A marketing firm asked a random set of married women and married men how much they were willing to spend for jewelry as a present for their spouse. Can the firm conclude, ata = 0.05, that the two groups have a different willingness to spend? Women Men Sample size 9. 14 Sample mean $80 $115 Sample standard deviation $24 $28 A) No, because the test value -0.29 is inside the noncritical region -2.262 < t< 2.262 B) Yes, because the test value -1.17 is inside the noncritical region -2.262

A marketing firm asked a random set of married women and married men how much they were willing to spend for jewelry as a present for their spouse. Can the firm conclude, ata = 0.05, that the two groups have a different willingness to spend? Women Men Sample size 9. 14 Sample mean $80 $115 Sample standard deviation $24 $28 A) No, because the test value -0.29 is inside the noncritical region -2.262 < t< 2.262 B) Yes, because the test value -1.17 is inside the noncritical region -2.262

Name: A marketing firm asked a random set of married women and married men
Uploaded: 2024-03-20T07:07:12.000Z
Channel: Bartleby
Description: A marketing firm asked a random set of married women and married men how muchthey were willing to spend for jewelry as a present for their spouse. Can the firmconclude, ata = 0.05, that the two groups have a different willingness to spend?WomenMenSa

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

The average American consumes 86 liters of alcohol per year. Does the average college student consume more alcohol per year? A researcher surveyed 11 randomly selected college students and found that they averaged 95.8 liters of alcohol consumed per year with a standard deviation of 20 liters. What can be concluded at the the a = 0.05 level of significance? a. For this study, we should use t-test for a population mean O b. The null and alternative hypotheses would be: Ho: μ = H₁: μ > O 86 O 86 c. The test statistic (t) = d. The p-value = e. The p-value is ? a f. Based on this, we should Select an answer the null hypothesis. g. Thus, the final conclusion is that ... 1.625 (please show your answer to 3 decimal places.) (Please show your answer to 4 decimal places.) O The data suggest the population mean is not significantly more than 86 at a = 0.05, so there is statistically insignificant evidence to conclude that the population mean amount of alcohol consumed by college students is…
Workers at a certain soda drink factory collected data on the volumes (in ounces) of a simple random sample of 21 cans of the soda drink. Those volumes have a mean of 12.19 oz and a standard deviation of 0.14 oz, and they appear to be from a normally distributed population. Use the sample data to test the claim that the population of volumes has a standard deviation less than 0.15 oz. Use a 0.01 significance level. Complete parts (a) through (d) below. a. Identify the null and alternative hypotheses. Choose the correct answer below. O A. Ho: o0.15 oz H,: o0.15 oz H:o=0.15 oz O C. Ho: o=0.15 oz H,: o<0.15 oz O D. Ho: o =0.15 oz H:o#0.15 oz b. Compute the test statistic. x² = ] (Round to three decimal places as needed.) c. Find the P-value. P-value= (Round to four decimal places as needed.) O Time Remaining: 00:15:21 Next
The term "Freshman 15" refers to the claim that college students typically gain 15 lb during their freshman year at college. Assume that the amounts of weight that male college students gain during their freshman year are normally distributed with a mean of 2.7 lb and a standard deviation of 11.4 lb. Find the probability that a randomly selected male college student gains 15 lb or more during his freshman year. What does the result suggest about the claim of the "Freshman 15"? ... The percentage of freshmen that gain 15 pounds or more in their first year of university is %. (Round to one decimal place as needed.)
A sample of 9 people were surveyed and asked how much money they wanted to spend on buying a new car with a mean of 30.55 and a standard deviation of 13.62 vs. how much they actually spent on buying a new car and the mean was 31.44 and a standard Deviation of 18.34. Test the claim that people spend more money than they say they want to at a 5% significance level. Assume everything is normally distributed.
A test of mental acuity involves measuring the subject's cognitive skills. Sixty randomly selected subjects take the test and produce a mean score of 75 with a standard deviation of 6. Find the P-value At the 0.01 level of significance, test the claim that the true mean score for all subjects is more than 73.
i need C and D
We will _______ if the test statistic lands in the unlikely region. Group of answer choices A. reject the alternative hypothesis B. fail to reject the null hypothesis C. cannot determine D. reject the null hypothesis and accept the alternative hypothesis
A study was conducted for an aviation safety agency in which they surveyed thousands of passengers at airports across Europe. In this study, the "mass" for a passenger includes the weight of the passenger and all of the passenger's carry-on items, including infants without their own seats. For the 5,904 adult male passengers measured in summer, the sample mean and standard deviation for this mass were 88.8 kg and 15.9 kg, respectively. (a) Construct a 95% confidence interval (in kilograms) for the population mean mass for adult male passengers in summer. (Round your answers to three decimal places.) ___to__ kg (b) Write a sentence or two interpreting the confidence interval you found in part (a). With 95% confidence, we can estimate that the population mean mass of male passengers traveling in the summer is contained within the interval. We can estimate that 95% of the male passengers traveling in the summer in the sample have a mass that is contained within the interval.…
A random sample of 15 hourly wages for restaurant servers (including tips) was drawn from a normal population. The sample mean and sample standard deviation were $14.9 and $6.75 respectevely.Can we infer at the 5% significance level that the mean wage for restaurant servers (including tips) is greater than 12?