The monthly advertising expenditure of the company is normally distributed with a standard deviation of $200. If a sample of 36 randomly selected months yields a mean advertising expenditure of $1360 monthly, what is a 94% confidence interval for the mean of the company’s monthly advertising expenditure?
Q: Suppose that shoe sizes of American women have a bell-shaped distribution with a mean of 8.47 and a…
A: Empirical rule: The probability that the observation under the normal curve lies within 1 standard…
Q: In an article in the Journal of Management, Joseph Martocchio studied and estimated the costs of…
A: (a) Obtain the probability that the average amount of paid time lost during a three-month period for…
Q: When Tallulah runs the 400 meter dash, her finishing times are normally distributed with a mean of…
A:
Q: The historical reports from two major networks showed that the mean number of commercials aired…
A: There are two independent samples which are first station and second station. We have to test…
Q: Suppose that grade point averages of undergraduate students at one university have a bell-shaped…
A: Given, grade point averages of undergraduate students at one university have a bell-shaped…
Q: Suppose that IQ scores have a bell-shaped distribution with a mean of 103 and a standard deviation…
A:
Q: According to the Center for Disease Control and Prevention (CDC), the mean life expectancy in 2015…
A:
Q: a store have a normal distribution with a mean of $33000 and a standard deviation of $3500.
A: Given: Mean, μ=33000Standard deviation,σ=3500
Q: According to the Center for Disease Control and Prevention (CDC), the mean life expectancy in 2015…
A: Given Data Population Mean,μ = 76.3 Population Standard Deviation,σ = 15.0 sample…
Q: An article in the San Jose Mercury News stated that students in the California state university…
A: Sample size (n)=43 Sample mean=5.1 Sample standard deviation (s)=1.2 Population Mean μ=4.5 We…
Q: Construct a 95% confidence interval about the mean age.
A: Given Data : Sample Size, n = 32 Sample Mean, x̄ = 37.3 sample…
Q: In a study of employee stock ownership plans, satisfaction by employees is measured and found to be…
A: Given,mean(μ)=4.89standard deviation(σ)=0.63sample size(n)=450
Q: A random sample of 51 four-year-olds attending day care centers provided a yearly tuition average of…
A:
Q: The salaries of professional baseball players are heavily skewed right with a mean of $3.2 million…
A: We have given that Sample sizes n1= 40 , n2= 35 Sample means xbar1= 3.2 , xbar2=1.9 Standard…
Q: The average length of time it takes to read a book can differ from one generation to another.…
A: Population standard deviation ()= 2.2population mean ()= 17sample mean = 16.1sample size (n) = 50It…
Q: pharmaceutical company wanted to estimate the population mean of monthly sales for their 250 sales…
A: Sample size n=40Sample mean=10000Standard deviation=1000
Q: Suppose the Board of Trustees of a wealthy private university has decided that it wants to waive…
A: Normal random variable x follows the distribution with bell shaped curve. The normal distribution is…
Q: Suppose that shoe sizes of American women have a bell-shaped distribution with a mean of 8.36 and a…
A:
Q: Suppose that IQ scores have a bell-shaped distribution with a mean of 103 and a standard deviation…
A:
Q: ppose that shoe sizes of American women have a bell-shaped distribution with a mean of 8.42 and a…
A: Given data,Mean μ=8.42sd σ=1.52P(6.9<X<9.94)=?
Q: A grocery store's receipts show that Sunday customer purchases have a skewed distribution with a…
A: Here, Given that the distribution is skewed so the probabilities, which are the properties of the…
Q: polling organization reported data from a survey of 2000 randomly selected Canadians who carry debit…
A: The following information has been provided: Hypothesized Population Mean (\mu)(μ) = 1010…
Q: The time needed for college students to complete a certain paper and pencil maze follows a Normal…
A: From the given information ,population mean u= 70sample mean x=…
Q: A researcher is conducting a study to determine the mean trough dosage of medication for a…
A: The following information is provided that is required to compute the sample size: Required…
Q: Step 3 of 3: Draw a conclusion and interpret the decision. Answer Tables E Keypad Keyboard Shortcuts…
A: Given Data : For first born children x̄1 = 24.4 σ1 = 1.5 n1 = 225 For…
Q: Consider a dataset containing the heights of 38 females who said they prefer to sit in the front of…
A: Type I Error: α=Preject H0|H0 is truth Type II Error: β=Preject Ha|Ha is true 1-α=PFaild to reject…
Q: Suppose the dealer incentive per vehicle for Honda’s Acura brand is thought to be bell-shaped and…
A: Approximately 99.7% of the observations falls within 3 standard deviations from the mean.
