Chapter 7
The Raising Canes in Snellville claims that 45% of its sales are to high school students. To test this theory, you take a random sample of 100 customers over the course of a week. Determine if the sampling distribution would be approximately normal and find the mean and standard deviation of the sampling distribution.
Random sample
100 x 10 = 1000 … Raising Canes has more than 1000 customers in a week
100(0.45) = 45 and 100(1 – 0.45) = 55
Df=n-1
The recommended daily allowance (RDA) of calcium for women between the ages of 18 and 24 years is 1200 milligrams (mg). Researchers who were involved in a large-scale study of women’s bone health suspected that their participants had significantly lower calcium intakes than the RDA. To test this suspicion, the researchers measured the daily calcium intake of a random sample of 36 women from the study who fell in the desired age range. The sample mean was 1056.2 mg and the standard deviation was 306.7 mg. Do these data give convincing evidence at the α = 0.01 significance level that the researchers’ suspicion is correct?
Write the Null and Alternative Hypothesis
Ho X= 1200
Ha X < 1200
How do you determine the shape (approximately normal or not) of the sampling distribution of sample *means*?
a. The sample must be randomly selected
b. The sample must be less than 10% of the population
c. EITHER: i) The population is approximately normal OR ii) The sample size is ≥30
How do you determine the shape (approximately normal or not) of the sampling distribution of sample *proportions*?
a. The sample must be randomly selected
b. The sample must be less than 10% of the population
c. The sample must be big enough to produce 10 successes and 10 failures aka np≥10 and n(1-p)≥10
A 2022 study of 45 randomly selected adults found that 23% of them had been in a car accident before. Suppose the actual percentage who have been in car accidents is 31%. *proportions problem*
A. What is the shape of the sampling distribution? If it is approximately normal, explain why.
Approximately normal because…
Random sample
45(10) = 450 Population of adults is more than 450
45(0.31) = 13.95 and 45(1 – 0.31) = 31.05
The t distribution is similar to the Normal distribution, but it has what that makes it different?
lower peak and larger tails to allow for more variability.
A research article claims that the national average vertical jump for teens is 15 inches. They wonder if the average at their school is higher, so they got a list of every student’s name in the school and selected a random sample of 20 students to measure their vertical leaps. The following is their data. Assume α= 0.05
State the Parameter
The average vertical jump for teens.
What is the formula for calculating the standard deviation of the distribution of sample *means*?
SD/ Square root N
What is the formula for calculating the standard deviation of the distribution of sample *proportions*?
Square root
p(1-p)/N
Walmart claims that the average amount spent by a customer in a visit is $92 with a standard deviation of $37. Suppose you poll 60 random customers about the amount of their purchase as your sample.
A. What is the shape of the sampling distribution? If it is approximately normal, explain why.
Approximately normal because…
Random sample
60(10) = 600 Population of Walmart customers is more than 600
n > 30
What two parts do we need in order to find the Confidence Interval of the Mean?
Mean and the Margin of Error.
Write the Null and Alternative Hypothesis
The mean is less than 1479
Ho: x= 1479
Ha: x < 1479
What is another way of combining weighted averages or finding unknown values in multiple, intersecting datasets
What is the probability of having a sample of 45 adults with 23% or less having been in an accident?
12.3%
The proportion of Brookwood High School students who are involved in at least one extracurricular activity is known to be 76%. With random samples of size 75 students, answer the following questions.
a. Describe the shape of the distribution of the sample proportions and provide evidence.
The sampling distribution of the sample proportions would be approximately normal.
Random Sample, more than 75x10=750 students in the population, 75(0.76) and 75(1-0.76) both greater than 10
When intrepting a Conclusion in Context what four parts must we mention?
1. Confidence Level
2. The Inteval
3. mean/proportion
4. what we are referencing.
What is a type 1 Error?
Bob was eating Chips ahoy cookies watching the Super Bowl. He noticed that he could see different amounts of chocolate chips on each cookie. He counted the number of visible chips on several cookies and recorded them below
14 18 20 19 25 10 16 21 17 20 12 22 18 19 10 15 16 19 18 17 22 21 15 17 23 11 12 14 16 18 20 22 21 19 18 17
Construct a 90% confidence interval for the mean number of visible chocolate chips.
a. Constructing a confidence interval for using Student’s t distribution.
Mean = 17.56 Choc Chips
Interval (16.5, 18.6)
A UFO club claims that more than 29% of U.S. adults have had vivid dreams about UFOs. You decide to test this claim and ask a random sample of 80 U.S. adults whether they have had vivid dreams about UFOs. Of those surveyed, 40 people reply yes.
How is the "random" condition met for this hypothesis test?
80 adults were chosen randomly
Google claims that the average Chromebook battery life is normally distributed with an average of 5.5 hours and a standard deviation of 45 minutes (0.75 hours). Describe the following about the distribution of samples of 6 Chromebook batteries.
What is the probability that one of the samples of 6 Chromebooks would give you an average of ≤ 5 hours?
25% chance that a sample of 6 Chromebooks would have an average battery life of less than 5 hours.
Blimpy asked students the average number of minutes they spent studying for the quiz tomorrow. Below is his data:
95 81 1 0 62 25 29 27 28 35 18 15 18 39 28 28 35 24 18 10 55 50 40 38 22 16 14 18
Construct a 95% confidence interval for the mean number of minutes studied.
a. Check the conditions for constructing a confidence interval for
Not normal. Can't do. Skewed Right
A professor believes that more than half (50%) of the students at MSU usually come to class late.
If we do not reject Ho, that means??
There is no evidence that the percent of students at MSU that are late to class is not 50%.
Izzy recorded the piercings students had in their head and recorded them on the chart below:
piercings 2 3 4 5 6 7 8
frequency 12 10 24 16 15 6 4
Use this as a sample to find a 99% confidence interval for the mean number of piercings recorded.
a. Check the conditions for constructing a confidence interval for using Student’s t distribution.
Random
n=87 greater than or equal to 30
87x10=870
Do more than 50% of US adults feel they get enough sleep?
In Gallup's December 2004 Lifestyle poll, they published a random survey of 1,003 US adults. 532 of those surveyed said that they do get enough sleep.
Conduct a one -proportion z-test to see if there is evidence that more than 50% of US adults feel that they get enough sleep. Use α=0.05 . And assume all three conditions are met.
Find the P-Value.
0.0271