Chapter 8 - Hypothesis Testing
Claim: The mean pulse rate left parenthesis in beats per minute right parenthesis of students in a large math class is less than 90. A simple random sample of the students has a mean pulse rate of 71.8.
The sample is not unusual if the claim is true. The sample is unusual if the claim is false. Therefore, there is sufficient evidence to support the claim.
Previously, 38% of parents of children in high school felt it was a serious problem that high school students were not being taught enough math and science. A recent survey found that 129 of 300 parents of children in high school felt it was a serious problem that high school students were not being taught enough math and science. Do parents feel differently today? Use the alpha equals 0.01 level of significance.
p = .38, p /= .38 p-val = .075 No, there is not sufficient evidence, meaning that we do not reject the null hypothesis.
Several years ago, 44% of college students felt the cost of college textbooks was too high. In a recent poll of 1,125 college students, 493 said they felt the cost of textbooks was too high. Construct a 90% confidence interval to assess whether this represents evidence that students' opinions about the cost of textbooks have changed.
p = .44, p/= .44 lower = .41 upper = .47 Since the interval contains the proportion stated in the null hypothesis, there is insufficient evidence that parents' attitudes toward the quality of education have changed.
Use technology to find the P-value for a right-tailed test about a mean with n = 13 and test statistic t = 3.494 .
p-val = 0.022 statcrunch -> stat -> calc -> T -> DF = 12
Test the hypothesis using the classical approach and the P-value approach. p = 0.7 p < 0.7 n = 150 x = 95 a = 0.01
z = -1.78 z-critical = 2.33 Do not reject the null hypothesis p-val = 0.037538 Do not reject the null hypothesis
Pattern for Stating Conclusions
At the _____% level of significance, this data (does / does not ) provide significant evidence to conclude ____________________________________ whatever is written in the alternative H1
Type I Error
Contradicting the null hypothesis when null hypothesis is true)
The claim is that the proportion of adults who smoked a cigarette in the past week is less than 0.20, and the sample statistics include n = 1013 subjects with 172 saying that they smoked a cigarette in the past week. Find the value of the test statistic.
H0: p = 0.20 q = 0.80 p-hat = 172/1013 z = (172/1013 - 0.2)/sqrt((.2*.8) / 1013) = -2.40
The claim is that the proportion of peas with yellow pods is equal to 0.25 (or 25%). The sample statistics from one experiment include 650 peas with 176 of them having yellow pods. Find the value of the test statistic.
H0: p = 0.25 q = 0.75 np = nq >= 5 so test stat is z p-hat = 176/650 z = (176/650 - 0.25) / sqrt((.25*.75)/650) = 1.22
According to a report, the standard deviation of nbsp monthly cell phone bills was $5.13 three years ago. A researcher suspects that the standard deviation of nbsp monthly cell phone bills is different today. Determine the null and alternative hypotheses.
H0: sigma = 5.13 H1: sigma /= 5.13
n = 23, t = -3.013 DF = 22
2*0.0032 = 0.0064
Using Confidence Intervals to Reject Ho.
A quick alternative for deciding whether to Reject Ho or Not Reject Ho in a two-tailed test: Compute the confidence interval for u using the sample data. If the confidence interval contains u_0 -> Do NOT Reject H0 If confidence interval lands away from u_0 -> The data contradicts H0 -> Reject H0
Using Confidence Intervals to Reject Ho
A quick alternative for deciding whether to Reject Ho or Not Reject Ho in a two-tailed test: - Compute the confidence interval for p using the sample data. - If the confidence interval contains H0 -> Do NOT Reject H0 - If confidence interval lands away from H0 -> The data contradicts H0 -> Reject H0
Reject H0
At the ....% level of significance, this data does provide significant evidence to conclude whatever is in the alternative H1.
47% of adults with children under the age of 18 previously reported that their family ate dinner together 7 nights a week. Suppose in a recent poll, 552 of 1200 adults with children under the age of 18 reported that their family ate dinner together 7 nights a week. Is there sufficient evidence that the proportion of families with children under the age of 18 who eat dinner together 7 nights a week has decreased at the alpha equals 0.1 significance level?
No, there is not sufficient evidence, because the test statistic is greater than the critical value, meaning that we do not reject the null hypothesis.
In hypothesis testing:
a, probability of a making a Type I error, is called the level of significance a and is chosen by the researcher before data is collected. The lower the a level the stronger the evidence from the sample has to be in order to "reject Ho"
Three years ago, the mean price of a single-family home was $243 comma 713. A real estate broker believes that the mean price has decreased since then.
There is sufficient evidence to conclude that the mean price of a single-family home has decreased.
Assume that the significance level is alpha equals 0.05. Use the given information to find the P-value and the critical value(s). The test statistic of z equals negative 0.74 is obtained when testing the claim that p less than 0.6.
p < 0.6, left-tailed area left of z = -0.74 is .2296 z = -1.65
H0: u = 5 H1: u > 5
right-tailed u
Hypothesis Testing: Overall Strategy
- Consider a stated hypothesis - Collect sample data & compare it to the hypothesis - Decide if the data is strong enough evidence to convince us (and others) - State our conclusion about the claim
"Rare Event" Rule
- Statistical Significance measures whether the result observed in the sample is unusual under the assumption that the null hypothesis is true. - "Rare Event Rule" When this type of rare event is observed in the data, rather than continuing to assume the null hypothesis is true, the more likely conclusion is that the assumption in Ho is probably false.
