9.1 -
What are the two possible explanations when a basketball player who claims to be an 80% free throw shooter makes only 64% of free throws taken in 50 attempts?
The player's claim is true (p=0.80). He really is an 80% free-throw shooter and his poor performance happened by chance OR the player's claim is false (p<0.80). Hew really makes less than 80% of his free-throws.
When the P-value is less than the significance level α in a significance test, we say that the result is _________ at the α level.
statistically significant
What is the general formula for a standardized test statistic?
(statistic - parameter)/(standard error)
A significance test for a population proportion is often referred to as a _________.
One sample z-test for p
If the sample size is small and the shape of the population distribution is unknown, the Normal/Large Sample condition is met if the graph of the sample data shows no _______ or ______.
strong skew or outliers
At the Hawaii Pineapple Company, managers are interested in the size of the pineapples grown in the company's fields. Last year, the mean weight of the pineapples harvested from one large field was 31 ounces. A new irrigation system was installed in this field after the growing season. Managers wonder if the mean weight of pineapples grown in the field this year will be greater than last year. So, they select an SRS of 50 pineapples from this year's crop. State appropriate hypotheses for performing a significance test. Be sure to define the parameter of interest.
H0: μ = 31 ozs
Ha: μ > 31 ozs
μ = the true mean weight (in ounces) of all pineapples grown in this field this year
Define a type II error.
Ha is true, but we do not find convincing evidence for Ha and we fail to reject H0.
A report claimed hat 13% of students typically walk to school. DeAnna thinks that the proportion is greater than 0.13 at her large elementary school. She surveys random sample of 100 students and finds that 17 typically walk to school. DeAnna would like to carry out a test at the alpha = 0.05 level of H0: p = 0.13 Ha: p > 0.13 where p = the proportion of all students at her school who typically walk to school. Check if the conditions for performing the significance test are met.
Random? Yes "random sample of 100 students"
LCC? Yes 13≥10 and 87≥10
The P-value in a two-sided test about a population proportion requires you to find the area in _______ of a standard normal distribution.
both tails
The Survey of Study Habits (SSHA) is a psychological test with scores that range from 0 to 200, with higher scores indicating more positive attitudes and better study habits. The mean score for U.S. college students is about 115. A teacher suspects that older students have better attitudes toward school. She gives the SSHA to an SRS of 45 students at her college who are at least 30 years old. The teacher wants to perform a test of H0: μ = 115 versus Ha: μ > 115 where μ is the mean SSHA score of students at her college who are at least 30 years old. Check that the conditions for performing the test are met.
Random? Yes "SRS of 45 students"
Normal/Large sample? Yes 45 ≥ 30
In the study of pineapple weights from the previous question, the sample mean weight from this year's crop is 31.4 ounces and the sample standard deviation is 2.5 ounces. A significance test yields a P-value of 0.1317. Interpret the P-value.
Assuming that the mean weight of all pineapples grown in the field this year is 31 ounces, there is about a 0.1317 probability of getting a sample mean of 31.4 ounces or greater by chance alone.
The probability of making a type I error in a significance test is equal to the _______.
significance level alpha, α
From the previous question, calculate the test statistic.
z = 1.19
Suppose you want to perform a test of H0: p = 0.7 versus Ha: p ≠ 0.7. An SRS of 80 from the population of interest yields 59 successes. Calculate the standardized test statistic and P-value. Assume that the conditions for inference are met.
z = 0.73
P-value = 2(1-0.7673) = 2(0.2327) = 0.4654
Continuing the previous question. H0: μ = 115 versus Ha: μ > 115 and where μ is the mean SSHA score of students at her college who are at least 30 years old. From the random sample of 45 students, the teacher gets an average score of 120 and a standard deviation of 13 calculate the standardized test statistic and the p-value.
t = 2.58 use df = 40 from table B
p-value is between 0.005 and 0.01
Based on the previous question with the P-value of 0.1317, what conclusion would you make.
Since 0.1317 > 0.05, we fail to reject H0 and we do not have convincing evidence that the mean weight of all pineapples grown in the field this year was greater than last year (31 ounces).
You are thinking about opening a restaurant and are searching for a good location. From your research, you know that the mean income of those living near the restaurant must be over $85,000 to support the type of upscale restaurant to wish to open. You decide to select a simple random sample of 50 people living near one potential site. Based on the mean income of this sample you will perform a test of H0: μ = $85,000 Ha: μ > $85,000 where μ is the mean income of the population of people who live near the restaurant. Describe a type I and type II error in this setting.
Type I: The true mean income of all the nearby households is $85,000, but we find convincing evidence that it is greater.
Type II: The true mean income of all the nearby households is greater than $85,000 but we do not find convincing evidence that it is.
From the previous question, find the p-value and give the full conclusion. (Both sentences)
p-value = 1- 0.8830 = 0.117
Assuming H0 is true (p = 0.13), there is about a 0.117 probability of getting a sample proportion of 0.17 or greater by chance alone.
Since the 0.117>0.05, we fail to reject H0 and we do not have convincing evidence that the proportion of all students who walk to school is greater than 0.13.
A state's DMV claims that 60% of teens pass their driving test on the first attempt. An investigative reporter examines and SRS of the DMV records for 125 teens; 86 of them passed the test on their first try. He wonders if the true proportion of teens who pass on the first try is different from the DMV's claim. A 95% confidence interval is calculated as (0.607, 0.769). Use the confidence interval (not a significance test) to make a decision about the DMV's claim?
All the values in the interval are greater than 0.60, so is not a plausible that 60% of teen pass their driving test on the first try. Therefore, we reject the DMVs claim and we do not have convincing evidence that the true proportion of teens who pass the driving test on their first try is different from 60%.
A machine is supposed to fill bags with an average of 198 ounces of candy. The manager of the candy factory wants to be sure that the machine does not consistently underfill or overfill the bags. SO the manager selects a random sample of 75 bags of candy produced that day and weighs each bag. The mean weight of these bags is 19.28 oz and the standard deviation is 0.81 oz. Is there convincing evidence that the alpha = 0.01 level that the mean amount of candy in all bags filled that day differs from 19 ounces?
A 99% confidence interval for μ is 19.033 to 19.527 ounces. Explain how the interval is consistent with the test.
t = 2.99
Use df = 60 and table B, P-value is between 0.002 and 0.005
The 99% confidence interval does not contain the null value (μ=19 oz), so we reject the null hypothesis at the alpha = 0.01 level.