stats chapter 12

Given two dependent random samples with the following results: Population 13636484833332020313131311919Population 22626454538382828171739392828 Use this data to find the 90%90% confidence interval for the true difference between the population means. Let d=(Population 1 entry)−(Population 2 entry)d=(Population 1 entry)−(Population 2 entry). Assume that both populations are normally distributed. Copy Data Step 1 of 4: Find the mean of the paired differences, x‾dx‾d. Round your answer to one decimal place. Step 2 of 4: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places. Step 3 of 4: Find the standard deviation of the paired differences to be used in constructing the confidence interval. Round your answer to one decimal place.

-0.4 1.943 9.5

A researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances for SUVs equipped with the tires. The mean braking distance for SUVs equipped with tires made with compound 1 is 4141 feet, with a population standard deviation of 12.112.1. The mean braking distance for SUVs equipped with tires made with compound 2 is 4646 feet, with a population standard deviation of 9.09.0. Suppose that a sample of 6666 braking tests are performed for each compound. Using these results, test the claim that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used. Let μ1μ1 be the true mean braking distance corresponding to compound 1 and μ2μ2 be the true mean braking distance corresponding to compound 2. Use the 0.050.05 level of significance. Step 3 of 5 : Find the p-value associated with the test statistic. Round your answer to four decimal places. Step 4 of 5 : Make the decision for the hypothesis test.

0.0036 reject null

In studying his campaign plans, Mr. Singleton wishes to estimate the difference between men's and women's views regarding his appeal as a candidate. He asks his campaign manager to take two random independent samples and find the 98%98% confidence interval for the difference. A random sample of 780780 male voters and 696696 female voters was taken. 207207 men and 285285 women favored Mr. Singleton as a candidate. Find this confidence interval. Step 1 of 4 : Find the values of the two sample proportions, pˆ1p^1 and pˆ2p^2. Round your answers to three decimal places. Step 2 of 4 : Find the critical value that should be used in constructing the confidence interval. Step 3 of 4 : Find the value of the standard error. Round your answer to three decimal places. Step 4 of 4 : Construct the 98%98% confidence interval. Round your answers to three decimal places.

0.265, 0.409 2.33 0.024 Lower endpoint: −0.200, Upper endpoint: −0.088

A telephone service representative believes that the proportion of customers completely satisfied with their local telephone service is different between the Northeast and the Midwest. The representative's belief is based on the results of a survey. The survey included a random sample of 920920 northeastern residents and 840840 midwestern residents. 54%54% of the northeastern residents and 44%44% of the midwestern residents reported that they were completely satisfied with their local telephone service. Find the 90%90% confidence interval for the difference in two proportions. Step 1 of 3 : Find the critical value that should be used in constructing the confidence interval. Step 2 of 3 : Find the value of the standard error. Round your answer to three decimal places. Step 3 of 3 : Construct the 90%90% confidence interval. Round your answers to three decimal places.

1.64485 0.024 0.061, 0.139

A researcher compares the effectiveness of two different instructional methods for teaching pharmacology. A sample of 5151 students using Method 1 produces a testing average of 81.681.6. A sample of 7676 students using Method 2 produces a testing average of 76.476.4. Assume the standard deviation is known to be 12.2412.24 for Method 1 and 11.1911.19 for Method 2. Determine the 90%90% confidence interval for the true difference between testing averages for students using Method 1and students using Method 2. Step 1 of 2: Find the critical value that should be used in constructing the confidence interval. Step 2 of 2: Construct the 90%90% confidence interval. Round your answers to one decimal place.

1.645 1.67 to 8.72

Given two independent random samples with the following results: n1=11x‾1=80s1=28 n2=9x‾2=99s2=18 Use this data to find the 90%90% confidence interval for the true difference between the population means. Assume that the population variances are equal and that the two populations are normally distributed. Copy Data Step 1 of 3 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places. Step 2 of 3 : Find the standard error of the sampling distribution to be used in constructing the confidence interval. Round your answer to the nearest whole number. Step 3 of 3 : Construct the 90%90% confidence interval. Round your answers to the nearest whole number.

1.734 10.788 -38 to 0

A researcher compares the effectiveness of two different instructional methods for teaching anatomy. A sample of 144144 students using Method 1 produces a testing average of 88.588.5. A sample of 181181 students using Method 2 produces a testing average of 72.272.2. Assume the standard deviation is known to be 13.7413.74 for Method 1 and 16.5416.54 for Method 2. Determine the 90%90% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2. Step 2 of 2 : Construct the 90%90% confidence interval. Round your answers to one decimal place.