P-value -> tells us how "rare" this event is
...a "rare" event -> tells us to reject the null hypothesis
Assume that a simple random sample has been selected and test the given claim. Use the P-value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.
Fail to reject H0. There is insufficient evidence to warrant rejection of the claim that the mean age of actresses when they win an acting award is 34 years. p-val = .1799 a = 0.01 p-val > a
P-value Decision Rule
If P-value is less than a -> Reject H0 If P-value is NOT less than a -> Fail to Reject H0
Identify the type I error and the type II error that corresponds to the given hypothesis. The proportion of settled medical malpractice suits is 0.29
Reject the claim that the proportion of settled malpractice suits is 0.29 when the proportion is actually 0.29.
A simple random sample of 43 adults is obtained from a normally distributed population, and each person's red blood cell count (in cells per microliter) is measured. The sample mean is 5.24 and the sample standard deviation is 0.55. Use a 0.01 significance level and the given calculator display to test the claim that the sample is from a population with a mean less than 5.4 comma which is a value often used for the upper limit of the range of normal values. What do the results suggest about the sample group?
TS = -1.908 P-val = .0316 Fail to reject H0. There is not sufficient evidence to support the claim that the sample is from a population with a mean less than 5.4. There is not enough evidence to conclude that the sample is from a population with a mean less than 5.4, so it is possible that the population has counts that are too high.
statistical hypothesis
a statement or claim regarding a characteristic of a population(s).
hypothesis test
a statistical procedure, based on sample evidence and probability, for testing the validity of a claim about the population.
A 0.01 significance level is used for a hypothesis test of the claim that the proportion of ballots cast for a political candidate is less than 0.5.
p = 0.5 p < 0.5 left-tailed a = 0.01 normal distribution critical val: -2.33
The mean waiting time at the drive-through of a fast-food restaurant from the time an order is placed to the time the order is received is 85.2 seconds. A manager devises a new drive-through system that she believes will decrease wait time. As a test, she initiates the new system at her restaurant and measures the wait time for 10 randomly selected orders. The wait times are provided in the table. Complete parts (a) and (b) below.
Yes, the conditions are satisfied. The P-value is greater than the level of significance so there is not sufficient evidence to conclude the new system is effective.
Steps in Hypothesis Testing
1. A claim is made. 2. Evidence (sample data) is collected in order to test the claim. 3. The data is analyzed in order to support claim or contradict the claim.
In this chapter, there are three ways to set up the null and alternative hypothesis.
1. Equal versus not equal (two-tailed test) Ho: parameter = some value H1: parameter /= some value 2. Equal versus less than (left-tailed test) Ho: parameter = some value H1: parameter < some value 3. Equal versus greater than (a right-tailed test) Ho: parameter = some value H1: parameter > some value
Do Not reject H0
At the ....% level of significance, this data does not provide significant evidence to conclude whatever is in the alternative H1.
A refrigerator manufacturer claims that the mean life of its refrigerators is less than 17 years. You are asked to perform a hypothesis test to test this claim. How would you write the null and alternative hypothesis
H0: u = 17 H1: u < 17
According to the report, the mean nothing monthly cell phone bill was $49.95 three years ago. A researcher suspects that the mean nothing monthly cell phone bill is higher today.
H0: u = 49.95, H1: u > 49.95
21% of adults with children under the age of 18 previously reported that their family ate dinner together 7 nights a week. Suppose in a recent poll, 222 of 1110 adults with children under the age of 18 reported that their family ate dinner together 7 nights a week. Is there sufficient evidence that the proportion of families with children under the age of 18 who eat dinner together 7 nights a week has decreased at the alpha equals 0.1 significance level?
No, there is not sufficient evidence, because the test statistic is greater than the critical value, meaning that we do not reject the null hypothesis.
Claim: The mean pulse rate left parenthesis in beats per minute right parenthesis of students in a large calculus class is greater than 62. A simple random sample of the students has a mean pulse rate of 62.3.
The sample is not unusual if the claim is true. The sample is not unusual if the claim is false. Therefore, there is not sufficient evidence to support the claim.
When unknown: t-distribution
The test statistic for the mean if (1) the population is normal or nearly normal, with no outliers or (2) n is large (n > 30) The test statistic is a t-score (called a t-test) with degrees of freedom, df = n - 1
Make a decision about the given claim. Do not use any formal procedures and exact calculations. Use only the rare event rule. Claim: A coin favors heads when tossed, and there are 15 heads in 18 tosses.
There does appear to be sufficient evidence to support the claim because there are substantially more heads than tails.
According to the Federal Housing Finance Board, the mean price of a single-family home two years ago was $299 comma 500. A real estate broker believes that because of the recent credit crunch, the mean price has increased since then. The null hypothesis is not rejected.
There is not sufficient evidence to conclude that the mean price of a single-family home has increased from its level two years ago of $299 comma 500.
Claim: A coin favors heads when tossed, and there are 10 heads in 18 tosses
There is not sufficient evidence to support the claim because there are not substantially more heads than tails.
