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 23.9 purchasing customers with a standard deviation of 1.9 customers. On 6 daysfollowing 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 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 3 of 3: Draw a conclusion and interpret the decision.

The objective of the question is to test the claim that the mean number of customers who make a…

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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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?
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…
It has been found that with the standard dosage of a drug, the change in people’s uric acid levelis normally distributed with a mean of 240 and a standard deviation of 60. What proportionof people receiving the standard dosage of this drug experience changes in uric acid levelbetween 210 and 270?
A student earned a 79 on a precalculus exam, and she earned a 92 on a statistics exam. The precalculus exam had a mean score of 72.3 with a standard deviation of 8.4. On the other hand, the statistics exam had a mean score of 87.4 with a standard deviation of 11.4. Determine which exam the student earned a better score on with respect to her peers.
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
You may need to use the appropriate technology to answer this question. Data were collected on the top 1,000 financial advisers. Company A had 239 people on the list and another company, Company B, had 121 people on the list. A sample of 16 of the advisers from Company A and 10 of the advisers from Company B showed that the advisers managed many very large accounts with a large variance in the total amount of funds managed. The standard deviation of the amount managed by advisers from Company A was s₁ = $583 million. The standard deviation of the amount managed by advisers from Company B was s₂ = $484 million. Conduct a hypothesis test at a = 0.10 to determine if there is a significant difference in the population variances for the amounts managed by the two companies. What is your conclusion about the variability in the amount of funds managed by advisers from the two firms? State the null and alternative hypotheses. 2 2 H₂:0 2 Find the value of the test statistic. (Round your answer…
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.

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