13.5 19.1

Eddie Clauer sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet. Random samples of sales receipts were studied for mail-order sales and internet sales, with the total purchase being recorded for each sale. A random sample of 55 sales receipts for mail-order sales results in a mean sale amount of $82.40$⁢82.40 with a standard deviation of $29.25$⁢29.25. A random sample of 1414 sales receipts for internet sales results in a mean sale amount of $75.50$⁢75.50 with a standard deviation of $18.75$⁢18.75. Using this data, find the 90%90% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases. Assume that the population variances are not equal and that the two populations are normally distributed. Step 1 of 3 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places. Step 2 of 3 : Find the standard error of the sampling distribution to be used in constructing the confidence interval. Round your answer to two decimal places. Step 3 of 3 : Construct the 90%90% confidence interval. Round your answers to two decimal places.

2.132 14.01 Lower endpoint: −22.97−22.97, Upper endpoint: 36.77

A systems analyst tests a new algorithm designed to work faster than the currently-used algorithm. Each algorithm is applied to a group of 8686 sample problems. The new algorithm completes the sample problems with a mean time of 18.1018.10 hours. The current algorithm completes the sample problems with a mean time of 19.0319.03 hours. Assume the population standard deviation for the new algorithm is 4.6394.639 hours, while the current algorithm has a population standard deviation of 5.1035.103 hours. Conduct a hypothesis test at the 0.050.05 level of significance of the claim that the new algorithm has a lower mean completion time than the current algorithm. Let μ1μ1 be the true mean completion time for the new algorithm and μ2μ2 be the true mean completion time for the current algorithm. Step 1 of 4 : State the null and alternative hypotheses for the test. Step 2 of 4 : Compute the value of the test statistic. Round your answer to two decimal places. Step 3 of 4 : Determine the decision rule for rejecting the null hypothesis H0H0. Round the numerical portion of your answer to three decimal places Step 4 of 4 : Make the decision for the hypothesis test..

= < -1.25 Reject H0 if z<−1.645 fail to reject null

A researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances for SUVs equipped with the tires. The mean braking distance for SUVs equipped with tires made with compound 1 is 7272 feet, with a population standard deviation of 13.413.4. The mean braking distance for SUVs equipped with tires made with compound 2 is 7575 feet, with a population standard deviation of 5.25.2. Suppose that a sample of 7070 braking tests are performed for each compound. Using these results, test the claim that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used. Let μ1μ1 be the true mean braking distance corresponding to compound 1 and μ2μ2 be the true mean braking distance corresponding to compound 2. Use the 0.050.05 level of significance. Step 1 of 5 : State the null and alternative hypotheses for the test. Step 2 of 5 : Compute the value of the test statistic. Round your answer to two decimal places. Step 3 of 5 : Find the p-value associated with the test statistic. Round your answer to four decimal places. Step 4 of 5 : Make the decision for the hypothesis test. Step 5 of 5 : State the conclusion of the hypothesis test.

= < -1.75 0.0401 reject null there is sufficient evidence

A study was designed to compare the attitudes of two groups of nursing students towards computers. Group 1 had previously taken a statistical methods course that involved significant computer interaction. Group 2 had taken a statistic methods course that did not use computers. The students' attitudes were measured by administering the Computer Anxiety Rating Scale (CARS). A random sample of 1010 nursing students from Group 1 resulted in a mean score of 60.760.7 with a standard deviation of 6.96.9. A random sample of 1515 nursing students from Group 2 resulted in a mean score of 66.366.3 with a standard deviation of 5.85.8. Can you conclude that the mean score for Group 1 is significantly lower than the mean score for Group 2? Let μ1μ1 represent the mean score for Group 1 and μ2μ2 represent the mean score for Group 2. Use a significance level of α=0.1α=0.1 for the test. Assume that the population variances are equal and that the two populations are normally distributed. Step 1 of 4 : State the null and alternative hypotheses for the test. Step 2 of 4 : Compute the value of the t test statistic. Round your answer to three decimal places. Step 3 of 4 : Determine the decision rule for rejecting the null hypothesis H0H0. Round your answer to three decimal places. Step 4 of 4 : State the test's conclusion.

= < -2.193 Reject H0 if t<−1.319 reject null

A researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances for SUVs equipped with the tires. The mean braking distance for SUVs equipped with tires made with compound 1 is 6969 feet, with a population standard deviation of 10.910.9. The mean braking distance for SUVs equipped with tires made with compound 2 is 7777 feet, with a population standard deviation of 14.814.8. Suppose that a sample of 3434 braking tests are performed for each compound. Using these results, test the claim that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used. Let μ1μ1 be the true mean braking distance corresponding to compound 1 and μ2μ2 be the true mean braking distance corresponding to compound 2. Use the 0.010.01 level of significance. Step 1 of 5 : State the null and alternative hypotheses for the test. Step 2 of 5 : Compute the value of the test statistic. Round your answer to two decimal places.

