Statistics Assignment - Speech or Presentation Example

Econ 210 Homework Ten: Probabilities and Hypothesis Testing Number of children per family (x) 0 2 3 Total Number of families (f) 55 a = 80 31550500fx080630150860a. The value of a is 80. a = 500 – (55 + 315 + 50) = 500 – 420 = 80b. The total number of children included in this study is 860.Total number of children = (0)(55) + (1)(80) + (2)(315) + (3)(150)= 0 + 80 + 630 + 450= 860c. The probability that the selected family has 3 children is 0.10 (or 10%).P(x = 3) = 50/500 = 0.10 or 10%2. p(infected) = 0.

60Therefore, p(non-infected) = 1 – p(infected) = 1 – 0.60 = 0.40This is case of binomial distribution. Where, n = 5 and p = 0.40The probability function of Y is Or 3. n = 5 and p(correct) = 1/5 = 0.20, P(20 Correct) = ?This is case of binomial distribution.The probability that the student will answer all 20 questions correctly is about zero. In other words, it is nearly impossible to answer all 20 questions correctly by randomly guess.4. Since, Charlie inserts his finger into the socket on n separate independent occasions; therefore, receiving a jolt of electricity that will trigger massive myocardial infarction resulting in instant death in one occasion is independent to other.

In other words, receiving a jolt of electricity on one occasion will not affect the other (following) occasions and hence, probability (p) will remain the same for other occasions.Therefore, the probability that Charlie receives the life-ending shock on that last nth insertion is p.However, if receiving a jolt of electricity on one-occasion affects the following occasions than probability (p) will not remain the same for other occasions (because Charlie will be not alive for following occasions).

In that circumstance, the probability that Charlie receives the life-ending shock on that last nth insertion will be. 5. n = 100, = 7 years and s = 2 yearsa. The null and alternate hypotheses are (Mean has not changed from the value of 7.5 years of 20 years ago) (Mean has changed from the value of 7.5 years of 20 years ago)The test is two-tailed test.b. The test statistic isc. The p-value for the test is 0.0124.p-value = 2 * P(z < -2.50) = 2*0.0062 = 0.0124Decision: Reject null hypothesis, H0, as p-value = 0.0124 < 0.05.

Conclusion: These data evidence that μ has changed from the value of 7.5 years of 20 years ago.