Binomial Distributions
A survey found that 60% of drivers prefer using GPS over printed directions. Suppose that, on a certain highway, the actual percentage is exactly 60%. You randomly select 10 drivers. Let x = the number of drivers among the 10 who prefer GPS. Then x is a binomial random variable with n = 10, p = 0.60.
1. Calculate P(x = 7)
2. How many of the 10 randomly selected drivers would you expect to prefer GPS?
1. 10 C 7 * (0.6)^7 * (1-0.6)^3 = 0.215
2. 10 * 0.60 = 6
A nutritionist wants to estimate the average number of grams of sugar in popular breakfast cereals. She randomly selects a sample of 40 cereals and finds that the mean sugar content is 10.2 grams, with a standard deviation of 2.5 grams.
Construct a 95% confidence interval for the true mean
Interpret the results
x +- t* (s/ sqrt(n))
df = 40 - 1 = 39
10.2 +- 2.0227 (2.5/sqrt(40))
(9.4, 10.99)
We are 95% confident that the true mean of sugar content in popular breakfast cereals is between 9.4 and 10.99 grams.
A fitness app company claims that its users walk at least 10,000 steps per day on average. A skeptical researcher believes the true average is actually less than 10,000 steps. She selects a random sample of 35 users, and the sample shows an average of 9720 steps per day with a standard deviation of 620 steps.
At the 0.05 significance level, test whether there is evidence that the average number of steps per day is less than 10,000.
null : mean = 10,000
alternative : mean < 10,000
Test Statisic : (9720 - 10000)/ (620 / sqrt(36)) = -2.672
P value : tdist(35) , P(-infinity < x < -2.672) = 0.005
0.005 < 0.05 Reject the null
A poll shows that 80% of pet owners say their pets sleep in their bedroom at night. Assume this percentage is exactly accurate for a certain city. You randomly survey 15 pet owners. Let x = the number of pet owners whose pets sleep in their bedroom. n = 15, p = 0.80
1. Calculate P(x= 12)
2. How many pet owners would you expect to say their pets sleep in their bedroom?
1. 15 C 12 * (0.8)^12 * (1-0.80^3 = 0.25
2. 15 * 0.80 = 12
A college career center surveys 300 graduating seniors and finds that 198 of them have already received at least one job offer.
Is this distribution normal?
Construct a 99% confidence interval for the true proportion of all graduating seniors at the college who have received a job offer.
Interpret
p hat = 198/300 = 0.66
300(0.66) = 198 > 10
300(1-0.66) = 102 > 10
Z* = 2.576
0.66 +- 2.576(sqrt((0.66)(1-0.66)/300)
(0.5895, 0.73)
We are 99% confident that the true proportion of graduating seniors who have already received a job offer is between 58.95% and 73%
A phone manufacturer claims that only 20% of its customers use more than 10 GB of data per month. A data analyst believes this proportion has increased.
They take a random sample of 200 customers and find that 52 of them used more than 10 GB of data last month
At the 0.01 significance level, test whether the proportion of customers using more than 10 GB has increased.
null : p = 0.20
alternative : p > 0.20
Test statistic: 2.121
P value : 0.017
0.017 > 0.01 Fail to Reject null