A chain of taco restaurants claims that the population mean of the wait times in their drive-thru for all customers is 3.14 minutes. You work for a competitor and you want to test that claim. To do so, you select a random sample of 40 of the chain's drive-thru customers and record the wait time in the drive-thru for each. Assume it is known that the population standard deviation of the wait times in the drive-thru for the taco chain's restaurants is 2.79 minutes. Based on your sample, follow the steps below to construct a 95% confidence interval for the population mean of the wait times in the drive-thru for all customers. Then state whether the confidence interval you construct contradicts the restaurant chain's claim. (If necessary, consult a list of formulas.) (a) Click on "Take Sample" to see the results from your random sample of 40 customers. (b) (c) Take Sample Sample size: 0 Point estimate: 0 Population standard deviation: 0 Critical value: Compute 0.00 Number of customers 0.00 40 Enter the values of the sample size, the point estimate for the population mean, the population standard deviation, and the critical value you need for your 95% confidence interval. (Choose the correct critical value from the table of critical values provided.) When you are done, select "Compute". 2.00 Sample mean 4.00 Standard error: Margin of error: 4.26 95% confidence interval: 95% confidence interval: 5.00 Based on your sample, graph the 95% confidence interval for the population mean of the wait times in the drive-thru for all customers. • Enter the lower and upper limits on the graph to show your confidence interval. • For the point (◆), enter the restaurant chain's claim of 3.14 minutes. 6.00 Sample standard deviation 8.00 2.44 Critical values 20.005 = 2.576 20.010 =2.326 20.025 = 1.960 20.050 = 1.645 20.100 = 1.282 X 10.00 10.00 Population standard O No, the confidence interval does not contradict the claim. The restaurant chain's claim of 3.14 minutes is outside the 95% confidence interval. deviation O Yes, the confidence interval contradicts the claim. The restaurant chain's claim of 3.14 minutes is inside the 95% confidence interval. 2.79 Does the 95% confidence interval you constructed contradict the restaurant chain's claim? Choose the best answer from the choices below. O No, the confidence interval does not contradict the claim. The restaurant chain's claim of 3.14 minutes is inside the 95% confidence interval. O Yes, the confidence interval contradicts the claim. The restaurant chain's claim of 3.14 minutes is outside the 95% confidence interval.

A professor believes that, for the introductory art history classes at his university, the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. He collects data from a random sample of 250 students in evening classes and finds that they have a mean test score of 76.8. He knows the population standard deviation for the evening classes to be 7.2 points. A random sample of 200 students from morning classes results in a mean test score of 77.8. He knows the population standard deviation for the morning classes to be 1.9 points. Test his claim with a 90% level of confidence. Let students in the evening classes be Population 1 and let students in the morning classes be Population 2.   Step 2 of 3 :   Compute the value of the test statistic. Round your answer to two decimal places.

Two friends, Karen and Jodi, work different shifts for the same ambulance service. They wonder if the different shifts average different numbers of calls. Looking at past records, Karen determines from a random sample of 39 shifts that she had a mean of 4.7 calls per shift. She knows that the population standard deviation for her shift is 1.5 calls. Jodi calculates from a random sample of 31 shifts that her mean was 5.4 calls per shift. She knows that the population standard deviation for her shift is 1.2 calls. Test the claim that there is a difference between the mean numbers of calls for the two shifts at the 0.02 level of significance. Let Karen's shifts be Population 1 and let Jodi's shifts be Population 2.   Step 2 of 3 :   Compute the value of the test statistic. Round your answer to two decimal places.

A professor believes that, for the introductory art history classes at his university, the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. He collects data from a random sample of 250 students in evening classes and finds that they have a mean test score of 88.1. He knows the population standard deviation for the evening classes to be 4.2 points. A random sample of 150 students from morning classes results in a mean test score of 89.1. He knows the population standard deviation for the morning classes to be 7.7 points. Test his claim with a 95 % level of confidence. Let students in the evening classes be Population 1 and let students in the morning classes be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to two decimal places.

A professor believes that, for the introductory art history classes at his university, the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. He collects data from a random sample of 250 students in evening classes and finds that they have a mean test score of 76.8. He knows the population standard deviation for the evening classes to be 7.2 points. A random sample of 200 students from morning classes results in a mean test score of 77.8. He knows the population standard deviation for the morning classes to be 1.9 points. Test his claim with a 90% level of confidence. Let students in the evening classes be Population 1 and let students in the morning classes be Population 2. Step 1 of 3 :  State the null and alternative hypotheses for the test. Fill in the blank below. H0: μ1=μ2 Ha: μ1__μ2

In preparation for the upcoming school year, a teacher looks at raw test scores on the statewide standardized test for the students in her class. Instead of looking at the scores relative to the norms in the state, the teacher wants to understand the scores relative to the students who will be in the class. To do so, she decides to convert the test scores into z-scores relative to the mean and standard deviation of the students in the class. The mean test score in her upcoming class is 49, and the standard deviation is 20.7. The teacher now wants to identify those students who may need extra help. She decides to look at students who have z-scores below z = -2.00 Identify the test score corresponding to a z-score of below z = -2.00. Round to the nearest whole number.

Two friends, Karen and Jodi, work different shifts for the same ambulance service. They wonder if the different shifts average different numbers of calls. Looking at past records, Karen determines from a random sample of 32 shifts that she had a mean of 4.8 calls per shift. She knows that the population standard deviation for her shift is 1.4 calls. Jodi calculates from a random sample of 40 shifts that her mean was 4.2 calls per shift. She knows that the population standard deviation for her shift is 1.1 calls. Test the claim that there is a difference between the mean numbers of calls for the two shifts at the 0.01 level of significance. Let Karen's shifts be Population 1 and let Jodi's shifts be Population 2.   Step 2 of 3 :   Compute the value of the test statistic. Round your answer to two decimal places.

Russell is doing some research before buying his first house. He is looking at two different areas of the city, and he wants to know if there is a significant difference between the mean prices of homes in the two areas. For the 39 homes he samples in the first area, the mean home price is $⁢176,400. Public records indicate that home prices in the first area have a population standard deviation of $⁢32,930. For the 32 homes he samples in the second area, the mean home price is $⁢181,500. Again, public records show that home prices in the second area have a population standard deviation of $⁢23,835. Let Population 1 be homes in the first area and Population 2 be homes in the second area. Construct a 95% confidence interval for the true difference between the mean home prices in the two areas. Round the endpoints of the interval to the nearest whole number, if necessary. Lowest Endpoint =  upper Endpoint=

You are interested in seeing whether emotions impact decision making. You have three groups--a happy group, a sad group, and a neutral group. For the happy group there are 5 participants and the mean risky decision making score 5.0 with a standard deviation of 0.7.  For the sad group there are 5 participants and the mean risky decision making score 5.4 with a standard deviation of 1.1. For the neutral group there are 5 participants and the mean risky decision making score 5.8 with a standard deviation of 2.8.  The sum of squares between samples is equal to 1.6. The sum of squares within samples is equal to 38.0. You calculated the f-ratio above. The corresponding p-value for the f-ratio you calculated is p = 0.78. Is this finding significant?

A developmental psychologist is interested in studying how long babies gaze at a photograph of a human face. To test this question, she shows a picture to a sample of six babies and records the length of their gaze in seconds. The data are as follows: 12, 17, 18, 20, 20, 29. Calculate the mean, median, mode, range, variance and standard deviation for the sample.