Question

A large software company gives job applicants a test of programming ability and the mean for that test has been 160 in the past. Twenty-five job applicants are randomly selected from one large university and they produce a mean score and standard deviation of 183 and 12, respectively. Use a 0.05 level of significance to test the claim that this sample comes from a population with a mean score greater than 160. Use the P-value method of testing hypotheses.

Answer 1

1.

Let u denotes the population mean. Hypotheses : To test null hypothesis Ho := 160 against alternative hypothesis H1 :p> 160 Let denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, T = 183.000, s2 = 144.000000, s = 12.00, n = 25 The test statistic can be written as: vnc - 160) ta S which under H, follows a t distribution with n-1 df. Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if tobs > to.05,n-1 = 1.710882 0.4- 0.3- 0.2 0.1 0.0 - -2 t Now, Value of the test statistic: The value of the test statistic tobs 25(183.0000 – 160) 12.0000 9.583333 9.58 associated degrees of freedom = 24 The critical value tcritical = t0.05, n-1 = t0.05,24 = 1.710882 1.71 P-value = P(t24 > 9.583333) = 1- P(t24 < 9.583333) =1-1.000000 = 0.000000 0.0000 Conclusion : Since p-value < 0.05 and tobs > tcritical = 1.710882, so we reject Ho 0.05 level of significance Hence, we can conclude that population mean is significantly greater than 160

2.

There is sufficient evidence to support the supplier's claim that no more than 1% are defective.

Let p denotes the population proportion. Hypotheses : To test null hypothesis H. :P < 0.01 against alternative hypothesis H1 :p > 0.01 Let p denotes the sample proportion and n denotes the sample size. Here, p=0.0300, n = 600, po = 0.01, standard error = 0.01 *(1 – 0.01) 600 0.004062 The test statistic can be written as : p - po) Po*(1-P) n 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 ZSTAT > 20.01 = 2.326348 0.4 0.3- density 0.2- 0.1 - 0.0 - -2 Now, Value of test statistic: The value of the test statistic (0.0300 – 0.01) ZSTAT = 4.923660 4.924 0.01*(1-0.01) 600 The critical value Zcritical = 20.01 = 2.326348 2.326 P - value = P(Z > ZSTAT) = P(Z > 4.923660) = 1- P(Z < 4.923660) =1-1.000000 =0.000000 0.0000 Conclusion : Since p-value < 0.01 and 2STAT > Zcritical = 2.326348, so we reject H, at 0.01 level of significance Hence, we can conclude that the population proportion is significantly greater than 0.01