H0: u = 130 H1: u < 130
left-tailed population mean
In 2005, 33% of university undergraduate students had at least one tattoo. A health practitioner obtains a random sample of 1066 university undergraduates and finds that 351 have at least one tattoo. Has the proportion of university undergraduate students with at least one tattoo changed since 2005? Use the alpha = 0.1 level of significance. Complete parts (a) through (d) below.
population proportion p = .33 p /= .33 p-val = .960 Do not reject H0 because the P-value is greater than the alpha = 0.05 level of significance There is not sufficient evidence
Complete parts (a) through (c) below. (a) Determine the critical value(s) for a right-tailed test of a population mean at the alpha = 0.01 level of significance with 10 degrees of freedom. (b) Determine the critical value(s) for a left-tailed test of a population mean at the alpha = 0.10 level of significance based on a sample size of n = 20. (c) Determine the critical value(s) for a two-tailed test of a population mean at the alpha = 0.05 level of significance based on a sample size of n = 18.
t = +2.764 t = -1.328 t = +- 2.11
Claim: The mean IQ score of statistics professors is less than 129. Sample data: n = 22, x overbar equals125, sequals11 .The significance level is alpha equals0.05.
t = -1.706 crit val = -1.721 Fail to reject the null hypothesis and do not support the claim that u < 129
H0: p = 0.9 H1: p /= 0.9
two-tailed population proportion
Critical Values & Rejection Region
- The critical value is the z-score that corresponds to a probability in the tail(s). - The rejection region is the range of values for which the sample data is "convincing enough" to reject the null hypothesis - Strong Evidence to REJECT Ho: If the sample proportion is too many standard deviations from the proportion stated in the null hypothesis-- then we reject the null hypothesis.
Using a table of critical t-values of the t distribution, find the range of values for the P-value for testing a claim about the mean body temperature of healthy adults for a left-tailed test with n equals 12 and test statistic t equals negative 3.086 .
0.005 < P-value < 0.01
n = 14, t = 1.816 DF = 13
0.05 < P-value < 0.10
Advantages of P-value Approach
1. One advantage of using P-values over the classical approach in hypothesis testing is that P-values provide information regarding the strength of the evidence. 2. Another is that P-values are interpreted the same way regardless of the type of hypothesis test being performed. The lower the P-value, the stronger the evidence is against the null hypothesis. p-value < .10 ............ some evidence of H1 p-value < .05 ............ significant evidence of H1 p-value < .01 ............ strong evidence of H1
When testing gas pumps for accuracy, fuel-quality enforcement specialists tested pumps and found that 1288 of them were not pumping accurately (within 3.3 oz when 5 gal is pumped), and 5692 pumps were accurate. Use a 0.01 significance level to test the claim of an industry representative that less than 20% of the pumps are inaccurate. Use the P-value method and use the normal distribution as an approximation to the binomial distribution.
5692 + 1288 = 6980 p = 0.2 p < 0.2 z = -3.2317 p-val = 0.0006 Because the P-value is less than the significance level, reject the null hypothesis. There is sufficient evidence support the claim that less than 20% of the pumps are inaccurate.
Determine whether the statement is true or false. If it is false, rewrite it as a true statement. In a hypothesis test, you assume the alternative hypothesis is true.
False. In a hypothesis test, you assume the null hypothesis is true.
Test Statistic for p
For hypothesis tests about a population proportion, p The test statistic for p is When 1. The sample is obtained from random sampling 2. sample size, n < 5% N (this means we can treat observations as independent binomial trials) 3. npo ≥ 5 and nqo ≥ 5 x-bar(this means it is OK to use the normal to approximate the binomial)
p-val method
For the P-value method, reject H0 if the P-value is less than or equal to the significance level alpha. Fail to reject H0 if the P-value is greater than alpha. Then restate the decision in simple, nontechnical terms, and address the original claim.
In other words:
H0: The null hypothesis, H0, is always the statement of "status quo" or "no change" and contains a statement of equality. The null hypothesis is the one that is assumed to be true until we have evidence to contradict it. H1: The claim we seek evidence for is always the alternative or research hypothesis, H1
Careful!
H0: You can never prove a claim that is stated in the null hypothesis. You can only say there is "not enough evidence" to contradict it. H1: When you wish to show evidence to support a claim, you set up that claim in the alternative hypothesis. The alternative, H1 hypothesis is called the research hypothesis.
(a) Determine the null and alternative hypotheses Six years ago, 11.8% of registered births were to teenage mothers. A sociologist believes that the percentage has increased since then.
H0: p = 0.118, H1: p > 0.118
A 0.05 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender selection, the proportion of baby girls is different from 0.5.
H0: p = 0.5 H1: p /= 0.5 two-tailed 0.05 -1.96, 1.96 (a/2 scores)
Determine the null and alternative hypotheses. Three years ago, the mean price of a single-family home was $243 comma 704 . A real estate broker believes that the mean price has decreased since then.
H0: u = 234,704, H1: u < 234,704
worse error
In hypothesis testing, the Type I error is the error considered the "worse" type of error, so it is the probability of a Type I error that is controlled for.
Type II Error
Leaving the null hypothesis unchallenged, when null hypothesis is false)
A certain drug is used to treat asthma. In a clinical trial of the drug, 15 of 267 treated subjects experienced headaches (based on data from the manufacturer). The accompanying calculator display shows results from a test of the claim that less than 8% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.01 significance level to complete parts (a) through (e) below. 1-PropZTest prop , 0.08 z = -1.434704888 p = 0.0756856054 p-hat = 0.0561797753 n = 267
Left-tailed test z = -1.43 P-value = 0.0757 H0: p = 0.8 Fail to reject the null hypothesis because the P-value is greater than the significance level a There is not sufficient evidence to support the claim that less than 8% of treated subjects experienced headaches.
