100BHWS2

100BHWS2 - H : = . 5 vs H 1 : > . 5, the p-value...

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STAT 100B HW II Solution Problem 1: Suppose we roll a die 1000 times, and the average of the 1000 numbers is 3.4. Is there evidence that the die is not fair? A: ¯ X = 3 . 4. μ 0 = E[ X ] = 6 k =1 k/ 6 = 3 . 5. σ 2 = Var[ X ] = 6 k =1 ( k - μ ) 2 / 6 = 2 . 9167. Z = ( ¯ X - μ 0 ) / ( σ/ n ) = - 1 . 8516. If we test H 0 : μ = 3 . 5 vs H 1 : μ 6 = 3 . 5 at 5% level, we should reject the null hypothesis if | Z | > 2. Because | Z | < 2, we do not have strong evidence that the die is not fair. Problem 2: Suppose we generate 1000 random numbers using a random number generator in the computer. The random number generator is supposed to output random numbers that are independent and uniformly distributed over [0 , 1]. If the average of these 1000 number is .53. Is there evidence that there is something wrong with this random number generator? A: ¯ X = . 53. μ 0 = E[ X ] = R 1 0 xdx = 1 / 2. E[ X 2 ] = R 1 0 x 2 dx = 1 / 3. σ 2 = Var[ X ] = E[ X 2 ] - E[ X ] 2 = 1 / 12. Z = ( ¯ X - μ 0 ) / ( σ/ n ) = 3 . 28. If we test
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Unformatted text preview: H : = . 5 vs H 1 : &gt; . 5, the p-value is much smaller than 2 . 5%. So there is strong evidence that there is something wrong. Problem 3: Suppose we randomly sample 10 people from the population of adult males, and we measure their heights. Suppose the heights (in the unit of cm) are 182 166 187 196 173 188 192 164 165 185. Please nd the 95% condence interval for the average height of the population. Let us assume that the heights of the population follow a normal distribution. A: X = 10 i =1 X i = 179 . 8. s 2 = 10 i =1 ( X i- X ) 2 / (10-1) = 140 . 84. t 9 , 2 . 5% = 2 . 262. So the 95% condence interval is [ X-t 9 , 2 . 5% s/ n, X + t 9 , 2 . 5% s/ n ] = [171 . 3 , 188 . 3]. 1...
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