Use the following information to answer the question. According to the website ww.costofwedding.com, the average cost of flowers for a wedding is $698. Recently, in a random sample of 40 weddings in the U.S. it was found that the average cost of the flowers was $734, with a standard deviation of $102. On the basis of this, a 95% confidence interval for the mean cost of flowers for a wedding is $701 to $767. Choose the statement that is the best interpretation of the confidence interval. a) We are extremely confident that the mean cost of flowers at a wedding is between $701 and $767. b) In about 95% of all samples of size 40, the resulting confidence interval will contain the mean cost of flowers at weddings. c) That probability that flowers at a wedding will cost less than $767 is nearly 100%. d) That probability that the flowers at a wedding will cost more than $698 is greater than 5%.

Q: In a survey, 8 people were asked how much they spent on their child's last birthday gift. The…

A: The sample size, i.e., the sampled number of people, n=8.The sample mean, x¯=$37.The sample standard…

Q: Find the margin of error at a 90% confidence level.

A:

Q: In a survey, 28 people were asked how much they spent on their child's last birthday gift. The…

A:

Q: In a survey, 28 people were asked how much they spent on their child's last birthday gift. The…

A:

Q: In a survey, 26 people were asked how much they spent on their child's last birthday gift. The…

A:

Q: An economist is interested in studying the incomes of consumers in a particular country. The…

A: It is given that the standard deviation is 1,000. Margin of error, E is 50. The confidence level is…

Q: In a survey, 24 people were asked how much they spent on their child's last birthday gift results…

A:

Q: o= $1000. a. State the null and alternative hypotheses. Ho: 4- Select your answer V3472 H: A- Select…

A: Given : Population mean =$3472 Population standard deviation =$1000

Q: The former manager of the coffee shop in a local bookstore claimed that the mean daily sales were…

A: Givensample size(n)=11Mean(x)=2956.14standard deviation(s)=370α=0.01

Q: In a survey, 28 people were asked how much they spent on their child's last birthday gift. The…

A: Since population standard deviation is unknown, Use t-distribution to find t-critical value. Find…

Q: In a survey, 13 people were asked how much they spent on their child's last birthday gift. The…

A: GivenMean(x)=42standard deviation(s)=7sample size(n)=13

Q: Beer Drinking. The mean annual consumption of beer per person in the US is 22.0 gallons . A random…

A: P value has find by using R command.

Q: A study reports that recent college graduates from New Hampshire face the highest average debt of…

A: Given, μ=$31813σ=$5000 n=40 Debt Debt 24042 31921 19157 31531 26757 34212 31923 14629…

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: there sufficient evidence at a=0.01 to conclude that their health care expenditure differs from the…

A:

Q: at the university spend on groceries each month. From a sample of 29 random students, you find that…

A: Given that  Sample size n =29 Sample mean =871.35 Standard deviation =235.06 NOTE:- According to…

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: A polling organization reported data from a survey of 2000 randomly selected Canadians who carry…

A: From the provided information, Sample size n = 2000 Sample mean x̄= 9.13 Standard deviation s = 7.20…

Q: A marketing researcher wants to estimate the mean savings ($) realized by shoppers who showroom.…

A: Sample size Sample mean Sample standard deviation

Q: C. Find a 95% confidence interval for the average life time of all the television sets produced by a…

A:

Q: estimate the mean age of all students currently enrolled. In a random sample of 20 students the mean…

A: Solution

Q: A study was conducted to estimate hospital costs for accident victims who wore seat belts.…

A: It is given that Sample mean = 9155Sample standard deviation = 5635Sample size = 21Confidence level…

Q: In a survey, 30 people were asked how much they spent on their child's last birthday gift. The…

A:

Q: A polling organization reported data from a survey of 2000 randomly selected Canadians who carry…

A:

Q: In a sample of 46 students, the mean scholarship amount received was $566, with a standard deviation…

A: From the provided information, Sample size (n) = 46 Sample mean (x̄) = 566 Sample standard deviation…

Q: In a sample of 31 students, the mean scholarship amount received was $562, with a standard deviation…

A: Given data,n=31x=562s=17.04df=n-1=30α=0.10t-critical value at α=0.10 and df=30 is tc=1.697

Q: aking the family to an amusement park has become ncreasingly costly according to the industry…

A: We have given that,  Sample mean (x̄) = $ 232, sample standard deviation (s) = 42 and sample size…

Q: The prices of used books at a large college bookstore are normally distributed. If a sample of 23…

A:

Q: A medical researcher surveyed 9 hospitals and found that the standard deviation for the cost for…

A:

Q: In a sample of 32 students, the mean scholarship amount received was $887, with a standard deviation…

A: Given information: n=32x¯=887s=27.4 df=n-1=32-1=31 Confidence level is 90% Significance level is…

Q: A survey of credit card holders revealed that Americans carried an average credit card balance of…

A: M1=$3900M2=$3300n1=400n2=450σ1=$880σ2=$810

Q: In a survey, 31 people were asked how much they spent on their child's last birthday gift. The…

A:

Q: A research team surveyed 251 adult Ohioans and found the mean monthly grocery cost for a single…

A: A research team surveyed 251 adult Ohioans and found the mean monthly grocery cost for a single…

Q: company manufactures a certain over-the-counter drug. The company samples 80 pills and finds that…

A: We have given that  Sample size n = 80 Sample mean = 325.5 Standard deviation s=10.3 90%confidence…

Q: Insurance Company A claims that its customers pay less for car insurance, on average than customers…

A: Formula :  test statistic for equal variance t test is

Q: A study of 100 patients with back pain reported that the mean duration of pain was 12 months. It is…

A: In question, Given that a study of 100 patients with back pain reported that the mean duration of…

Q: A polling organization reported data from a survey of 2000 randomly selected Canadians who carry…

A:

Q: In a survey, 29 people were asked how much they spent on their child's last birthday gift. The…

A: Answer  Given  Confidence interval CI = 80%   α=1-0.80    =0.2 sample size [n] =29  degree of…

Q: In a survey, 13 people were asked how much they spent on their child's last birthday gift. The…

A: From the provided information, Sample size (n) = 13 Sample mean (x̄) = 50 Sample standard deviation…

Q: The amount of money raised by Clayton State teams for the Statistics Games is normally distributed,…

A:

Q: The national fire protection association reported that house fires in the United States result in…

A:

Q: Suppose a study was conducted of community colleges in California to determine the average cost of…

A: As the population standard deviation is not known, we will use t distribution. The t distribution is…

Q: In a survey, 21 people were asked how much they spent on their child's last birthday gift. The…

A: Margin of error is the error due to random sampling. The margin of error decreases with increase in…

Q: The standard deviation, S, for the annual medical cost of 24 households was found to equal $1689…

A: Suppose σ is the annual medical cost for all households.