A simple random sample of size nequals 200 drivers were asked if they drive a car manufactured in a certain country. Of the 200 drivers surveyed, 108 responded that they did. Determine if more than half of all drivers drive a car made in this country at the alpha equals 0.05 level of significance. Complete parts (a) through (d).
Population proportion p = 0.5 p > 0.5 Do not reject H0 because the P-value is greater than the alpha = 0.05 level of significance
A safety administration conducted crash tests of child booster seats for cars. Listed below are results from those tests, with the measurements given in hic (standard head injury condition units). The safety requirement is that the hic measurement should be less than 1000 hic. Use a 0.05 significance level to test the claim that the sample is from a population with a mean less than 1000 hic. Do the results suggest that every child booster seat will meet the specified requirement?
Reject H0. There is sufficient evidence to conclude the mean safety rating for this type of child booster seats is less than 1000 hic. There is strong evidence that the mean is less than 1000 hic, but individually the child booster seats will vary around that mean, so some individual booster seats may be above 1000 hic.
Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. In a manual on how to have a number one song, it is stated that a song must be no longer than 210 seconds. A simple random sample of 40 current hit songs results in a mean length of 242.9 sec and a standard deviation of 56.85 sec. Use a 0.05 significance level and the accompanying Minitab display to test the claim that the sample is from a population of songs with a mean greater than 210 sec. What do these results suggest about the advice given in the manual?
Reject H0. There is sufficient evidence to support the claim that the sample is from a population of songs with a mean length greater than 210 sec The results suggest that the advice of writing a song that must be no longer than 210 seconds is not sound advice.
The average daily volume of a computer stock in 2011 was mu equals35.1 million shares, according to a reliable source. A stock analyst believes that the stock volume in 2014 is different from the 2011 level. Based on a random sample of 40 trading days in 2014, he finds the sample mean to be 29.1 million shares, with a standard deviation of sequals 11.4 million shares. Test the hypotheses by constructing a 95% confidence interval. Complete parts (a) through (c) below.
Reject the null hypothesis because mu equals35.1million shares does not fall in the confidence interval.
A certain drug is used to treat asthma. In a clinical trial of the drug, 19 of 265 treated subjects experienced headaches (based on data from the manufacturer). The accompanying calculator display shows results from a test of the claim that less than 11% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level to complete parts (a) through (e) below.
Reject the null hypothesis because the P-value is less than or equal to the significance level, alpha. There is sufficient evidence to support the claim that less than 11% of treated subjects experienced headaches.
A survey of 1,611 randomly selected adults showed that 506 of them have heard of a new electronic reader. The accompanying technology display results from a test of the claim that 36% of adults have heard of the new electronic reader. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.05 significance level to complete parts (a) through (e).
Reject the null hypothesis because the P-value is less than or equal to the significance level, alpha. There is sufficient evidence to warrant rejection of the claim that 36% of adults have heard of the new electronic reader.
Identify the type I error and the type II error that correspond to the given hypothesis. The percentage of adults who have a job is greater than 88 %.
Reject the null hypothesis that the percentage of adults who have a job is equal to 88 % when that percentage is actually equal to 88 %. Fail to reject the null hypothesis that the percentage of adults who have a job is equal to 88 % when that percentage is actually greater than 88 %.
Test the hypothesis, using (a) the classical approach and then (b) the P-value approach. Be sure to verify the requirements of the test. p = 0.7 p > 0.7 n = 100 x = 85 a = 0.05
Reject the null hypothesis, because the test statistic is greater than the critical value. Reject the null hypothesis, because the P-value is less than alpha.
Assume that a researcher wants to use sample data to test the claim that the sample is from a population with a mean less than 1.8 min. The researcher will use a 0.05 significance level to test that claim. If the researcher wants to use the confidence interval method of testing hypotheses, what level of confidence should be used for the confidence interval?
Since this is a one-tailed test, the researcher should use a confidence level of 0.90. confidence level = 1 - 2a
In a clinical trial, 37 out of 950 patients taking a prescription drug complained of flulike symptoms. Suppose that it is known that 2.8% of patients taking competing drugs complain of flulike symptoms. Is there sufficient evidence to conclude that more than 2.8% of this drug's users experience flulike symptoms as a side effect at the alpha equals 0.05 level of significance?
Since P-value < alpha, reject the null hypothesis and conclude that there is sufficient evidence that more than 2.8% of the users experience flulike symptoms.
Explain what "statistical significance" means.
Statistical significance means that the result observed in a sample is unusual when the null hypothesis is assumed to be true.
