A new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in thestore on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements areplayed. They found that for 13 days following no advertisements, the mean was 23.8 purchasing customers with a standard deviation of 1.4 customers. On 8 daysfollowing advertising, the mean was 24.4 purchasing customers with a standard deviation of 2.2 customers. Test the claim, at the 0.10 level, that the mean number ofcustomers who make a purchase in the store is lower for days following no advertising compared to days following advertising. Assume that both populations areapproximately normal and that the population variances are equal. Let days following no advertisements be Population 1 and let days following advertising bePopulation 2.Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.DE ANDaugustion lanUZDDEJME MUSULMA2255-2005²TEACHE102-000-000----D

The sample size for no advertising group is 13, sample mean is 23.8 and sample standard deviation is…

Answered: A new small business wants to know if…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements are played. They found that for 7 days following no advertisements, the mean was 22.1 purchasing customers with a standard deviation of 1.2 customers. On 10 days following advertising, the mean was 24.1 purchasing customers with a standard deviation of 1.6 customers. Test the claim, at the 0.05 level, that the mean number of customers who make a purchase in the store is lower for days following no advertising compared to days following advertising. Assume that both populations are approximately normal and that the population variances are equal. Let days following no advertisements be Population 1 and let days following advertising be Population 2. Step 2 of 3: Compute…
One year Terry had the lowest ERA (earned-run average, mean number of runs yielded per nine innings pitched) of any male pitcher at his school, with an ERA of 2.98. Also, Karla had the lowest ERA of any female pitcher at the school with an ERA of 3.37. For the males, the mean ERA was 4.456 and the standard deviation was 0.557. For the females, the mean ERA was 4.166 and the standard deviation was 0.635. Find their respective z-scores. Which player had the better year relative to their peers, Terry or Karla? (Note: In general, the lower the ERA, the better the pitcher.) Terry had an ERA with a z-score of Karla had an ERA with a z-score of (Round to two decimal places as needed.) Which player had a better year in comparison with their peers? OA. Karla had a better year because of a lower z-score. B. Terry had a better year because of a lower z-score. OC. Terry had a better year because of a higher z-score. OD. Karla had a better year because of a higher z-score.
A new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements are played. They found that for 13 days following no advertisements, the mean was 23.9 purchasing customers with a standard deviation of 1.9 customers. On 6 days following advertising, the mean was 24.7 purchasing customers with a standard deviation of 1.6 customers. Test the claim, at the 0.01 level, that the mean number of customers who make a purchase in the store is lower for days following no advertising compared to days following advertising. Assume that both populations are approximately normal and that the population variances are equal. Let days following no advertisements be Population 1 and let days following advertising be Population 2. Step 1 of 3: State the…
It seems these days that college graduates who are employed full-time work more than 40-hour weeks. Data are available that can help us decide if this is true. A survey was recently sent to a group of adults selected at random. There were 15 respondents who were college graduates employed full-time. The mean number of hours worked per week by these 15 respondents was 43 hours, with a standard deviation of 8 hours. Assume that the population of hours worked per week by college graduates employed full-time is normally distributed with mean μ. Can we conclude that μ is greater than 40 hours? Use the 0.05 level of significance. State the null hypothesis H0 and the alternative hypothesis H1 Find the t-test df & Solve the t-test value Find the critical value Can we conclude, at the 0.05 level of significance, that the mean number of hours worked per week by college graduates is greater than 40 hours? Yes or No?
A psychology student conducted a study on using a chief executive officer's facial structure to predict a firm's financial performance. The facial width-to-height ratio (WHR) for each in a sample of 47 CEOs at publicly traded firms was determined. The sample resulted in x=2.23 and s=0.35. The student wants to predict the financial performance of a firm based on the value of the true mean facial WHR of CEOs. The student wants to use the value of μ=2.1. Do you recommend he use this value? Conduct a test of hypothesis for μ to help you answer the question. Specify all the elements of the test, including H0, Ha, test statistic, p-value, and your conclusion. Test at α=0.01.
