The Gallup Poll asked a random sample of 1785 adults if they attended church during the past week. Let ô be the proportion o people in the sample who attended church. A newspaper report claims that 40% of all U.S. adults went to church last week. Suppose this claim is true. Verify the sampling distribution of p is approximately Normal. Because np = 1785(0.40) = 714.0 ≥ 10 and n(1 p) = 1785(0.60) = 1071 > 10, the sampling distribution of p is approximately Normal. Because 40> 30, the sampling distribution of p is approximately Normal. Because 1785 > 30, the sampling distribution of p is approximately Normal. Because p = 0.40, the distribution of p is skewed to the right. Because np = 1785(0.40) = 714.0 10 and n(1- ·P) = 1785(0.60) = 1071 ≥ 10, the sampling distribution of p is

Holt Mcdougal Larson Pre-algebra: Student Edition 2012
1st Edition
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL
Chapter11: Data Analysis And Probability
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The Gallup Poll asked a random sample of 1785 adults if they attended church during the past week. Let p be the proportion of
people in the sample who attended church. A newspaper report claims that 40% of all U.S. adults went to church last week.
Suppose this claim is true.
Verify the sampling distribution of p is approximately Normal.
Because np = 1785(0.40) = 714.0 ≥ 10 and n(1 − p) = 1785(0.60) = 1071 ≥ 10, the sampling distribution of p is
approximately Normal.
Because 40> 30, the sampling distribution of ô is approximately Normal.
Because 1785 > 30, the sampling distribution of p is approximately Normal.
0.40, the distribution of p is skewed to the right.
Because p =
Because np = 1785(0.40) = 714.0 10 and n(1 − p) = 1785(0.60) = 1071 ≥ 10, the sampling distribution of p is
not approximately Normal.
Transcribed Image Text:The Gallup Poll asked a random sample of 1785 adults if they attended church during the past week. Let p be the proportion of people in the sample who attended church. A newspaper report claims that 40% of all U.S. adults went to church last week. Suppose this claim is true. Verify the sampling distribution of p is approximately Normal. Because np = 1785(0.40) = 714.0 ≥ 10 and n(1 − p) = 1785(0.60) = 1071 ≥ 10, the sampling distribution of p is approximately Normal. Because 40> 30, the sampling distribution of ô is approximately Normal. Because 1785 > 30, the sampling distribution of p is approximately Normal. 0.40, the distribution of p is skewed to the right. Because p = Because np = 1785(0.40) = 714.0 10 and n(1 − p) = 1785(0.60) = 1071 ≥ 10, the sampling distribution of p is not approximately Normal.
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