Formal Process: Hypothesis Testing
Step 1: State the null and alternative hypothesis: H0 and H1 Step 2: Classical Approach Graph the Rejection Region (using the critical values for a & the direction of the test) Step 3: Compute value of the Test Statistic 2 different formulas (z-score or t-score) Step 4: P-value Approach: Compute the P-value: Does this data's test statistic happen with high probability or low probability? Step 5: Make a Decision: "Reject Ho" or "Do not Reject Ho" Step 6: State a conclusion. (In a complete sentence with the specifics of the tested hypothesis)
P-values
The P-value is the probability of getting a sample with z-score as extreme(or more) as the result we got from the current sample. (...how "rare" is this data?) The p-value is computed as the area in the tail(s) beyond the computed test statistic.
alternative hypothesis
The alternative hypothesis, denoted, H1 (read "H-one"), is a claim to be tested. The claim that we are trying to gather evidence for determines the H1 alternative hypothesis. H1 is sometimes called the research hypothesis
When n < 30
The hypothesis testing procedures we are using are robust, which means that minor departures from normality will not adversely affect the results of the test. (see acceptance corridors in QQ plots) - However, for small samples, if the data has outliers, this procedure should not be used.
null hypothesis
The null hypothesis, denoted H_o (read "H-naught"), is a statement to be tested. The null hypothesis is assumed true until evidence indicates otherwise.
Claim: The mean IQ score of students in a large statistics class is less than 130. A simple random sample of the students has a mean IQ score of 129.5.
The sample is not unusual if the claim is true. The sample is unusual if the claim is false. Therefore, there is not sufficient evidence to support the claim.
Claim: The mean respiration rate left parenthesis in breaths per minute right parenthesis of students in a large statistics class is greater than 10. A simple random sample of the students has a mean respiration rate of 10.6.
The sample is not unusual if the claim is true. The sample is unusual if the claim is false. Therefore, there is not sufficient evidence to support the claim.
Claim: The mean respiration rate left parenthesis in breaths per minute right parenthesis of students in a large statistics class is less than 30. A simple random sample of the students has a mean respiration rate of 11.3.
The sample is not unusual if the claim is true. The sample is unusual if the claim is false. Therefore, there is sufficient evidence to support the claim
According to the report, the mean monthly cell phone bill was $48.49 three years ago. A researcher suspects that the mean monthly cell phone bill is higher today. The null hypothesis is rejected.
There is sufficient evidence to conclude that the mean monthly cell phone bill is higher than its level three years ago of $48.49.
Six years ago, 11.3% of registered births were to teenage mothers. A sociologist believes that the percentage has increased since then.
There is sufficient evidence to conclude that the percentage of teenage mothers has increased.
Four Outcomes of Decision Making
Two Correct Decisions - You correctly accuse me of "stacking the deck", when I really did stack the deck. - You don't accuse me, when I actually didn't stack the deck. Two ERROR Decisions - You accuse me of "stacking the deck", but I didn't!! - You don't accuse me, when I actually did!!!
A survey of 1,652 randomly selected adults showed that 585 of them have heard of a new electronic reader. The accompanying technology display results from a test of the claim that 35% of adults have heard of the new electronic reader. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.01 significance level to complete parts (a) through (e). Sample proportion: 0.354116 Test statistic, z: 0.3508 Critical z: plus or minus 2.5758 P-Value: 0.7258
Two-tailed test z= .35 P-value = .7258 H0: p = .35 Fail to reject the null hypothesis because the P-value is greater than the significance level a There is not sufficient evidence to warrant rejection of the claim that 35% of adults have heard of the new electronic reader.
Assume a significance level of alpha equals 0.05 and use the given information to complete parts (a) and (b) below. Original claim: The proportion of male golfers is less than 0.3. The hypothesis test results in a P-value of 0.199.
a. Fail to reject Upper H 0 because the P-value is greater than alpha. b. There is not sufficient evidence to support the claim that the proportion of male golfers is less than 0.3.
Assume a significance level of alpha equals 0.01 and use the given information to complete parts (a) and (b) below. Original claim: Women have heights with a mean equal to 160.3 cm. The hypothesis test results in a P-value of 0.1024.
a. Fail to reject Upper H 0 because the P-value is greater than alpha. b. There is not sufficient evidence to warrant rejection of the claim that the mean height of women is equal to 160.3 cm.
A 0.05 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender selection, the proportion of baby girls is different from 0.5. Assume that sample data consists of 136 girls in 256 births, so the sample statistic of 17/32 results in a z score that is 1 standard deviation above 0. Complete parts (a) through (h) below.
a. H_0: p = 0.5 H_1: p =/ 0.5 b. a = 0.05 c. np = nq = 128 >= 5, q = 1 - p = 0.5 d. because of H_1: p =/ 0.5, two tailed e. z = (p-q)/sqrt(pq/n) = (17/32 - 0.5) / sqrt(.5*.5 / 256) = 1 f. z = 1 is positive, so P-value is 0.1587 (area to right) 2(.1587) = .3174 g. z = 1.96 and - 1.96 h. The critical region corresponds to the values of the test statistic that cause one to reject the null hypothesis. Since the significance level a is the probability of making the mistake of rejecting the null hypothesis when it is true, the area of the critical region is equal to alpha equals 0.05.
