9.5c Complete the steps of a one-mean hypothesis testing with Population SD known - P-Value approach

Drew wondered if the average age of students in AP Statistics classes in his high school is under 18. He randomly selected 10 AP Statistics students in his school and the data set is given below. Using the information from a previous survey, he has concluded that the population standard deviation for the age of AP Statistics students in his school is 1.23. Assume that the ages of students in AP Statistics classes in this high school are normally distributed. 17,19,18,17,15,18,16,17,17,16 Use a calculator to test whether the true age of AP Statistics students in his school is under 18. Identify the p-value from the calculator output, rounding to three decimal places.

.005

Give the numerical value of the parameter p in the following binomial distribution scenario.A softball pitcher has a 0.692 probability of throwing a strike for each pitch and a 0.308 probability of throwing a ball. If the softball pitcher throws 29 pitches, we want to know the probability that more than 17 of them are strikes.Consider strikes as successes in the binomial distribution. Do not include p= in your answer.

0.692 The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is a strike, so p=0.692.

A magazine editor was wondering if the number of ads in the magazine had gone up in the past few years. Historically, the magazine has had a mean of 86 ads per issue with a standard deviation of 12 ads per issue. The editor randomly selected 10 magazines from the last two years and counted the ads. The data are given below. Using the historical standard deviation as the population standard deviation, use a TI-83, TI-83 Plus, or TI-84 calculator to test whether the mean number of ads has increased. Identify the p-value from the calculator output, rounding to three decimal places. 80 94 105 82 94 86 101 85 91 92

0.94

A government official wanted to determine if the total income-tax revenue in a region had dropped since last year. The official surveyed 10 people in the region. Instead of using the standard deviation from the survey, the official decided to use the census data for the country to assume that the population standard deviation of income is $23.8 thousand. Last year, the mean regional income was $58.5 thousand. The government official conducts a one-mean hypothesis at the 5% significance level, to test if the average regional income decreased from the previous year of $58.5 thousand. (a) Which answer choice shows the correct null and alternative hypotheses for this test?

H0:μ=$58.5; Ha:μ<$58.5, which is a left-tailed test. The null hypothesis should be the government official's suspected average: H0:μ=$58.5. He wants to test if the average average income is less than $58.5 thousand, so the alternative hypothesis is Ha:μ<$58.5, which is a left-tailed test because of the less than sign. FEEDBACK

Caden is an editor for a local newspaper. The newspaper's editor-in-chief asked Caden to determine whether the average distance a reader lives from the newspaper's headquarters is greater than 25 miles. He sent a survey to a random selection of the newspaper's readers and received 10 responses. Using the results from the current survey and several previous surveys, Caden decided to assume that the population standard deviation for the distance the newspaper's readers live from its headquarters is 10.3 miles. Caden conducts a one-mean hypothesis at the 5% significance level, to test if the average distance a reader lives from the newspaper's headquarters is greater than 25 miles. (a) Which answer choice shows the correct null and alternative hypotheses for this test?

H0:μ=25; Ha:μ>25, which is a right-tailed test. The null hypothesis should be Caden's suspected average: H0:μ=25. He wants to test if the average distance a reader lives from the newspaper's headquarters is greater than 25 miles, so the alternative hypothesis is Ha:μ>25, which is a right-tailed test because of the greater than sign.

A hospital administrator noticed a large change in the weight of babies being born at the hospital in the last few months. She randomly selected 10 babies born in the last month. Historically, babies at the hospital have had a mean weight of 7.58 pounds and a standard deviation of 1.05 pounds. To test the weights, the administrator decided to use the historical standard deviation as the population standard deviation. The hospital administrator conducts a one-mean hypothesis at the 5% significance level, to test if there has been a change in the mean weight of babies being born at the hospital in the last few months from 7.58 pounds. (a) Which answer choice shows the correct null and alternative hypotheses for this test?

H0:μ=7.58; Ha:μ≠7.58, which is a two-tailed test. The null hypothesis should be the historical mean weight of babies born at this hospital: H0:μ=7.58. The study wants to know if the mean weight of babies in the past few months is different from 7.58 pounds now. This means that we just want to test if the mean is not 7.58 pounds now. So, the alternative hypothesis is Ha:μ≠7.58, which is a two-tailed test.

