Sampling Distributions (Means)
A teacher collects a random sample of 10 students’ quiz scores from a large class. The population standard deviation is unknown. The sample has a mean of 78 and a sample standard deviation of 5.3. Can the teacher assume the sample mean is normally distributed
No, because the sample size is less than 30
10 < 30
A school finds that 10% of its students are left-handed. A random sample of 40 students is selected. Can the sampling distribution of the sample proportion be approximated using a normal distribution?
n*p = 40(0.10) = 4 < 10
n*(p-1) = 40(.90)= 36 >10
No, both of these conditions need to be greater than 10 to make it normally distributed.
A 99% confidence interval for the average daily water usage in a neighborhood is calculated to be (230 gallons, 250 gallons).
Interpret this interval in context.
We are 99% confident that the average daily water usage in a neighborhood is between 230 and 250 gallons.
A study finds that 62% of 250 participants support renewable energy investments. A 95% confidence interval is calculated as (0.56, 0.68).
Interpret the meaning of this interval in context.
We are 95% confident that the proportion of the 250 participants who support renewable energy investments is between 56% and 68%
A political researcher wants to estimate the proportion of voters who support a new education policy. She wants the estimate to be within ±4% of the true proportion with 95% confidence.
What is the minimum sample size needed?
E = p(1-p) (Z*/E)^2
= (0.5)(1-0.5) (1.96/0.04)^2
= 600.25
=601
A bakery reports that the weights of its loaves of bread are normally distributed with a mean of 1.05 pounds and a standard deviation of 0.08 pounds. If a random sample of 64 loaves is selected, what is the probability that the sample mean weight is between 1.03 and 1.07 pounds?
normaldist(1.05, 0.08/sqrt64)
P(1.03 < x < 1.07) = 0.954
In a large population, 48% of voters support a new public transportation initiative. If a random sample of 80 voters is selected, what is the standard deviation (standard error) of the sampling distribution of the sample proportion?
standard error for proportion: sqrt((0.48)(1-0.48)/80) = 0.0558
A sample of 12 college students has an average study time of 18.2 hours per week with a sample standard deviation of 3.7 hours. Construct a 90% confidence interval for the true mean number of study hours per week among all college students.
df = 12-1 = 11
t* = 1.796
18.2 +- 1.796 (3.7/sqrt(12)) = (16.28, 20.12)
A factory checks 150 products from a batch and finds that 18 of them are defective. Find a 90% confidence interval for the true proportion of defective items in the batch.
p = 18/150 = 0.12
Z* = 1.645
0.12 +- 1.645 (sqrt(p(1-p))/150) = (0.076, 0.164)
In a past study, 30% of people said they preferred online shopping over in-store shopping. A market researcher wants to estimate the current proportion to within ±3% with 99% confidence.
What is the minimum sample size required?
p = 0.30
Z* = 2.576
E = 0.03
= (0.3)(1-0.3)(2.576/0.03)^2
= 1548.35
=1549
The time it takes a cashier to ring up a customer’s order at a grocery store is normally distributed with a mean of 3.2 minutes and a population standard deviation of 0.6 minutes. If a random sample of 36 customers is selected, what is the probability that their average checkout time is more than 3.4 minutes?
normaldist(3.2, 0.6/sqrt36)
P(x > 3.4) = 0.023
A recent survey found that 60% of adults in a city own a pet. If a random sample of 100 adults is selected, what is the probability that more than 65% in the sample own a pet?
normaldist(0.60, sqrt((0.60)(1-0.60)/100)
P(x > 0.65) = 0.154
The average time to assemble a bike at a factory is believed to be 45 minutes. A manager takes a random sample of 36 bikes and finds the sample mean time is 47 minutes. Assume the population standard deviation is 6 minutes. Construct a 95% confidence interval for the true average time it takes to assemble a bike.
Z* = 1.96
47 +- Z* (6/sqrt(36)) = (45.04, 48.96)
In a poll of 500 customers, 385 said they were satisfied with a company’s service. What is the margin of error for a 95% confidence interval of the true proportion of satisfied customers?
Z* = 1.96
Standard Error = sqrt((385/500 (1-385/500)/500)
E = 1.96 * 0.1882
= 0.0369
A nutritionist wants to estimate the average amount of sodium in a popular brand of soup. From past data, the population standard deviation is 120 mg. How large of a sample is needed to estimate the mean sodium level to within ±25 mg with 95% confidence?
n = (sZ*/E)^2
= (120(1.96)/25)^2
= 88.5
= 89