A 0.05 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender selection, the proportion of baby girls is greater than 0.5. Assume that sample data consists of 120 girls in 225 births, so the sample statistic of eight fifteenths results in a z score that is 1 standard deviation above 0. Complete parts (a) through (h) below.
a. H_0: p = 0.5 H_1: p > 0.5 b. a = 0.05 c. Notice that the claim makes a statement about the population proportion. This means that the normal distribution is the correct sampling distribution to use. Also, notice that np = nq = 112.5 >= 5 ,where q = 1-p = 0.5. Thus, both requirements to use the normal distribution are satisfied. d. Because the alternative hypothesis is H_1: p > 0.5, the critical region lies in the extreme right region under the curve. Therefore, the test is right-tailed. e. z = (p-q)/sqrt(pq/n) = (8/15-0.5)/sqrt(.5*.5* / 225) = 1 f. z = 1 -> p-value = .1587 (right tail area) g. a = 0.05 -> z = 1.65 The critical region corresponds to the values of the test statistic that cause one to reject the null hypothesis. Since the significance level a is the probability of making the mistake of rejecting the null hypothesis when it is true, the area of the critical region is equal to alpha equals 0.05.
Assume a significance level of alpha equals 0.05 and use the given information to complete parts (a) and (b) below. Original claim: The proportion of male golfers is less than 0.5. The hypothesis test results in a P-value of 0.018. a. State a conclusion about the null hypothesis. (Reject Upper H 0 or fail to reject Upper H 0.) b. Without using technical terms, state a final conclusion that addresses the original claim. Which of the following is the correct conclusion?
a. Reject Upper H 0 because the P-value is less than alpha. b. There is sufficient evidence to support the claim that the proportion of male golfers is less than 0.5.
Assume that the significance level is alpha equals 0.05. Use the given statement and find the P-value and critical value(s). The test statistic of z = - 2.13 is obtained when testing the claim that p equals 1/4
alt. hyp. is p /= 1/4, two-tailed area between z = -2.13 is 2*0.0166 = 0.0332 since two tailed, a/2 = 0.025 and z = -1.96, 1.96
In statistics, what does df denote? If a simple random sample of 25 speeds of cars on California Highway 405 is to be used to test the claim that the sample values are from a population with a mean greater than the posted speed limit of 65 mi/h, what is the specific value of df?
df denotes the number of degrees of freedom. For this sample, df = 24.
Several years ago, 43% of parents who had children in grades K-12 were satisfied with the quality of education the students receive. A recent poll asked 1,235 parents who have children in grades K-12 if they were satisfied with the quality of education the students receive. Of the 1,235 surveyed, 536 indicated that they were satisfied. Construct a 95% confidence interval to assess whether this represents evidence that parents' attitudes toward the quality of education have changed.
np0(1-p0) = 1235(0.43)(1-.43) = 303 lower = .43 upper = .49 Since the interval contains the proportion stated in the null hypothesis, there is insufficient evidence that parents' attitudes toward the quality of education have changed.
The _________ hypothesis is a statement that the value of a population parameter is equal to some claimed value.
null
The _______ _______ is a statement of no change comma no effect comma or no difference.
null hypothesis
Suppose a mutual fund qualifies as having moderate risk if the standard deviation of its monthly rate of return is less than 5%. A mutual-fund rating agency randomly selects 24 months and determines the rate of return for a certain fund. The standard deviation of the rate of return is computed to be 3.58%. Is there sufficient evidence to conclude that the fund has moderate risk at the alpha equals 0.05 level of significance? A normal probability plot indicates that the monthly rates of return are normally distributed.
o = 0.05 o < 0.05 x^2 = 11.791 p-val = .026 Since the P-value is less than the level of significance, reject the null hypothesis. There is sufficient evidence to conclude that the fund has moderate risk at the 0.05 level of significance.
A simple random sample of 49 men from a normally distributed population results in a standard deviation of 9.1 beats per minute. The normal range of pulse rates of adults is typically given as 60 to 100 beats per minute. If the range rule of thumb is applied to that normal range, the result is a standard deviation of 10 beats per minute. Use the sample results with a 0.10 significance level to test the claim that pulse rates of men have a standard deviation equal to 10 beats per minute. Complete parts (a) through (d) below.
o = 10 o /= 10 x^2 = 39.749 p-val = .4084 Do not reject H0, because the P-value is greater than the level of significance. There is insufficient evidence to warrant rejection of the claim that the standard deviation of men's pulse rates is equal to 10 beats per minute.
A genetic experiment involving peas yielded one sample of offspring consisting of 435 green peas and 133 yellow peas. Use a 0.01 significance level to test the claim that under the same circumstances, 23% of offspring peas will be yellow. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution.
p = .23 p /= .23 z = .24 p-val = .8103 Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha. There is not sufficient evidence to warrant rejection of the claim that 23% of offspring peas will be yellow.
In a previous poll, 24% of adults with children under the age of 18 reported that their family ate dinner together 7 nights a week. Suppose that, in a more recent poll, 283 of 1200 adults with children under the age of 18 reported that their family ate dinner together 7 nights a week. Is there sufficient evidence that the proportion of families with children under the age of 18 who eat dinner together 7 nights a week has decreased at the alpha equals 0.05 significance level?
p = .24, p < .24 p-val = .368 No, there is not sufficient evidence because the P-value is greater than the level of significance. Therefore, do not reject the null hypothesis.
Five years ago, 9.8% of high school students had tried marijuana for the first time before the age of 13. A school resource officer (SRO) thinks that the proportion of high school students who have tried marijuana for the first time before the age of 13 has decreased since then.
p = 0.098, p < 0.098 There is sufficient evidence to conclude that the proportion of high school students has decreased. The SRO committed a type I error because he rejected the null hypothesis when, in fact, it is true.