A hospital administrator noticed a large change in the weight of babies being born at the hospital in the last few months. She randomly selected 10 babies born in the last month. Historically, babies at the hospital have had a mean weight of 7.63 pounds and a standard deviation of 1.02 pounds. To test the weights, the administrator decided to use the historical standard deviation as the population standard deviation. The hospital administrator conducts a one-mean hypothesis at the 5% significance level, to test if change in the weight of babies being born at the hospital in the last few months from 7.63 pounds. (a) Which answer choice shows the correct null and alternative hypotheses for this test?

H0:μ=7.63; Ha:μ≠7.63, which is a two-tailed test. The null hypothesis should be the historical mean weight of babies born at this hospital: H0:μ=7.63. The study wants to know if the mean weight of babies in the past few months is different from 7.63 pounds. This means that we just want to test if the mean is not 7.63 pounds. So, the alternative hypothesis is Ha:μ≠7.63, which is a two-tailed test.

Caden is an editor for a local newspaper. The newspaper's editor-in-chief asked Caden to determine whether the average distance a reader lives from the newspaper's headquarters is greater than 25 miles. He sent a survey to a random selection of the newspaper's readers and received 10 responses. Using the results from the current survey and several previous surveys, Caden decided to assume that the population standard deviation for the distance the newspaper's readers live from its headquarters is 10.3 miles. Caden conducts a one-mean hypothesis at the 5% significance level, to test if the average distance a reader lives from the newspaper's headquarters is greater than 25 miles. (a) H0:μ=25; Ha:μ>25, which is a right-tailed test. (b) The distance (in miles) that the readers lived from the newspaper's headquarters is given below. Use a TI-83, TI-83 Plus, or TI-84 calculator to test whether the mean is greater than 25 miles. Identify the test statistic, z, and p-value from the calculator output, rounding to two decimal places. 21 49 32 35 28 42 9 16 25 27

Test statistic = 1.04 p-value = 0.15

A hospital administrator noticed a large change in the weight of babies being born at the hospital in the last few months. She randomly selected 10 babies born in the last month. Historically, babies at the hospital have had a mean weight of 7.63 pounds and a standard deviation of 1.02 pounds. To test the weights, the administrator decided to use the historical standard deviation as the population standard deviation. The hospital administrator conducts a one-mean hypothesis at the 5% significance level, to test if change in the weight of babies being born at the hospital in the last few months from 7.63 pounds. (a) H0:μ=7.63; Ha:μ≠7.63, which is a two-tailed test. (b) Weight (in pounds) of babies born in the past few months is given below. Use a TI-83, TI-83 Plus, or TI-84 calculator to test whether the mean weight of babies born at the hospital in the last month is different than the historical average of 7.63 pounds. Identify the test statistic, z, and p-value from the calculator output, rounding to two decimal places. 6.3 8.2 7.4 7.1 7.5 6.7 9.2 5.8 7.8 6.0

Test statistic =-1.33 p-value=0.18

A hospital administrator noticed a large change in the weight of babies being born at the hospital in the last few months. She randomly selected 10 babies born in the last month. Historically, babies at the hospital have had a mean weight of 7.58 pounds and a standard deviation of 1.05 pounds. To test the weights, the administrator decided to use the historical standard deviation as the population standard deviation. The hospital administrator conducts a one-mean hypothesis at the 5% significance level, to test if there has been a change in the mean weight of babies being born at the hospital in the last few months from 7.58 pounds. (a) H0:μ=7.58; Ha:μ≠7.58, which is a two-tailed test. (b) Weight (in pounds) of babies born in the past few months is given below. Use a TI-83, TI-83 Plus, or TI-84 calculator to test whether the mean weight of babies born at the hospital in the last month is different than the historical average of 7.58 pounds. Identify the test statistic, z, and p-value from the calculator output, rounding to two decimal places. 7.5 5.7 10.2 5.8 6.8 5.0 7.3 8.2 7.4 6.8

Test statistic(z)=-1.54 p-value(p)= .12

Drew wondered if the average age of students in AP Statistics classes in his high school is under 18. He randomly selected 10 AP Statistics students in his school and the data set is given below. Using the information from a previous survey, he has concluded that the population standard deviation for the age of AP Statistics students in his school is 1.23. Assume that the ages of students in AP Statistics classes in this high school are normally distributed. 17,19,18,17,15,18,16,17,17,16 A calculator was used to determine the p-value for this hypothesis test. The p-value was 0.005. If the level of significance was 0.01, interpret the results.