Q: Suppose that IQ scores have a bell-shaped distribution with a mean of 104 and a standard deviation…
A: Given, Mean, mu= 104 Standard deviation, sigma= 17 We know that, Z score= (X-mu)/sigma
Q: A recent study of 28 city residents showed that the mean of the number of years they lived at their…
A: The formula of confidence interval for the population mean is given by confidence…
Q: The salaries of professional baseball players are heavily skewed right with a mean of $3.2 million…
A: The sample standard deviations are s1=2 and s2=1.5.
Q: A parent interest group is looking at whether birth order affects scores on the ACT test. It was…
A:
Q: Suppose that shoe sizes of American women have a bell-shaped distribution with a mean of 8.47 and a…
A:
Q: Suppose that for a certain individual credit card purchases follow a normal distribution with mean…
A: The normal distribution has a property called “68-95-99 rule” that 68% of the values lie in the…
Q: According to the Center for Disease Control and Prevention (CDC), the mean life expectancy in 2015…
A: Given Data Population Mean,μ = 71.8 Population Standard Deviation,σ = 15.0 sample…
Q: A recent survey of 8 social network sites has a mean of 13.1 million visitors for a specific month.…
A: Given that, Mean = 13.1 Standard deviation = 3.1 Sample size = 8
Q: The Journal of Visual Impairment & Blindness (May–June 1997) published a study of the lifestyles of…
A: Let "X" be the number of hours of sleep for this group of students. Find z-value: Refer Standard…
Q: An article appeared in an Australian newspaper. It described a study of academic performance and…
A:
Q: When 40 people used Weight Watchers for one year, their mean weight loss was 3.0 lbs and their…
A: The test hypotheses are given below: H0:μ=0 lbsHa:μ>0 lbs From the given information, the…
The monthly advertising expenditure of the company is
standard deviation of $200. If a sample of 36 randomly selected months yields a
advertising expenditure of $1360 monthly, what is a 94% confidence interval for the mean of
the company’s monthly advertising expenditure?
Step by step
Solved in 2 steps with 2 images
- Suppose we want a 99% confidence interval for the average amount spent on books by freshmen in their first year at college. The amount spent has a normal distribution with standard deviation $31. How large should the sample be if the margin of error is to be less than $3?.A study of 100 patients with back pain reported that the mean duration of pain was 12 months. It is known that the standard deviation of duration of back pain is 6.8 months. Assuming that the duration of this problem is normally distributed in the population; determine a 95% confidence interval for the mean duration of back pain in the population.20 TV remote controls were followed over the years to find the mean lifespan of this model. The sample gave a mean of 7.2 years, and the standard deviation of 0.6 years. The distribution of the population is unknown. Construct a 90% confidence interval for the mean lifespan of these remote controls.
