StatCrunch: Data Analysis and Statistical Software

2017 data

Chosen data: sleeping

Mean: 8.80 hrs.

Standard error: 0.027

Survey and time of accomplishment.

Three months

Value got is 8.80 hrs.

P ( <8.80)

First converting to Z

Z= n is sample of persons, is deviation, a lower mean to be 6.44hrs, and as an exact mean. 8.80

Z = 6.44-8.80 = -2.36

P(z<-2.36) is 0.0091: the probability of getting a value less than 8.80hrs is very small.

9 people

Their times: 8, 8.5, 8, 9,8.8, 8.6, 8.8, 9.1, 8.6.

average = sum/total number

77.4/9 = 8.6 hours

=

The sum

P(people you get greater than 8.5)

P(z>2.36)

Probability is 0.0091

500 adults, and probability of not owning card is 29%

p = 0.29 n = 500 q = (1-0.29)

= 0.71 500 (0.71)(0.29) = 10.295

Since n*p*q  (500)(0.29)(0.71)  10.295 which is  10,

p ˆ is normally distributed.

The question request for P(0.25  p ˆ  0.30).

Is p ˆ normally distributed? Well, Yes it is, based part (a) answer.

Therefore:  p ˆ  p  0.29 and  p ˆ ? 0.02029

Answer: P(0.25  p ˆ  0.30) =0.7682

p ˆ is 125/500 = 0.25

P( p ˆ  0.25). note p ˆ normally distributed

 of p ˆ  p  0.29

 p = 0.02029

P( p ˆ  0.25) =0.0243

Since 0.0243 is less than 0.05, it would be unusual for a given random sample of 500 adult persons to result in 125 or fewer people of those who do not own a card.