An advertiser was about to launch an advertisement campaign for a new product. He conducted a random survey to find the preference for the new product among the respondents. Based on his previous experience, he found the standard deviation for the population as 4. Because the population mean was not known to him, he randomly selected 25 responses and found that the sample mean preference for the new product was 10. Which of the following parameters did he use to calculate a confidence interval at the 95% confidence level for this population mean? z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576

Q: If n=30, (x-bar)=38, and s=8, construct a confidence interval at a 90% confidence level. Assume the…

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Q: Suppose a sample of a restaurant’s sales taken from 49 business days shows average daily sales of…

A: Given, n=49x=6300σ=1400α=1-0.95=0.05 The degrees of freedom is obtained as- df=n-1=49-1=48 The…

Q: A random sample of 230 people living in Seattle found that 83 of them had a pet. A random sample of…

A: Solution: Given information: n1= 230 Sample size of people iving in Seattle n2= 172  Sample size of…

Q: A random sample of size n₁ = 23, taken from a normal population with a standard deviation 01 = 6,…

A: Hey there! Thank you for posting the question. Since there are multiple questions posted, we will…

Q: Dana is a math professor who usually teaches in-person but has recently started teaching online.…

A: Given that, mean , μ=76.5standard deviation,σ=19.1and sample mean(class-average),x =68.4 and sample…

Q: The number of television sets coming off a production line each day is known to have a standard…

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Q: If n=12, (x-bar)=33, and s=16, construct a confidence interval at a 80% confidence level. Assume the…

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Q: The admissions officer at a small college compares the scores on the Scholastic Aptitude Test (SAT)…

A: Since, the population variances are unknown the Student's t-test will be applied. The variances are…

Q: sample of 35 students was taken from the local university. These students spent an average of $180…

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Q: An admission officer of a small liberal arts college wishes to compare the mean admission test…

A: Given: n1= 15 ,x̄1= 495.0 ,s1= 55.0 n2= 17 ,x̄2= 545.0 ,s2= 50.0 At 95.0 % confidence level with…

Q: An agronomist would like to estimate the average weight of all Indian River County grapefruits. The…

A: The sample size is 262, sample mean is 203 and population standard deviation is 20.

Q: Scientists tested each in a sample of 6 alleles of antigen-produced protein for level of peptide.…

A: The sample size n=6 The sample mean x¯=1.54 The sample standard deviation s=0.13 The confidence…

Q: . If she wants to make a 95% confidence interval for the mean with the margin of error of 5 cm, at…

A: here from given information , σ = 17 cm MOE = 5cm here Z-critical for 95 % confidence level = 1.96…

Q: At a college, a popular achievement test is used for advanced college placement. Counselors selected…

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Q: How important is it for a waiter or waitress to complement the customers' dinner selection. A waiter…

A: For customers that were not complemented The average tip was 16% so the proportion is p1⏞=0.16 The…

Q: A market researched for a consumer electronics company wants to study the television viewing habits…

A: Given: n=40x¯=15.3S=3.8 Degrees of freedom: df=n-1=40-1=39 The critical-value is obtained by using…

Q: You are interested in estimating the the mean weight of the local adult population of female…

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Q: The number of television sets coming off a production line each day is known to have a standard…

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Q: A random sample of high school students is used to estimate the mean time all high school students…

A: Solution: From the given information, a 99% confidence interval is 1.7 hours to 3.1 hours.

Q: The marketing director concluded, with 95% confidence, that between $113.36 and $130.64 contains the…

A: Given: Sample size,n=40Sample mean x¯=122Sample standard deviation s=27confidence level,c =95% =…

Q: In a study, Group 1 was found to have an average of 53 and a population standard deviation of 13 out…

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Q: Gary observed the fuel efficiency of his SUV. After the first 100 times he filled up the tank, he…

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Q: A group of students go to a lake that you fished in, and caught 28 adult striped bass fish. The mean…

A: Obtain the 95% confidence interval for the population mean of adult striped bass for that lake. The…

Q: You would like to construct a 99% confidence interval to estimate the population mean score on a…

A: Solution: Let X be the score. From the given information, x-bar=485 and S=70. Here, it needed to…

Q: the sample proportion of females is  x=118  n=164 P = .720  #1 Make a 95% confidence interval for…

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Q: If n=14, (x-bar)=46, and s=18, construct a confidence interval at a 80% confidence level. Assume the…

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

Q: You will need your z and t charts for this  Death penalty: A group claims that the overall feelings…

A: The question is about confidence interval.Given :No. of interviewed people ( n ) = 643No. of people…

Q: On average, the level of radioactivity was 198.2 gammas. The standard deviation was 3.3 gammas.…

A: Given information Sample size n = 10 Sample mean x̅ = 200 Population standard deviation = 3.3…

Q: A toy manufacturer wants to know how many new toys children buy each year. A sample of 242 children…

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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: each week is 1.3. The study found that for a sample of 2549 teenagers the mean number of softdrinks…

A: Given : Sample Size , n= 2549 Sample Mean , x̄ =5.4 population standard deviation…

Q: a sample of 25 automobiles shows that the mean price was $20,500 with standard deviation of $2,500.…

A: Givensample size = n = 25Mean price = $20,500Standard deviation = $2500 The distribution is normal.…

Q: To determine the views of students on a particular campus about fraternities, an 11-point attitude…

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Q: A teacher gives a test to his class of 25 students. Since he wants to quickly obtain an estimate of…

A: The class of 25 students is considered as the population. Sample size n =12 Sample mean x¯=82 Sample…

Q: A small business wanted to estimate the mean income from their customers. They took a sample of 256…

A: We have to find  the 90% confidence interval estimate for the population mean income of their…

Q: The dean randomly selected 33 students who took the calculus final exam and looked at their scores…

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Q: If n=21, (x-bar)=37, and s=3, construct a confidence interval at a 90% confidence level. Assume the…

A: Given Information: Sample size n=21 Sample mean x¯=37 Sample standard deviation s=3 Confidence level…

Q: A new drug to treat high cholesterol is being tested by pharmaceutical company. The cholesterol…

A: Mean difference (M) = 15.91 mg/dL Standard deviation of the mean difference (s) = 15.2714 mg/dL…

Q: A researcher is attempting to calculate a 95% confidence interval for the mean average wingspan of…

A: The random variable wingspan follows normal distribution. We have to find the 95% confidence…

Q: A researcher wants to compare the effectiveness of two different study techniques for a test. The…

A: To calculate the 95% Confidence Interval for the difference in mean test scores for the two…

Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.