In 1997, a survey of 980 households showed that 153 of them use e-mail. Use those sample results to test the claim that more than 15% of households use e-mail. Use a 0.05 significance level. Use this information to answer the following questions.
p = 0.15 p > 0.15 z = .54 p-val: .295 There is not sufficient evidence to support the claim that more than 15% of households use e-mail. No, the conclusion is not valid today because the population characteristics of the use of e-mail are changing rapidly.
A recent broadcast of a television show had a 15 share, meaning that among 6000 monitored households with TV sets in use, 15% of them were tuned to this program. Use a 0.01 significance level to test the claim of an advertiser that among the households with TV sets in use, less than 25 % were tuned into the program. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution.
p = 0.25 p < 0.25 z = -17.89 p-val = 0 Reject H0. There is sufficient evidence to support the claim that less than 25% of the TV sets in use were tuned to the program.
In a study of pregnant women and their ability to correctly predict the sex of their baby, 58 of the pregnant women had 12 years of education or less, and 34.5% of these women correctly predicted the sex of their baby. Use a 0.05 significance level to test the claim that these women have an ability to predict the sex of their baby equivalent to random guesses. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, and conclusion about the null hypothesis. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution. Do the results suggest that their percentage of correct predictions is different from results expected with random guesses?
p = 0.5 p /= 0.5 z = -2.36 p-val = .0183 Reject, is not, is
In a recent poll, 801 adults were asked to identify their favorite seat when they fly, and 508 of them chose a window seat. Use a 0.01 significance level to test the claim that the majority of adults prefer window seats when they fly. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution
p = 0.5 p > 0.5 z = 7.60 p-val = .000 Reject the null hypothesis because the P-value is less than or equal to the significance level, alpha. There is sufficient evidence to support the claim that the majority of adults prefer window seats when they fly.
A clinical trial was conducted using a new method designed to increase the probability of conceiving a girl. As of this writing, 950 babies were born to parents using the new method, and 888 of them were girls. Use a 0.01 significance level to test the claim that the new method is effective in increasing the likelihood that a baby will be a girl. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution.
p = 0.5 p > 0.5 z = 26.80 p-val = 0 Reject the null hypothesis because the P-value is less than or equal to the significance level, alpha. There is sufficient evidence to support the claim that the new method is effective in increasing the likelihood that a baby will be a girl.
In February 2008, an organization surveyed 1034 adults aged 18 and older and found that 528 believed they would not have enough money to live comfortably in retirement. Does the sample evidence suggest that a majority of adults in a certain country believe they will not have enough money in retirement? Use the alpha equals 0.01 level of significance.
p = 0.5, p > 0.5 Since P-value > alpha, do not reject the null hypothesis and conclude that there is not sufficient evidence that a majority of adults in the United States believe they will not have enough money in retirement.
In a recent poll of 750 randomly selected adults, 588 said that it is morally wrong to not report all income on tax returns. Use a 0.05 significance level to test the claim that 75% of adults say that it is morally wrong to not report all income on tax returns. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution.
p = 0.75 p /= 0.75 z = 2.15 p-val = .0316 Reject H0. There is sufficient evidence to warrant rejection of the claim that 75% of adults say that it is morally wrong not to report all income on tax returns.
Trials in an experiment with a polygraph include 96 results that include 22 cases of wrong results and 74 cases of correct results. Use a 0.05 significance level to test the claim that such polygraph results are correct less than 80% of the time. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution.
p = 0.80 p < 0.80 z = -.70 (use 74/96 in calc) p-val = .2420 Fail to reject H0. There is not sufficient evidence to support the claim that the polygraph results are correct less than 80% of the time.
Assume that the significance level is alpha equals 0.05. Use the given information to find the P-value and the critical value(s). The test statistic of z equals 2.11 is obtained when testing the claim that p greater than 0.4.
p > 0.4, right-tailed area right of z = 2.11 is 0.0174 critical value z = 1.65
For the given claim, complete parts (a) and (b) below. Claim: At least 20% of Internet users pay bills online. A recent survey of 341 Internet users indicated that 19% pay their bills online.
p >= 0.2 H0: p = 0.2 H1: p < 0.2
In a Harris poll, adults were asked if they are in favor of abolishing the penny. Among the responses, 1282 answered "no," 463 answered "yes," and 379 had no opinion. What is the sample proportion of yes responses, and what notation is used to represent it?
p-hat = .218 The symbol p-hat is used to represent a sample proportion.
For the given claim, complete parts (a) and (b) below. Claim: High school teachers have incomes with a standard deviation that is less than $20 comma 750. A recent study of 149 high school teacher incomes showed a standard deviation of $18 comma 750.
sigma < 20750 H0: sigma = 20750 H1: sigma < 20750
The claim is that the white blood cell counts of adult females are normally distributed, with a standard deviation equal to 2.97. A random sample of 17 adult females has white blood cell counts with a mean of 6.78 and a standard deviation of 2.73. Find the value of the test statistic.
sigma = 2.97 use x^2 stat x^2 = (n-1)*s^2 / sigma^2 x^2 = (17-1)*2.73^2 / 2.97^2 = 13.519
A simple random sample of 25 filtered 100 mm cigarettes is obtained, and the tar content of each cigarette is measured. The sample has a mean of 19.6 mg and a standard deviation of 3.25 mg. Use a 0.05 significance level to test the claim that the mean tar content of filtered 100 mm cigarettes is less than 21.1 mg, which is the mean for unfiltered king size cigarettes. What do the results suggest, if anything, about the effectiveness of the filters?
t = -2.308 p-val = .015 Reject H0. There is sufficient evidence to support the claim that the mean tar content of filtered 100 mm cigarettes is less than 21.1 mg. The results suggest that the filters are effective.
