In a time-use study, twenty randomly selected managers spend a mean of 2.4 hours each day on paperwork. The standard deviation of the twenty scores is 1.3 hours. Construct the 95% confidence interval for the mean paperwork time of all managers.
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In a time-use study, twenty randomly selected managers spend a
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- A graduate student is interested in how viewing different types of scenes affects working memory. For his study, he selects a random sample of 36 adults. The subjects complete a series of working memory tests before and after walking in an urban setting. Before the walk, the mean score on the test of working memory was 9.1. After the walk, the mean score was 1.4 higher. The graduate student has no presupposed assumptions about how viewing different types of scenes affects working memory, so he formulates the null and alternative hypotheses as: H00 : μDD = 0 H11 : μDD ≠ 0 Assume that the data satisfy all of the required assumptions for a repeated-measures t test. The graduate student calculates the following statistics for his hypothesis test: Mean difference (MDD) 1.4 Estimated population standard deviation of the differences (s) 1.6 Estimated standard error of the mean differences (sMDMD) 0.2667 Degrees of freedom (df) 35 The t statistic 5.25 The critical values of t…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 7 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $16. For 12 randomly selected customers of Company B, you find that they pay a mean of $160 per month with a standard deviation of $14. Assume that both populations are approximately normal and that the population variances are equal to test Company A's claim at the 0.10 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.2. A sample of six college basketball player had an average weight of 217 pounds with a sample standard deviation of 10 pounds. Find the 95% confidence interval of the true mean weight of all basketball player.
- 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 13 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $19. For 9 randomly selected customers of Company B, you find that they pay a mean of $157 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A's claim at the 0.05 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Ho: M₁ = μ₂ Ha:M₁ •H₂A sample of 76 female workers and another sample of 48 male workers from a state produced mean weekly earnings of $743.50 for the females and $777.63 for the males. Suppose that the population standard deviations of the weekly earnings are $80.05 for the females and $88.68 for the males. The null hypothesis is that the mean weekly earnings are the same for females and males, while the alternative hypothesis is that the mean weekly earnings for females is less than the mean weekly earnings for males. Directions: • Label your answers with the correct statistical symbols. • If you use the Ti, identify which function and values you used to calculate. If you solve by hand, show all steps 2.5 The significance level for the test is 1%. What is/are the critical value(s)? 2.6. What is the value of the test statistic, rounded to three decimal places? 2.7. What is the p-value for this test, rounded to four decimal places? 2. 8. Using the p-value approach, do you reject or fail to reject the null…A learn-to-type software program claims that it can improve your typing skills. To test the claim and possibly help yourself out, you and eight of your friends decide to try the program and see what happens. Use the table below to construct an 85%85% confidence interval for the true mean change in the typing speeds for people who have completed the typing program. Let Population 1 be the typing speed before taking the program and Population 2 be the typing speed after taking the program. Round the endpoints of the interval to one decimal place, if necessary. Typing Speeds (in Words per Minute) Before After 42 33 41 54 31 50 30 48 41 31 49 52 34 54 48 49 39 48
- Saving gas Congress regulates corporate fuel economyand sets an annual gas mileage for cars. A companywith a large fleet of cars hopes to meet the 2011 goal of30.2 mpg or better for their fleet of cars. To see if thegoal is being met, they check the gasoline usage for50 company trips chosen at random, finding a mean of32.12 mpg and a standard deviation of 4.83 mpg. Is this strong evidence that they have attained their fuel econ-omy goal? a) Write appropriate hypotheses.b) Are the necessary assumptions to make inferencessatisfied?c) Describe the sampling distribution model of mean fueleconomy for samples like this.d) Find the P-value.e) Explain what the P-value means in this context.f) State an appropriate conclusion.According to the U.S. Department of Transportation’s Air Travel Consumer Report, the nation’s 12 largest airlines recorded an on-time arrival percentage of 77.4% in 2001. Of interest is to estimate the mean delay time for the 22.6% of all flights that did not arrive on time during 2013. A simple random sample of 28 late arriving flights was selected, and the mean delay time of this sample of 28 flights was 14.2 minutes, with a sample standard deviation of s= 6.4 minutes. Use this information to calculate and interpret a 98% confidence interval for the mean delay time for all flights that did not arrive on time during 2013.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 15 people who buy insurance from Company A, the mean cost is $154 per month with a standard deviation of $13. For 11 randomly selected customers of Company B, you find that they pay a mean of $159 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.02 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 study is conducted to compare the performance of students with more than one personal electronic gadget and those with only one. A number of them were taken as subjects for the study. The mean grades of these students and the standard deviations are shown below: For one gadget: The mean is 83, standard deviation is 12, and sample size is 7. For more than one gadget: The mean is 79, standard deviation is 13, and sample size is 5. Is it possible to conclude that there is no significant difference in the mean grades of the two types of students at a 95% confidence interval? Note: This is a two-tailed test. Also take note that this is finding the difference between two population means.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.Managers of an industrial plant want to determine which of two types of fuel, gas or electric, is more cost efficient (measured in cost per unit of energy). Independent random samples were taken of plants using electricity and plants using gas. These samples consisted of 10 plants using electricity, which had a mean cost per unit of $53.50 and standard deviation of $8.19 , and 11 plants using gas, which had a mean of $55.40 and standard deviation of $8.23 . Assume that the populations of costs per unit are normally distributed for each type of fuel, and assume that the variances of these populations are equal. Construct a 95% confidence interval for the difference −μ1μ2 between the mean cost per unit for plants using electricity, μ1 , and the mean cost per unit for plants using gas, μ2 . Then find the lower limit and upper limit of the 95% confidence interval. Carry your intermediate computations to at least three decimal places. Round your responses to at least…