= < -2.554

A systems analyst tests a new algorithm designed to work faster than the currently-used algorithm. Each algorithm is applied to a group of 7676 sample problems. The new algorithm completes the sample problems with a mean time of 15.9015.90 hours. The current algorithm completes the sample problems with a mean time of 18.2918.29 hours. Assume the population standard deviation for the new algorithm is 5.5605.560 hours, while the current algorithm has a population standard deviation of 4.0264.026 hours. Conduct a hypothesis test at the 0.10.1 level of significance of the claim that the new algorithm has a lower mean completion time than the current algorithm. Let μ1μ1 be the true mean completion time for the new algorithm and μ2μ2 be the true mean completion time for the current algorithm. Step 1 of 4 : State the null and alternative hypotheses for the test. Step 2 of 4 : Compute the value of the test statistic. Round your answer to two decimal places. Step 3 of 4 : Determine the decision rule for rejecting the null hypothesis H0H0. Round the numerical portion of your answer to two decimal places. Step 4 of 4 : Make the decision for the hypothesis test.

= < -3.0352 z< -1.28 reject null

Two teaching methods and their effects on science test scores are being reviewed. A random sample of 1111 students, taught in traditional lab sessions, had a mean test score of 73.773.7 with a standard deviation of 33. A random sample of 55 students, taught using interactive simulation software, had a mean test score of 85.385.3 with a standard deviation of 5.85.8. Do these results support the claim that the mean science test score is lower for students taught in traditional lab sessions than it is for students taught using interactive simulation software? Let μ1μ1 be the mean test score for the students taught in traditional lab sessions and μ2μ2 be the mean test score for students taught using interactive simulation software. Use a significance level of α=0.05α=0.05 for the test. Assume that the population variances are equal and that the two populations are normally distributed. Step 1 of 4 : State the null and alternative hypotheses for the test. Step 2 of 4 : Compute the value of the t test statistic. Round your answer to three decimal places. Step 3 of 4 : Determine the decision rule for rejecting the null hypothesis H0H0. Round your answer to three decimal places. Step 4 of 4 : State the test's conclusion.

= < -5.370 Reject H0 if t<−1.761 reject null hypothesis

A manufacturer claims that the calling range (in feet) of its 900-MHz cordless telephone is greater than that of its leading competitor. A sample of 1717 phones from the manufacturer had a mean range of 11001100 feet with a standard deviation of 3333 feet. A sample of 1010 similar phones from its competitor had a mean range of 10901090 feet with a standard deviation of 4242 feet. Do the results support the manufacturer's claim? Let μ1μ1 be the true mean range of the manufacturer's cordless telephone and μ2μ2 be the true mean range of the competitor's cordless telephone. Use a significance level of α=0.05α=0.05 for the test. Assume that the population variances are equal and that the two populations are normally distributed. Step 1 of 4 : State the null and alternative hypotheses for the test. Step 2 of 4 : Compute the value of the t test statistic. Round your answer to three decimal places. Step 3 of 4 : Determine the decision rule for rejecting the null hypothesis H0H0. Round your answer to three decimal places. Step 4 of 4 : State the test's conclusion.

= > 0.688 Reject H0 if t>1.708. fail to reject null

A salesman for a new manufacturer of cellular phones claims not only that they cost the retailer less but also that the percentage of defective cellular phones found among his products, ( p1p1 ), will be no higher than the percentage of defectives found in a competitor's line, ( p2p2 ). To test this statement, the retailer took a random sample of 100100 of the salesman's cellular phones and 125125 of the competitor's cellular phones. The retailer found that 77 of the salesman's cellular phones and 55 of the competitor's cellular phones were defective. Does the retailer have enough evidence to reject the salesman's claim? Use a significance level of α=0.1α=0.1 for the test. Step 1 of 6 : State the null and alternative hypotheses for the test. Step 2 of 6 : Find the values of the two sample proportions, pˆ1p^1 and pˆ2p^2. Round your answers to three decimal places. Step 3 of 6 : Compute the weighted estimate of p, p‾‾p‾. Round your answer to three decimal places. Step 4 of 6 Compute the value of the test statistic. Round your answer to two decimal places Step 5 of 6 : Determine the decision rule for rejecting the null hypothesis H0H0. Round the numerical portion of your answer to two decimal places. Step 6 of 6 : Make the decision for the hypothesis test.

