Definition/formulas
What do you use to find a z-score given the data point, μ, and σ?
z=(x-m)/r
In a z* confidence intervals, what do we use for mean and standard deviation?
We use the true mean and standard deviation, μ and σ
What actually does the z score represent?
A z score (or standard score or standardized value) is the number of standard deviations that a given value x is above or below the mean
How is the null hypothesis symbolized?
Null Hypothesis= Ho
19 8 2 9 -6
13 -8 15 15 -14 -11
Test to see if there are any outliers in these numbers.
No outliers
What are the steps to find a z*?
1st- (1-given percent in decimal form)/2
2nd- Use that number as the prop above
1-PA=PB
3rd look for the PB on the inside of table A and get the closest z-score
4th- use that z score to find x using x= z+
Find the z* for a 99% confidence interval.
(1-0.99)/2 = 0.005
1-0.005= 0.995
0.995--->
(2.57+2.58)/2 =
2.575
If the z-score is negative, is the value less than, greater than, or equal to the mean?
Less than
If the P-value is greater than, the decision is to...
Fail to reject the null hypothesis
In a group of 40 people, 10 are healthy and every person the of the remaining 30 has either high blood pressure, a high level of cholesterol or both. 5 have high blood pressure and 25 have high level of cholesterol.
If a person is selected randomly from this group, what is the probability that he/she has high blood pressure (event A)
1/8
What is the margin of error in a confidence interval?
±z*(σ/√n)
Or
±t*(s/√n)
Find the margin of error for an 85% confidence interval using random samples of size 36, mean of 800, and a standard deviation of 100.
z*=1.44
1.44(100/√36)=
1.44(16.666666)=±24
What does this notation mean: X~N(12,4)
X~N(,)
12 is the mean and 4 is the standard deviation
If the P-value is less than the level of significance, the decision is to...
Reject the null hypothesis
A card is drawn from a well shuffled pack of 52 cards. Find the probability of drawing a face or black card with replacement
38/52
When doing a hypothesis test, how do you know if it’s one-sided or two-sided?
We base it off the sign of the alternative hypothesis. One-sided uses < or >, two-sided uses ≠
Data follows a normal distribution with a standard deviation of 75 lbs and a sample mean of 500 lbs. Using a sample size of 25, what is the 80% confidence interval?
(1-0.8)/2=0.1
1-0.1=0.9
z*=1.28
500±(1.28)(75/√25)
500±(1.28)(15)
500±19.2
(480.8, 519.2)
Scores on test average 86 with a standard deviation of 9. What is the z-score for someone who earned a 73 on the test?
z=73-869=-1.44
A college football coach records the mean weight that his players can bench press as 263 pounds, with a standard deviation of 62 pounds. Three of his players thought that the mean weight was more than that amount. They asked 31 of their teammates for their estimated maximum lift on the bench press exercise. The actual different weights were (frequencies are in parentheses):
205(3); 215(3); ; 241(2); 252(3); 265(3); 275(2); 312(2); 316(5); 338(2); 341(2); 345(2);; 385(2).
a/ Conduct a hypothesis test using a 2.5% level of significance to determine if the bench press mean is more than 263 pounds.
Reject the Ho, enough evidence to show that the mean is greater than 263
z=286-2636231=2.065
pValue = Normcdf (2.065,1E99,0,1)= .0195
0.195<.025
Suppose Nancy has classes four days a week. She attends classes four days a week 85% of the time, three days 10% of the time, two days 4% of the time, and one day 1% of the time. Suppose one week is randomly selected.
Suppose one week is randomly chosen. Construct a probability distribution table (called a PDF table). What does the x*P(x) column sum to?
E(x)=7.3
What is the biggest difference between using the different test statistics for finding a confidence interval?
it depends if you are given sigma
The mean annual salary for registered nurses in the U.S. is $38,000. Assume that σ = $2700. Random samples of size 900 nurses are drawn from this population and the mean of each sample is determined. Find the 90% confidence interval.
38000± 148
(37852, 38148)
A bone mineral density test can be helpful in identifying the presence or likelihood of osteoporosis, a disease-causing bones to become more fragile and more likely to break. The result of a bone density test is commonly measured as a z score. The population of z scores is normally distributed with a mean of 0 and a standard deviation of 1, so these test results meet the requirements of a standard normal distribution.
A randomly selected adult undergoes a bone density test. Find the probability that this person has a bone density test score less than 1.68.
P(z<1.68)= NORMCDF (-1E99,1.68,0,1)= 0.0954
Suppose a baker claims that his bread height is more than 15 cm, on average. Several of his customers do not believe him. To persuade his customers that he is right, the baker decides to do a hypothesis test. He bakes 10 loaves of bread. The mean height of the sample loaves is 17 cm. The baker knows from baking hundreds of loaves of bread that the standard deviation for the height is 0.5 cm and the distribution of heights is normal. Use significance level 2.5% to find the test statistic and determine if the mean is greater than 15.
Fail to reject the Ho.
z=17-150.5/10=12.65
p value= Normcdf (12.65,1E99,0,1)= 5.7
.025<5.7
What are the labels in the normally distributed curve using the empirical rule?
m for mean, left is m-std dev, left is m+std dev.