- Suppose that IQ scores have a bell-shaped distribution with a mean of 103 and a standard deviation of 14. Using the empirical rule, what percentage of IQ scores are at least 145?An experiment was conducted on rodents to determine if the drug Haldol influenced the amount of "rough play." Five groups were included in the experiment. Subsequent behavior in terms of mean "rough play" behaviors was recorded. The output and box plot below, display the mean differences from baseline (i.e., after the drug was administered compared to before drug administration). What are the null and alternative hypotheses of this test? Summary of Roughplay differences Scd. Dev. Haldol dese Mean Freq. -.63158419 13.290618 19 Vehicle | 0.025 mg i 0.05 mg/ 1 0.1 mg/k -12.754389 13.783537 0.2 mg/k I 18 1.1481501 20.447041 5.0499951 16.035879 20 19 -20.S5 16.431064 20 Total -5.6631968 1B.558994 96 Analysis of Variance Frob F (p-value) Source SS df KS F Test- Statistic 8999.32832 4 2249.33203 9.63 0.0000 Between groups Within groups 91 260.682623 23722.1178 Tozal 32721.4461 95 344.436275 chif (4) 4.1710 Prob>chiz = 0.303 = Bartlett's test for equal variances: Roeghpiay diferences -50 Ho: 0…A polling organization reported data from a survey of 2000 randomly selected Canadians who carry debit cards. Participants in this survey were asked what they considered the minimum purchase amount for which it would be acceptable to use a debit card. Suppose that the sample mean and standard deviation were $9.14 and $7.70, respectively. (These values are consistent with a histogram of the sample data that appears in the report.)Do these data provide convincing evidence that the mean minimum purchase amount for which Canadians consider the use of a debit card to be appropriate is less than $10? Carry out a hypothesis test with a significance level of 0.01. (Use a statistical computer package to calculate the P-value. Round your test statistic to two decimal places and your P-value to four decimal places.) t = P-value = State your conclusion. Reject H0. We do not have convincing evidence that the mean minimum purchase amount for which Canadians consider it…
- In a school district, all sixth grade students take the same standardized test. The superintendant of the school district takes a random sample of 30 scores from all of the students who took the test. She sees that the mean score is 169 with a standard deviation of 7.0674. The superintendant wants to know if the standard deviation has changed this year. Previously, the population standard deviation was 13. Is there evidence that the standard deviation of test scores has decreased at the α=0.005 level? Assume the population is normally distributed. Step 4: Make a decision. Is it reject Null hypothesis or Fail to reject Null hypothesis? Step 5: What is the conclusion? Is there sufficient evidence or there is not sufficient evidence to show that the standard deviation of the test scores has decreased?The National Coalition on Healthcare suggests that the mean annual premium that a health insurer charges an employer for a health plan covering a family of four averaged $13,000 in 2009. A sample of 30 families of four yields a mean annual premium paid by their employer to be $13,500 with a sample standard deviation of $300. We are interested in whether the mean annual premium that a health insurer charges an employer for a health plan covering a family of four is different from $13,000 using a significance level of 0.10. Select one: a. tdata = 9.129, do not reject Ho. b. tdata = 9.129, reject Ho. c. tdata = -9.129, do not reject Ho. O d. tdata = 4.156, reject Ho.Suppose a survey of a random sample of 100 smokers, conducted by the Department of Health, suggests that the mean number of cigarettes a person smokes in a day in Smokelandia (Y) is 2.72 and the standard deviation (sy) is 0.58. The Department of Health is concerned about the results of the survey and wants to test whether the mean number of cigarettes a person smokes in a day is 2.51 or not. The test statistic associated with the above test is 3.6204 (Round your answer to two decimal places.) If the Department of Health uses a 5% significance level, the test statistic suggests that we the hypothesis that the mean number of cigarettes a person smokes in a day is 2.51. Suppose the Department of Health decides to reject the null hypothesis (H,) if |Y- 2.51|>0.08. The size of the above test is (Round your answer to two decimal places.) If u = 2.69, the power of the above test is (Round your answer to two decimal places.)
- Find the 90% confidence interval for the variance and standard deviation of the ages of students at Arcanum high school if a sample of 24 students has a standard deviation of 2.3 years. Assume the variable is 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.Bone mineral density (BMD) is a measure of bone strength. Studies show that BMD declines after age 45. The impact of exercise may increase BMD. A random sample of 59 women between the ages of 41 and 45 with no major health problems were studied. The women were classified into one of two groups based upon their level of exercise activity: walking women and sedentary women. The 39 women who walked regularly had a mean BMD of 5.96 with a standard deviation of 1.22. The 20 women who are sedentary had a mean BMD of 4.41 with a standard deviation of 1.02. Which of the following inference procedures could be used to estimate the difference in the mean BMD for these two types of women