The monthly advertising expenditure of the company is normally distributed with a standard deviation of $225. If a sample of 41 randomly selected months yields a mean advertising expenditure of $1650 monthly, what is a 94% confidence interval for the mean of the company’s monthly advertising expenditure?
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The monthly advertising expenditure of the company is
standard deviation of $225. If a sample of 41 randomly selected months yields a mean
advertising expenditure of $1650 monthly, what is a 94% confidence interval for the mean
of the company’s monthly advertising expenditure?
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- A study of 96 bolts of carpet showed that their average length was 77.5 yards. The standard deviation of the population is 1.6 yards. Which of the following is the 90% confidence interval for the mean length per bolt of carpet?A 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 study of 40 English professors showed that they spend, on average, 12.6 minutes correcting a student's term paper. The standard deviation was 2.5 minutes. Construct the 99% confidence interval of the mean for all term papers.
- The standard deviation, S, for the annual medical cost of 24 households was found to equal $1689 find a 95% confidence interval for the standard deviation of the annual medical cost for all households. Assume medical costs are normally distributed.In a recent sample of 89 used cars sales costs, the sample mean was $6,225 with a standard deviation of $3,151. Assume the underlying distribution is approximately normal. Which distribution should you use for this problem? Construct a 95% confidence interval for the population mean cost of a used car.The 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.)
- 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 12people who buy insurance from Company A, the mean cost is $153 per month with a standard deviation of $16. For 15 randomly selected customers of Company B, you find that they pay a mean of $160 per month with a standard deviation of $10. 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 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. H0: μ1=μ2 Ha: μ1_____μ2 Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places Step 3 of…Pilots who cannot maintain regular sleep hours due to their work schedule often suffer from insomnia. A recent study on sleeping patterns of pilots focused on quantifying deviations from regular sleep hours. A random sample of 28 commercial airline pilots was interviewed, and the pilots in the sample reported the time at which they went to sleep on their most recent working day. The study gave the sample mean and standard deviation of the times reported by pilots, with these times measured in hours after midnight. (Thus, if the pilot reported going to sleep at p.m., the measurement was -1.) The sample mean was 0.8 hours, and the standard deviation was 1.6 hours. Assume that the sample is drawn from a normally distributed population. Find a 90% confidence interval for the population standard deviation, that is, the standard deviation of the time (hours after midnight) at which pilots go to sleep on their work days. Then give its lower limit and upper limit.Beer Drinking. The mean annual consumption of beer per person in the US is 22.0 gallons . A random sample of 300 Washington D.C. residents yielded a mean annual beer consumption of 27.8 gallons. At the 10% significance level, do the data provide sufficient evidence to conclude that the mean annual consumption of beer per person for the nation’s capital differs from the national mean? Assume that the standard deviation of annual beer consumption for Washington D.C. residents is 55 gallons.
- A large financial company employs a human resources mamanger who is in charge of employe benefits. The manger wishes to estimate the average dental expenses per employee for the company. She selects a random sample of 60 employee records for the past year and determines that the sample mean of $492. Moreover, it is known from past studies that the population standard deviation for annual dental expenses is $74. Use this data to calculate a 95% cofidence interval for the mean expenses of all employees.Suppose we wish to calculate a 90% confidence interval for the average amount spent on books by freshmen in their first year at a major university. The interval is to have a margin of error of $2. Assume that the amount spent on books by freshmen has a normal distribution with a standard deviation of $30. How many observations are required to achieve this margin of error?In a study of birth order and intelligence, IQ tests were given to 18- and 19-year-old men to estimate the size of the difference, if any, between the mean IQs of firstborn sons and secondborn sons. The following data for 10 firstborn sons and 10 secondborn sons are consistent with the means and standard deviations reported in the article. It is reasonable to assume that the samples come from populations that are approximately normal. Can you conclude that the mean IQ of firstborn sons is greater than the mean IQ of secondborn sons? Let μ1 denote the mean IQ of firstborn sons and μ2 denote the mean IQ of secondborn sons. Use the α = 0.01 level and the P-value method with the table. Firstborn 128 101 128 112 121 105 122 98 106 108 Secondborn 121 125 110 107 114 93 80 94 91 83 Part(a): State the appropriate null and alternate hypotheses. H0: H1: This is a _____…
