Consider the hypothesis test given by 5.0: 5 .0: 10 pH p H In a random sample of 100 subjects, the sample proportion is found to be . 55.0ˆ p (a) Determine the test statistic. Show all work; writing the correct test statistic, without supporting work, will receive no
Chapter 6 Sampling Distributions True/False 1. If we have a sample size of 100 and the estimate of the population proportion is .10, we can estimate the sampling distribution of [pic]with a normal distribution. Answer: True Difficulty: Easy 2. A sample size of 500 is sufficiently large enough to conclude that the sampling distribution of [pic] is a normal distribution, when the estimate of the population proportion is .995. Answer: False Difficulty: Medium 3.
Now we all the 12 costs per units amount from table 2 and divided by 12 in order to get the average. 38 + 38 + 38 + 35 + 35 + 38 + 35 + 38 + 33 + 39 +35 +35 = 437 Now 437/12 = 36.45 The average materials cost per unit is $36.45 2. Average labor cost per unit I use the same formula as I did above to figure out the average labor cost per unit. Labor Probability Cost Random Numbers Interval 0.18 $ 22.00 0.0 < 0.18 0.08 $ 23.00 0.19 < 0.26 0.25 $ 24.00 0.27 < 0.51 0.19 $ 25.00 0.52 < 0.70 0.3 $ 28.00 0.71 <
The first class in a relative frequency table is 50–59 and the corresponding relative frequency is 0.2. What does the 0.2 value indicate? Answer: ____Convert 0.2 to a % by times by 100. So 0.2 = 20%, so 20% of the data is between 50 and 59______________ 3. When you add the values 3, 5, 8, 12, and 20 and then divide by the number of values, the result is 9.6.
Doing a sampling of the demographics in college around the United States in 2009 it can be seen that 65% of the students are women and 45% are male. In Dr Zak’s test it was distributed as 70% female and 30% male. Of the distribution among races it was as follows: 65% White, 14% Black, 11% Hispanic, and .7% Asian. When looking at the test subjects for Dr Zak it is seen that 89% of the students were White, .04%
Assignment #2 1) Improve the result from problem 4 of the previous assignment by showing that for every e> 0, no matter how small, given n real numbers x1,...,xn where each xi is a real number in the interval [0, 1], there exists an algorithm that runs in linear time and that will output a permutation of the numbers, say y1, ...., yn, such that ∑ ni=2 |yi - yi-1| < 1 + e. (Hint: use buckets of size smaller than 1/n; you might also need the solution to problem 3 from the first assignment!) 2) To evaluate FFT(a0,a1,a2,a3,a4,a5,a6,a7) we apply recursively FFT and obtain FFT( a0,a2,a4,a6) and FFT(a1,a3,a5,a7). Proceeding further with recursion, we obtain FFT(a0,a4) and FFT(a2,a6) as well as FFT(a1,a5) and FFT(a3,a7). Thus, from bottom up, FFT(a0,a1,a2,a3,a4,a5,a6,a7)
1. The graph approximates the points: E(r) σ Minimum Variance Portfolio 10.89% 19.94% Tangency Portfolio 12.88% 23.34% 10. The reward-to-variability ratio of the optimal CAL (using the tangency portfolio) is: 11. a. The equation for the CAL using the tangency portfolio is: Setting E(rC) equal to 12% yields a standard deviation of: 20.56% b. The mean of the complete portfolio as a function of the proportion invested in the risky portfolio (y) is: E(rC) = (l - y)rf + yE(rP) = rf + y[E(rP) - rf] = 5.5 + y(12.88 - 5.5) Setting E(rC) = 12% ==> y = 0.8808 (88.08% in the risky portfolio) 1 - y = 0.1192 (11.92% in T-bills) From the composition of the optimal risky portfolio: Proportion of stocks in complete portfolio = 0.8808 × 0.6466 = 0.5695 Proportion of bonds in complete portfolio = 0.8808 × 0.3534 = 0.3113 12.
In hypothesis testing, the smaller the p value the more important it is. It is usually compared with the level of significance value and if it is lower, the null hypothesis is rejected. If it is higher than the null hypothesis is not rejected. It is important to note that a very small p value is an indication that there is a greater chance of the null hypothesis being rejected.
AP Statistics Name _______________ 3/24/09 Wood/Myers Period _____ Test #12 (Chapter 13) Honor Pledge __________ Part I - Multiple Choice (Questions 1-10) – Circle the letter of the answer of your choice. 1. An SRS of size 100 is taken from a population having proportion 0.8 successes. An independent SRS of size 400 is taken from a population having proportion 0.5 successes. The sampling distribution for the difference in sample proportions has what standard deviation?
Was there a normal distribution? Defend your position by interpreting the normality tests. (0 marks as this question is part of data screening for the writing of the results in Task 10) The results show that the test is not significant as according to Smirnov and Shapiro-Wilk, the sig. =.200, .175, .183 and .122, which are greater than .05. Thus meaning that the assumption of normality was met According to skewness and kurtosis there is no problem with normality as both scores of -1.57 (skewness) and -0.39 (kurtosis) are both within the range of +/- 3.29.