Assume that the significance level is alpha equals 0.01. Use the given information to find the P-value and the critical value(s). With H1: p /= 6/7, the test statistic is z = -2.02
two-tailed area: 2*0.0217 = .0434 a/2 = 0.01/2 = 0.005 ->z = -2.58, 2.58
For the given claim, complete parts (a) and (b) below. Claim: The mean weight of beauty pageant winners is 105 pounds. A study of 25 randomly selected beauty pageants resulted in a mean winner weight of 110 pounds.
u = 105 H0: u = 105 H1: u /= 105
The claim is that the IQ scores of statistics professors are normally distributed, with a mean greater than 127. A sample of 10 professors had a mean IQ score of 131 with a standard deviation of 12. Find the value of the test statistic.
u = 127 t = (x-u)/s/sqrt(n) = (131-127)/12/sqrt(10) = 1.054
A coin mint has a specification that a particular coin has a mean weight of 2.5 g. A sample of 39 coins was collected. Those coins have a mean weight of 2.49468 g and a standard deviation of 0.01218g. Use a 0.05 significance level to test the claim that this sample is from a population with a mean weight equal to 2.5 g. Do the coins appear to conform to the specifications of the coin mint?
u = 2.5 u /= 2.5 t = -2.728 p-val = .0096 Reject H0. There is sufficient evidence to warrant rejection of the claim that the sample is from a population with a mean weight equal to 2.5g. No, since the coins seem to come from a population with a mean weight different from 2.5g.
Several years ago, the reported mean age of an inmate on death row was 38.3 years. A sociologist wondered whether the mean age of a death-row inmate has changed since then. He randomly selects 33 death-row inmates and finds that their mean age is 36.5, with a standard deviation of 9.2. Construct a 99% confidence interval about the mean age of death row inmates. What does the interval imply?
u = 38.3 u /= 38.3 (32.11, 40.89) Since the given mean age is in the interval, do not reject the null hypothesis.
A survey of 61,649 people included several questions about office relationships. Of the respondents, 26.6% reported that bosses scream at employees. Use a 0.05 significance level to test the claim that more than 1/4 of people say that bosses scream at employees. How is the conclusion affected after learning that the survey is an online survey in which Internet users chose whether to respond? Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution.
use .266*61649 in calc z = 9.18 p-val = 0 Reject H0. There is sufficient evidence to support the claim that more than 1/4 of people say that bosses scream at employees. If the sample is a voluntary response sample, the conclusion might not be valid.
Workers at a certain soda drink factory collected data on the volumes (in ounces) of a simple random sample of 25 cans of the soda drink. Those volumes have a mean of 12.19 oz and a standard deviation of 0.12oz, and they appear to be from a normally distributed population. If the workers want the filling process to work so that almost all cans have volumes between 11.96 oz and 12.56 oz, the range rule of thumb can be used to estimate that the standard deviation should be less than 0.15 oz. Use the sample data to test the claim that the population of volumes has a standard deviation less than 0.15 oz. Use a 0.05 significance level. Complete parts (a) through (d) below.
x^2 = (n-1)s^2 / o^2 = 24*.12^2 / .15^2 = 15.36 p-val = .0902 Do not reject H0, because the P-value is greater than the level of significance. There is insufficient evidence to conclude that the population standard deviation of can volumes is less than 0.15oz.
The piston diameter of a certain hand pump is 0.8 inch. The manager determines that the diameters are normally distributed, with a mean of 0.8 inch and a standard deviation of 0.005 inch. After recalibrating the production machine, the manager randomly selects 21 pistons and determines that the standard deviation is 0.0044 inch. Is there significant evidence for the manager to conclude that the standard deviation has decreased at the alpha equals 0.05 level of significance?
x^2 = 15.488 p-val = .252 Since the P-value is greater than the level of significance, do not reject the null hypothesis. There is not sufficient evidence to conclude that the standard deviation has decreased at the 0.05 level of significance.
A simple random sample of pulse rates of 25 women from a normally distributed population results in a standard deviation of 11.6 beats per minute. The normal range of pulse rates of adults is typically given as 60 to 100 beats per minute. If the range rule of thumb is applied to that normal range, the result is a standard deviation of 10 beats per minute. Use the sample results with a 0.05 significance level to test the claim that pulse rates of women have a standard deviation equal to 10 beats per minute. Complete parts (a) through (d) below.
x^2 = 32.294 p-val = .2397 Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that pulse rates of women have a standard deviation equal to 10 beats per minute.
Test the hypothesis, using (a) the classical approach and then (b) the P-value approach. Be sure to verify the requirements of the test. p = 0.2 p > 0.2 n = 250 x = 60 a = 0.01
z = 1.58 z-critical = 2.33 Do not reject the null hypothesis, because the test statistic is less than the critical value. p-val = 0.0571 Do not reject the null hypothesis, because the P-value is greater than alpha.