= > pˆ1=0.070 pˆ2=0.040 0.053 1.00 Reject H0 if z>1.28 fail to reject

An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 7575 type K batteries and a sample of 4848 type Q batteries. The type K batteries have a mean voltage of 9.439.43, and the population standard deviation is known to be 0.5750.575. The type Q batteries have a mean voltage of 9.569.56, and the population standard deviation is known to be 0.7530.753. Conduct a hypothesis test for the conjecture that the mean voltage for these two types of batteries is different. Let μ1μ1 be the true mean voltage for type K batteries and μ2μ2 be the true mean voltage for type Q batteries. Use a 0.050.05 level of significance. Step 1 of 4 : State the null and alternative hypotheses for the test. Step 2 of 4 : Compute the value of the test statistic. Round your answer to two decimal places. Step 3 of 4 : Determine the decision rule for rejecting the null hypothesis H0H0. Round the numerical portion of your answer to two decimal places. Step 4 of 4 : Make the decision for the hypothesis test.

= does not = -1.02 Reject Ho if |z| > 1.96 fail to reject null

An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 4949 type K batteries and a sample of 3838 type Q batteries. The type K batteries have a mean voltage of 9.169.16, and the population standard deviation is known to be 0.6690.669. The type Q batteries have a mean voltage of 9.369.36, and the population standard deviation is known to be 0.1940.194. Conduct a hypothesis test for the conjecture that the mean voltage for these two types of batteries is different. Let μ1μ1 be the true mean voltage for type K batteries and μ2μ2 be the true mean voltage for type Q batteries. Use a 0.010.01 level of significance. Step 1 of 4 : State the null and alternative hypotheses for the test Step 2 of 4 : Compute the value of the test statistic. Round your answer to two decimal places. Step 3 of 4 : Determine the decision rule for rejecting the null hypothesis H0H0. Round the numerical portion of your answer to three decimal places.

= does not = -1.988

An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 4747 type K batteries and a sample of 3131 type Q batteries. The type K batteries have a mean voltage of 9.199.19, and the population standard deviation is known to be 0.1320.132. The type Q batteries have a mean voltage of 9.519.51, and the population standard deviation is known to be 0.4920.492. Conduct a hypothesis test for the conjecture that the mean voltage for these two types of batteries is different. Let μ1μ1 be the true mean voltage for type K batteries and μ2μ2 be the true mean voltage for type Q batteries. Use a 0.020.02 level of significance. Step 3 of 4 : Determine the decision rule for rejecting the null hypothesis H0H0. Round the numerical portion of your answer to two decimal places. Step 4 of 4 : Make the decision for the hypothesis test.

Reject H0 if |z|>2.33. reject null

A researcher compares the effectiveness of two different instructional methods for teaching anatomy. A sample of 144144 students using Method 1 produces a testing average of 88.588.5. A sample of 181181 students using Method 2 produces a testing average of 72.272.2. Assume the standard deviation is known to be 13.7413.74 for Method 1 and 16.5416.54 for Method 2. Determine the 90%90% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2.

Since we are trying to determine a 90% confidence interval, it should be clear that we are looking for the z-value required to obtain an area of (1−α)=0.90 centered under the standard normal curve. Therefore, from a normal table, it can be found that the value of zα/2z for a 90% confidence interval is 1.645.

alternative hypothesis

There is a difference in average service outage time between cable and satellite subscribers.

null hypothesis

There is no difference in average service outage time between cable and satellite subscribers.

A researcher compares the effectiveness of two different instructional methods for teaching anatomy. A sample of 131131 students using Method 1 produces a testing average of 88.188.1. A sample of 142142 students using Method 2 produces a testing average of 87.187.1. Assume the standard deviation is known to be 11.4611.46 for Method 1 and 9.789.78 for Method 2. Determine the 95%95% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Step 2 of 2 : Construct the 95%95% confidence interval. Round your answers to one decimal place.

critical value 1.96 confidence interval -1.54 3.54

An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 7171 type K batteries and a sample of 5353 type Q batteries. The type K batteries have a mean voltage of 8.968.96, and the population standard deviation is known to be 0.3570.357. The type Q batteries have a mean voltage of 9.229.22, and the population standard deviation is known to be 0.6040.604. Conduct a hypothesis test for the conjecture that the mean voltage for these two types of batteries is different. Let μ1μ1 be the true mean voltage for type K batteries and μ2μ2 be the true mean voltage for type Q batteries. Use a 0.050.05 level of significance. Step 4 of 4 : Make the decision for the hypothesis test.

reject null

A researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances for SUVs equipped with the tires. The mean braking distance for SUVs equipped with tires made with compound 1 is 6363 feet, with a population standard deviation of 13.813.8. The mean braking distance for SUVs equipped with tires made with compound 2 is 6868 feet, with a population standard deviation of 10.010.0. Suppose that a sample of 7777 braking tests are performed for each compound. Using these results, test the claim that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used. Let μ1μ1 be the true mean braking distance corresponding to compound 1 and μ2μ2 be the true mean braking distance corresponding to compound 2. Use the 0.10.1 level of significance. Step 4 of 5 : Make the decision for the hypothesis test. Step 5 of 5 : State the conclusion of the hypothesis test.

reject null there is sufficient evidence