**Probability Distribution Calculation
**

1. A physicist needs to use α = .00001 for a certain particle accelerator experiment. She has correctly determined that she will need to use the z-score and a two-tailed test. What is the critical value of z? Show your work.

2. A random variable X has this known distribution:

X ∼ N(12,25)

(a) Calculate P (X >27). (2 points)

(b) Calculate P(11 < X <30). In other words, the probability that X is between 11 and 30.

3. It’s known that 12 percent of buses in a city fail smog tests. Regulators will randomly sample 31 for testing. What’s the chance that no more than 22 percent of buses fail the test?

4. Nike will equip 100 NBA players with new shoes to test if shoe wearers have average rebounds that differ from the population. They expect the shoe enables one more rebound on average. Rebounds have a known population distribution with μ = 3.4 and σ=3.1. Using a more stringent α = .01, and assuming Nike are right about the effect, what’s their chance of rejecting the null hypothesis with this study?

5. This table describes a sample of college students:

Assume a student is randomly chosen from this sample and let:

• A be students who are Seniors

• B be students who live on campus

(a) Calculate P(B)

(b) Calculate P(B|A)

6 A regression with calorie consumption as the dependent variable and height (in centimeters) as the independent variable finds a slope b of 5.4, with a standard error of 2.1. If relevant, the t distribution for this study would have 41 degrees of freedom.

(a) Calculate the 95% confidence interval of b.

(b) Calculate the p-value using a two-sided test

(c) Calculate the p-value for this hypothesis setup (if different):

H0 :b≤0,HA :b>0

7. Identify the fundamental statistics issue in the scenarios below. Please restrict yourself to no more than 2 sentences. (Note the issues are not all necessarily from the lecture on pitfalls.)

(a) A data analyst tries every letter in the alphabet and finds that NBA players with last names starting with “M” have significantly higher points per game, with p = .045. The general manager wants to use this surprising finding to select players.

(b) Sudden Infant Death Syndrome causes 91 deaths for every 100,000 births. An expert witness in a criminal case argues that the chance a family could have two infants die this way is extremely small. He calculates this probability as: (91/100,000) *(91/100,000)

(c) A new drug, Xonalfalam, has been subjected to 23 randomized clinical trials. The results have been published for only 2 of these trials, but both of those find a statistically significant benefit for the treatment group. On this basis, it is argued that the drug should gain FDA approval.

**Solution:**

Critical value of Z

Z_((1-α/2) )= Z_((1-0.00001/2) )

=Z_0.999995

4.4172

X ∼ N(12,25) implies that X is distributed with a mean of 12 and a variance of 25.

(a). P (X > 27) = P(Z>(27-12)/5)

= P(Z>3) = 1- P(Z<3)

= 1- 0.99865

0.0014

(b). P(11 < X < 30) = P((11-12)/5

= P(-0.2

= 0.99984 - 0.42074

0.5791

This is a binomial distribution problem.

P = 0.12, n = 31, 22% of 31 = 6.82

The chance that no more than 22 percent of buses fail the test = P(X ≤6)

P(X ≤6)= P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)

P(X ≤6)=0.9295

Hence, the chance that no more than 22 percent of buses fail the test is 0.9295

H_0: μ = 3.4 vs H_1:μ>3.4

n =100, μ = 3.4 and σ=3.1,

Z=(x ̅-μ)/(σ⁄√n) ~ N(0,1)

Z=(x ̅-3.4)/(3.1⁄√100)

(x ̅-3.4)/(3.1⁄√100)= (x ̅-3.4)/0.31

We reject H_0if P((x ̅-3.4)/0.31)≤0.01

(x ̅-3.4)/0.31= 2.3263

x ̅=3.4+0.721153

x ̅=4.12

Z= (4.12-3.4)/0.31

Z= 2.3225

P(Z) = 0.0101

Z= (4.5-3.4)/0.31

P(Z) = 0.0002

Hence, their chance of rejecting the null hypothesis with this study is high if and only if the average rebounds of players with new shoes is at less 4.5

Probability

(a). P(B) = 5166/9911

= 0.521239

(b). P(B|A) = P(B∩A)/P(A)

P(B ∩A) = 1739/9911

P(A)=2373/9911

P(B|A) = (1739/9911) * (9911/2373)

=1739/2373

= 0.7328

b= 5.4, se= 2.1, df=41

the critical value t_((1-α/2, n-1) )= t_((1-0.05/2, 41) )

t_((0.975, 41) )= 2.0195

C.I for β=b±t_((1-α/2, n-1) )*SE

=5.4±*2.0195*2.1

=5.4± 4.24095

C.I for β=(1.1591,9.6410)

(b). P-value =1-Pt(t,41)

Recall that t=b/se

t=5.4/2.1= 2.5714

P-value =2*[1-Pt(2.5714,41) ]

= 2*(1 – 0.9931)

0.0138

(c). P-value for a right tailed test

P-value =1-Pt(2.5714,41)

= 1 – 0.9931

= 0.0069

(a). the First letter of the last name of players has no relationship with players’ performance and as such, selecting players with such findings will be misleading as their performance was due to mere chance.

(b). The probability that two infants from a family will die is not as presented, rather, the 91 out of 100000 should be seen as a probability of success and binomial distribution applied.

(c). The two significant trials are too small for the new drug to gain FDA approval; the approval will have great consequences on the health of the users.