The average starting salary of students who graduated from colleges of Business in 2009 was $48,400. A sample of 100 graduates of 2010 showed an average starting salary of $50,000. Assume the standard deviation of the population is known to be $8000. We want to determine whether or not there has been a significant increase in the starting salaries.  Question: f. The null hypothesis is to be tested at the 5% level of significance. Compute the p-value. What is the p-value? (Round to the 4 decimal places)

A magazine article claims that the average cost for a take-out lunch is $9.22. You randomly select a sample of 38 take-out restaurants, and look up the price of their most popular lunch combo. You find an average of $10.08. Assume a population standard deviation of $2.13. a. State the null and alternative hypothesis for the claim that the average cost for a take-out lunch is $9.22. Ho: ? ______9.22 Ha: ? ______ 9.22   b. The _______  hypothesis is the claim. c. The goal of the test is to _______   the claim.

On the PSAT test taken by about 1.5 million high school juniors each year, the mean score on the reading portion of the test was 46.9, with a standard deviation of 10.9. The scores are normally distributed. If a student needed a score on this portion of the test that put them in the top 5% of scores nationally to qualify for a particular scholarship, what is the minimum score they must achieve?

The average starting salary of students who graduated from colleges of Business in 2009 was $48,800. A sample of 100 graduates of 2010 showed an average starting salary of $50,000. Assume the standard deviation of the population is known to be $7000. We want to determine whether or not there has been a significant increase in the starting salaries. a. State the null and alternative hypotheses to be tested. b. Compute the test statistic. c. The null hypothesis is to be tested at the 5% level of significance. Determine the critical value for this test. d. What do you conclude from the test? e. What does the test result mean in the context of this problem?

A study reported that finger rings increase the growth of bacteria on health-care workers’ hands. Research suggests that 31 percent of health-care workers who wear rings have bacteria on one or both hands. Suppose that independent random samples of 100 health-care workers wearing rings is selected. What is the standard deviation of the sampling distribution of the sample proportions of health-care workers having bacteria on one or both hands? A. 100 B. 0.0462 C. 0.69

Sales at a fast-food restaurant average $6,000 per day. The restaurant decided to introduce an advertising campaign to increase daily sales. To determine the effectiveness of the advertising campaign, a sample of 49 days of sales were taken. They found that the average daily sales were $6,300 per day. From past history, the restaurant knew that its population standard deviation is about $1,000. The restaurant wishes to test whether sales have increased as a result of the advertising campaign. If the level of significance is 0.05, what is the decision?   Multiple Choice   Fail to reject the null hypothesis.   Reject the null hypothesis and conclude the mean is higher than $6,000 per day.   Reject the null hypothesis and conclude the mean is lower than $6,000 per day.   Reject the null hypothesis and conclude that the mean is equal to $6,000 per day.

Watching TV: The 2012 general Social Survey asked a large number of people how much time they spent watching TV each day. The mean number of hours was 3.09 with a standard deviation of 2.9. Assume that in a sample of 37 teenagers, the sample standard deviation of daily TV time is 2.2 hours, and that the population of TV watching times is normally distributed. Can you conclude that the population standard deviation of TV watching times for teenagers differs from 2.9? Use the a=0.10 level of significance.

Watching TV: The 2012 general Social Survey asked a large number of people how much time they spent watching TV each day. The mean number of hours was 3.09 with a standard deviation of 2.73. Assume that in a sample of 32 teenagers, the sample standard deviation of daily TV time is 3.2 hours, and that the population of TV watching times is normally distributed. Can you conclude that the population standard deviation of TV watching times for teenagers is greater than 2.73 ? Use the a= 0.01 level of significance. Part 1 of 5 State the appropriate null and alternate hypotheses. Ho:0 2.73 H1:0 > . 2.73 This hypothesis test is a right-tailed test. Part: 1/5 Part 2 of 5 Find the critical value. Round the answer to three decimal places. For a-0.01, the critical value is 岛国

Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 1313 people who buy insurance from Company A, the mean cost is $151$⁢151 per month with a standard deviation of $16$⁢16. For 99 randomly selected customers of Company B, you find that they pay a mean of $158$⁢158 per month with a standard deviation of $19$⁢19. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.050.05 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3:  Compute the value of the test statistic. Round your answer to three decimal places.

A Two-Sample Hypothesis TestThe mean number of English courses taken in a two-year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 29 males and 16 females. The males took an average of three English courses with a standard deviation of 0.8. The females took an average of four English courses with a standard deviation of 1.0. Are the means statistically the same? (Assume a 5% level of significance.)Useful tools:Normal Distribution Calculatort-Distribution Calculator 1. Which of the following null and alternative hypotheses match this scenario? A---H0:Δμ=0H0:Δμ=0Ha:Δμ≠0Ha:Δμ≠0 B---H0:Δμ≠0H0:Δμ≠0Ha:Δμ=0Ha:Δμ=0 C---H0:Δμ=0H0:Δμ=0Ha:Δμ>0 2. Which type of test should be applied? A---The alternative hypothesis indicates a right-tailed test. B---The alternative hypothesis indicates a left-tailed test. C---The alternative hypothesis indicates a two-tailed test. 3. Which type of distribution should be…