The p-value, 0.005, is less than the level of significance, 0.01. Reject the null hypothesis that the mean age is 18. There is sufficient evidence to conclude that the mean age is under 18. Since the p-value, 0.005 is less than the level of significance, α=0.01, The null hypothesis should be rejected. Since the null hypothesis is rejected, we choose the alternative hypothesis--there is sufficient evidence to suggest the mean age is less than 18.

A hospital administrator noticed a large change in the weight of babies being born at the hospital in the last few months. She randomly selected 10 babies born in the last month. Historically, babies at the hospital have had a mean weight of 7.63 pounds and a standard deviation of 1.02 pounds. To test the weights, the administrator decided to use the historical standard deviation as the population standard deviation. The hospital administrator conducts a one-mean hypothesis at the 5% significance level, to test if change in the weight of babies being born at the hospital in the last few months from 7.63 pounds. (a) H0:μ=7.63; Ha:μ≠7.63, which is a two-tailed test. (b) z0=−1.33, p-value is = 0.18 (c) Which of the following are appropriate conclusions for this hypothesis test? Select all that apply.

We fail to reject H0. At the 5% significance level, the data do not provide sufficient evidence to conclude that the average weight of babies at this hospital changed from the historical average of 7.63 pounds. In this case, p=0.18 and α=0.05, so p>α. Therefore, we fail to reject the null hypothesis. This means that at the 5% significance level, the test results are not statistically significant, and do not provide evidence against the null hypothesis. The hospital administrator cannot conclude that at the 5% significance level that the average weight of babies at this hospital changed from the historical average of 7.63 pounds.

Caden is an editor for a local newspaper. The newspaper's editor-in-chief asked Caden to determine whether the average distance a reader lives from the newspaper's headquarters is greater than 25 miles. He sent a survey to a random selection of the newspaper's readers and received 10 responses. Using the results from the current survey and several previous surveys, Caden decided to assume that the population standard deviation for the distance the newspaper's readers live from its headquarters is 10.3 miles. Caden conducts a one-mean hypothesis at the 5% significance level, to test if the average distance a reader lives from the newspaper's headquarters is greater than 25 miles. (a) H0:μ=25; Ha:μ>25, which is a right-tailed test. (b) z0=1.04, p-value is = 0.15 (c) Which of the following are appropriate conclusions for this hypothesis test? Select all that apply.

We fail to reject H0. At the 5% significance level, the data do not provide sufficient evidence to conclude that the average distance a reader lives from the newspaper's headquarters is greater than 25 miles. In this case, p=0.15 and α=0.05, so p>α. Therefore, we fail to reject the null hypothesis. This means that at the 5% significance level, the test results are not statistically significant, and do not provide evidence against the null hypothesis. Caden cannot conclude that at the 5% significance level that the average distance a reader lives from the newspaper's headquarters is greater than 25 miles.

A government official wanted to determine if the total income-tax revenue in a region had dropped since last year. The official surveyed 10 people in the region. Instead of using the standard deviation from the survey, the official decided to use the census data for the country to assume that the population standard deviation of income is $23.8 thousand. Last year, the mean regional income was $58.5 thousand. The government official conducts a one-mean hypothesis at the 5% significance level, to test if the average regional income decreased from the previous year of $58.5 thousand. (a) H0:μ=$58.5; Ha:μ<$58.5, which is a left-tailed test. (b) z0=−0.78, p-value is = 0.22. (c) Which of the following are appropriate conclusions for this hypothesis test? Select all that apply.

We fail to reject H0. At the 5% significance level, the data do not provide sufficient evidence to conclude that the average regional income decreased from the previous year of $58.5 thousand. In this case, p=0.22 and α=0.05, so p>α. Therefore, we fail to reject the null hypothesis. This means that at the 5% significance level, the test results are not statistically significant, and do not provide evidence against the null hypothesis. The government official cannot conclude that at the 5% significance level that the average regional income decreased from the previous year of $58.5 thousand.

A hospital administrator noticed a large change in the weight of babies being born at the hospital in the last few months. She randomly selected 10 babies born in the last month and found the weights given below (in pounds). Historically, babies at the hospital have had a mean weight of 7.58 pounds and a standard deviation of 1.05 pounds. To test the weights, the administrator decided to use the historical standard deviation as the population standard deviation. The hospital administrator conducts a one-mean hypothesis at the 5% significance level, to test if there has been a change in the mean weight of babies being born at the hospital in the last few months from 7.58 pounds. (a) H0:μ=7.58; Ha:μ≠7.58, which is a two-tailed test (b) z0=−1.54, p-value is = 0.12. (c) Which of the following are appropriate conclusions for this hypothesis test? Select all that apply.

