Question 3 of 12, Step 1 of 1CorrectThe director of research and development is testing a new drug. She wants to know if there is evidence at the 0.02 level that the drug stays in the system for more than343 minutes. For a sample of 50 patients, the mean time the drug stayed in the system was 349 minutes. Assume the population standard deviation is 25. Make thedecision to reject or fail to reject the null hypothesis.囲 TablesE KeypadAnswerKeyboard ShortcutsO Reject Null HypothesisO Fail to Reject Null Hypothesis

Level of significance (α) = 0.02 Sample size (n) = 50 Sample mean (x¯) = 349 min population standard deviation (σ ) = 25 Claim - Drug stays in the system is more than 343 minutes. ( μ= 343 ) Null hypothesis (Ho) : μ =343 Alternative hypothesis (H1) : μ >343 [ Right tailed test ] Level of significance (α ) = 0.02 Test statistic (z) =(x¯-μ)σn (349-343)2550 =63.5355=1.6970 ≃1.70 Z- statistic = 1.70 Z- critical value - at alpha equals to 0.02 and in right tailed test Z critical value is 2.326 Z- Critical value = 2.326 Here, Z-statistic is greater than Z-critical value . ( Z-statistic=1.70 > Z-critical = 2.326 ) Therefore, We reject null hypothesis.

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