NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6. In an earlier study, the population proportion was estimated to be 0.36. How large a sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 95% confidence level with an error of at most 0.04? Round your answer up to the next integer.

A researcher surveyed 12 men who lost their fathers earlier in their lives.  His survey included the age of the subjects when their fathers died and their confidence that they would someday be happily married themselves (100 point scale – higher score = more confidence).  The results are shown below. Mx=15                My =60 SSx=348            SSy=5198 SPxy=933 Age Confidence Rating 12 34 8 30 11 89 21 69 15 55 7 38 18 78 23 66 22 89 19 79 9 35 15 58 The effect size is equal to _____________ . This means it is a ______________ effect.

NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6. In an earlier study, the population proportion was estimated to be 0.46. How large a sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 98% confidence level with an error of at most 0.03? Round your answer up to the next integer.

(12) adults said they would travel to outer space in their lifetime, given the chance. Estimate the true proportion of adults who would like to travel to outer space with 88% confidence (Round to two decimal places). Estimate population proportion: New York Times poll found that 329 out of 763 12

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What is the answer of the 7.14 question?

Time to Complete the Course Right 45 47 50 49 50 50 48 44 Left | 45 | 43 48 47 50 52 | 44 42 Assume a Normal distribution. What can be concluded at the the a = 0.01 level of significance level of significance? For this study, we should use t-test for the difference between two dependent population means a. The null and alternative hypotheses would be: p1 p2 Ні: p1 p2 b. The test statistic t v v = 1.199 (please show your answer to 3 decimal places.) c. The p-value = [1338 your answer to 4 decimal places.) d. The p-value is > va e. Based on this, we should fail to reject f. Thus, the final conclusion is that ... * (Please show v the null hypothesis. O The results are statistically insignificant at a = 0.01, so there is insufficient evidence to conclude that the population mean time to complete the obstacle course with a patch over

Suppose that having a widow's peak was a dominant trait (W) compared to not having a widow's peak (w). In a sample size of 30 students, 15 were observed to have a widow's peak and 15 did not have a widow's peak. The expected frequency of a widow's peak is 75% chance of getting it and a 25% chance of not getting it. Perform a chi square analysis to see if the data above agrees with the predicted outcome.