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The following hypothesis are given. Null = .40 Alternate not = ..pdf
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The following hypothesis are given. Null = .40 Alternate not = .40 A sample of 120 observations revealed that p=.30 The significance level is .05 Can the null be rejected? State the decision rule. Compute the value of the test statistic. What is the decision regarding the null hypothesis? Solution I ran the test in MINITAB and have following output: Test and CI for One Proportion Test of p = 0.4 vs p not = 0.4 Exact Sample X N Sample p 95% CI P-Value 1 36 120 0.300000 (0.219756, 0.390396) 0.025 Here p-value <<< alpha (0.05) hence we reject the null hypothesis. The decision rule is : reject null hypothesis if p-value < (0.05) The test statistic = Z = ((0.3)*(.7)/120) = The null hypothesis is rejected. Hope this helps. I am not near a calculator but you can calculate the value of test statistic as I have plugged values in the formula. Please rate this as a life saver. Thanks..
The following hypothesis are given. Null = .40 Alternate not = ..pdf
- The following hypothesis are given. Null = .40 Alternate not = .40 A sample of 120 observations revealed that p=.30 The significance level is .05 Can the null be rejected? State the decision rule. Compute the value of the test statistic. What is the decision regarding the null hypothesis? Solution I ran the test in MINITAB and have following output: Test and CI for One Proportion Test of p = 0.4 vs p not = 0.4 Exact Sample X N Sample p 95% CI P-Value 1 36 120 0.300000 (0.219756, 0.390396) 0.025 Here p-value <<< alpha (0.05) hence we reject the null hypothesis. The decision rule is : reject null hypothesis if p-value < (0.05) The test statistic = Z = ((0.3)*(.7)/120) = The null hypothesis is rejected. Hope this helps. I am not near a calculator but you can calculate the value of test statistic as I have plugged values in the formula. Please rate this as a life saver. Thanks.