +1.960 P-value = P(Z > Zobs) = 2 * P(Z > 1.298527) = 2* (1 - 0.902947) = 0.194106 0.1941 Conclusion : Since p-value > 0.05 and |zobsZcritical = 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 290.

a-3) We can not conclude that the population mean differs from 290.

b)

b-2) Reject H₀ since the p-value is less than the significance level.

Let u denotes the population mean. Hypotheses : To test null hypothesis Ho :fl =290 against alternative hypothesis H :fl + 290 Let i denotes the sample mean, o denotes the population standard deviation and n denotes the sample size. Here, T = 255.0000, o2 = 5041.0000,0 = 71.00, n = 85 The test statistic can be written as: ne – 290) which under H, follows a standard normal distribution. Decision rule / Rejection region : We reject H, at 0.01 level of significance if P-value < 0.01 or if |zobs) > 20.005 = 2.575829 0.4- 0.3 density 0.2- 0.1- 0.0 2. 0 Z Now, Value of test statistic: The value of the test statistic is Zobs 85 (255.0000 - 290) 71.00 -4.544846 -4.54 The critical value = 2.575829 +2.576 P-value = P(Z > Zobs) = 2* P(Z < -4.544846) = 2 *0.000003 = 0.000005 0.0000 Conclusion: Since p-value < 0.01 and |zobs | > Zcritical = 2.58, so we reject H, at 0.01 level of significance Hence, we can conclude that the the population mean is not equal to 290.

b-3) We conclude that the population mean differs from 290.