We fail to reject H0. At the 5% significance level, the data do not provide sufficient evidence to conclude that the average weight of babies at this hospital changed from the historical average of 7.58 pounds. In this case, p=0.12 and α=0.05, so p>α. Therefore, we fail to reject the null hypothesis. This means that at the 5% significance level, the test results are not statistically significant, and do not provide evidence against the null hypothesis. The hospital administrator cannot conclude that at the 5% significance level that the average weight of babies at this hospital changed from the historical average of 7.58 pounds.

A soda company sells one of its soft drinks in 12 ounce cans. In order to ensure that every can has at least 12 ounces in it, the machines in the factory are set to fill each can with an average of 12.1 ounces of soda. Every week, a quality-control technician tests 10 cans to make sure that the average amount of soda in the cans is still 12.1 ounces. If the conclusion of the test is that the number of ounces of soda in the cans is different from 12.1, the technician will declare that the process is out of control, and the machine will be stopped and calibrated. From previous tests, the technician found that the standard deviation of the number of ounces of soda in the cans is approximately 0.01. The number of ounces of soda in the 10 cans that were tested this week is given below. Assume that the standard deviation from the technician's previous tests is the population standard deviation. Use a TI-83, TI-83 Plus, or TI-84 calculator to test whether filling process is out of control. Identify the test statistic, z, from the calculator output, rounding to two decimal places. 12.12 12.09 12.11 12.09 12.13 12.1 12.07 12.06 12.11 12.13

z=0.32

Give the numerical value of the parameter p in the following binomial distribution scenario.A softball pitcher has a 0.721 probability of throwing a strike for each pitch and a 0.279 probability of throwing a ball. If the softball pitcher throws 19 pitches, we want to know the probability that more than 15 of them are strikes.Consider strikes as successes in the binomial distribution. Do not include p= in your answer.

p= 0.721 The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is a strike, so p=0.721.

Identify the parameters p and n in the following binomial distribution scenario. A softball pitcher has a 0.579 probability of throwing a strike for each pitch and a 0.421 probability of throwing a ball. If the softball pitcher throws 27 pitches, we want to know the probability that exactly 17 of them are strikes. (Consider strikes as successes in the binomial distribution.)

p=0.579 n=27 The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is a strike, so p=0.579. The total number of trials, or pitches, is n=27.

A government official wanted to determine if the total income-tax revenue in a region had dropped since last year. The official surveyed 10 people in the region. Instead of using the standard deviation from the survey, the official decided to use the census data for the country to assume that the population standard deviation of income is $23.8 thousand. Last year, the mean regional income was $58.5 thousand. The government official conducts a one-mean hypothesis at the 5% significance level, to test if the average regional income decreased from the previous year of $58.5 thousand. (a) H0:μ=$58.5; Ha:μ<$58.5, which is a left-tailed test. (b) Income (in thousands) of individuals in the region is given below. Use a TI-83, TI-83 Plus, or TI-84 calculator to test whether the mean is less than $58.5 thousand. Identify the test statistic, z, and p-value from the calculator output, rounding to two decimal places. 33.2 63.7 48.8 88.6 59.4 21.9 53.7 30.2 39 87.9

test statistic (z)=-.78 p-value (p)=.22

A soda company sells one of its soft drinks in 12 ounce cans. In order to ensure that every can has at least 12 ounces in it, the machines in the factory are set to fill each can with an average of 12.1 ounces of soda. Every week, a quality-control technician tests 10 cans to make sure that the average amount of soda in the cans is still 12.1 ounces. If the conclusion of the test is that the number of ounces of soda in the cans is different from 12.1, the technician will declare that the process is out of control, and the machine will be stopped and calibrated. From previous tests, the technician found that the standard deviation of the number of ounces of soda in the cans is approximately 0.01. The number of ounces of soda in the 10 cans that were tested this week is given below. Assume that the standard deviation from the technician's previous tests is the population standard deviation. Use a TI-83, TI-83 Plus, or TI-84 calculator to test whether filling process is out of control. Identify the p-value from the calculator output, rounding to three decimal places. 12.15 12.04 12.11 12.09 12.13 12.1 12.07 12.06 12.14 12.14

p value= 0.343