April and Heather began arguing about who did better on their tests, but they couldn't decide who did better given that they took different tests. April took a test in Science and earned a 74.4, and Heather took a test in English and earned a 69.7. Use the fact that all the students' test grades in the Science class had a mean of 71.3 and a standard deviation of 8.9, and all the students' test grades in English had a mean of 60.5 and a standard deviation of 8.5 to answer the following questions. a) Calculate the z-score for April's test grade. b) Calculate the z-score for Heather's test grade. c) Which person did relatively better? April Heather O They did equally well. Submit Question logi $22 *3 F4 F8 7. 8. %4
A Statistics instructor wanted to compare the scores on the same final exam given to students at a local community college and local university. The Statistics instructor feels that students at the community college would be better prepared and have higher exam scores on the final exam. After grading the final exams from both institutions, the Statistics instructor found that for a sample of 36 community college students, the sample had a mean score of 84 with a standard deviation of 3.4. A sample of 45 university students had a mean score 79 with a standard deviation of 2.6. Construct a 95% confidence interval for the difference between population mean exam scores. Round your final results to one decimal place. _____________ ≤ ??? − ????? ≤ ______________
A new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements are played. They found that for 11 days following no advertisements, the mean was 16.6 purchasing customers with a standard deviation of 1.1 customers. On 10 days following advertising, the mean was 17.7 purchasing customers with a standard deviation of 1.4 customers. Test the claim, at the 0.02 level, that the mean number of customers who make a purchase in the store is lower for days following no advertising compared to days following advertising. Assume that both populations are approximately normal and that the population variances are equal. Let days following no advertisements be Population 1 and let days following advertising be Population 2. Step 2 of 3 : Compute…
One year Terry had the lowest ERA (earned-run average, mean number of runs yielded per nine innings pitched) of any male pitcher at his school, with an ERA of 3.02. Also, Karen had the lowest ERA of any female pitcher at the school with an ERA of 3.24. For the males, the mean ERA was 4.364 and the standard deviation was 0.639. For the females, the mean ERA was 3.801 and the standard deviation was 0.959 Find their respective z-scores. Which player had the better year relative to their peers, Terry or Karen? (Note: In general, the lower the ERA, the better the pitcher.) Terry had an ERA with a z-score of Karen had an ERA with a z-score of (Round to two decimal places as needed) Which player had a better year in comparison with their peers? OA. Terry had a better year because of a higher z-score. OB. Karen had a better year because of a lower z-score. OC. Terry had a better year because of a lower z-score. O D. Karen had a better year because of a higher z-score. GIL
A nationwide job recruiting firm wants to compare the annual incomes for childcare workers in Utah and Oregon. Due to recent trends in the childcare industry, the firm suspects that the mean annual income of childcare workers in Utah is less than the mean annual income of childcare workers in Oregon. To see if this is true, the firm selected a random sample of 20 childcare workers from Utah and an independent random sample of 20 childcare workers from Oregon and asked them to report their mean annual income. The data obtained were as follows. Annual income in dollars 27060 , 42320 , 36445 , 37895, 44564 , 35325 , 30313 , 28698, 33706 , 48365 , 26946 , 25063 , 28339 , 29265 , 40051 , 37026 , 40150 , 42753 , 27820, 39471 Utah 29769 , 48859 , 32480 , 36288,45102 , 46219 ,51098 , 45516 , 38834 , 44659 , 40755 , 38006 , 39947 , 29801 , 32200, 38731 , 41370 , 28059 , 33753 , 35271 Oregon Send data to calc... v Send data to Excel The population standard deviation for the annual incomes of…
It is known that average house prices today is $371,040 higher compared to 10 years ago and median in the current time also $382,000 higher than house prices 10 years ago. Provide an interpretation of the change of mean and median based on the information above. Compare and contrast these values and comment on the distribution of the increases in house prices.
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.

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