Open StatCrunch Stat Calculators T (x 2.624) Compute. → → → →DF: 14 →P ≥ → = (0.0100963) 0.0101 P-value ≈ (c) Use to determine whether to rejectornot. Since (0.0100963) is less (0.05), we reject the null hypothesis. P-value than α (ii) left-tailed (a) Draw a distribution with the area that represents the value shaded. (b) Use StatCrunch to find the value for each test value, . Open StatCrunch Stat Calculators T 22 (x -2.321) Compute. → → → →DF: →P ≤ → = (0.01497327) 0.0150 P-value ≈ (c) Use to determine whether to rejector not. Since (0.01497327) is less (0.05), we reject the null hypothesis. P-value than α (iii) two-tailed (a) Draw a distribution with the area that represents the value shaded. (b) Use StatCrunch to find the value for each test value, . Open StatCrunch Stat Calculators T DF: → → → →Between, 16→ P (-1.562≤ x 1.562) Compute. ≤ → The is 1 0.86215185 = P-value – 0.137848 0.1378 P-value ≈ (c) Use to determine whether to rejectornot. Since (0.1378) is not less (0.05), we fail to reject the null hypothesis. P-value than α There is insufficient evidence to support the claim. Example 3: A survey of 15 large U.S. cities finds that the average commute time one way is 25.4minutes. A chamber of commerce executive feels that the commute in his city is lessand wants to publicize this. He randomly selects 25 commuters and finds the average is 22.1 minutes with a standard deviation of 5.3 minutes. At, is
Open StatCrunch → Stat → Calculators → T →DF: 14 →P (x ≥ 2.624) → Compute. P-value = (0.0100963) ≈ 0.0101 (c) Use to determine whether to rejectornot. Since P-value (0.0100963) is less than α (0.05), we reject the null hypothesis. (ii) left-tailed (a) Draw a distribution with the area that represents the value shaded. (b) Use StatCrunch to find the value for each test value, . Open StatCrunch → Stat → Calculators → T →DF: 22 →P (x ≤ -2.321) → Compute. P-value = (0.01497327) ≈ 0.0150 (c) Use to determine whether to rejector not. Since P-value (0.01497327) is less than α (0.05), we reject the null hypothesis. (iii) two-tailed (a) Draw a distribution with the area that represents the value shaded. (b) Use StatCrunch to find the value for each test value, . Open StatCrunch → Stat → Calculators → T →Between, DF: 16→ P (-1.562≤ x ≤ 1.562) → Compute. The P-value is 1 – 0.86215185 = 0.137848 P-value ≈ 0.1378 (c) Use to determine whether to rejectornot. Since P-value (0.1378) is not less than α (0.05), we fail to reject the null hypothesis. There is insufficient evidence to support the claim. Example 3: A survey of 15 large U.S. cities finds that the average commute time one way is 25.4minutes. A chamber of commerce executive feels that the commute in his city is lessand wants to publicize this. He randomly selects 25 commuters and finds the average is 22.1 minutes with a standard deviation of 5.3 minutes. At, is