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Police recorded the average speed of cars driving on a busy street by a school. For a sample of 16 speeds, it was determined that the average amount over the speed limit for the 16 speeds was 11.4 mph with a standard deviation of 6 mph. The 90% confidence interval estimate for this sample is 8.77 mph to 14.03 mph. a) What is the margin of error for this problem? b) What size sample is needed to reduce the margin of error to no more than + 17 a) The margin of error isOmph.

The police department of a U.S. city reported that the mean number (μ) of auto thefts per neighborhood per year is 6.50, with a standard deviation (σ) of 1.20. One suburban community reports a mean (x̄) number of auto thefts per neighborhood per year at 8.0, from a random sample of 25 neighborhoods.  Assume that you're working at the 0.05 level of significance. What is the approximate value of the calculated test statistic (Z)?

A chemist measures the haptoglobin concentrations (in grams per liter) in the blood of eight healthy adults. The mean and standard deviation of the haptoglobin concentrations (in grams per liter) are 1.615 and 1.387 respectively. Obtain a 90% confidence interval for the mean haptoglobin concentration in adults.

A major newspaper (that rhymes with Small Street Journal) reported an average household income for its subscribers is $119,155. Assume this estimate of the mean household income is based on a sample of 80 subscribers, and, based on past studies, the population standard deviation is known to be sigma=$30,000. Construct a 95% confidence interval for the population mean. Group of answer choices $60,356 - $177,954 $118,420 - $119,890 $112,447 - $125,863 $112,581 - $125,729

A researcher is interested in estimating the mean blood alcohol content (BAC) of people arrested for driving under the influence. The sample consists of 250 individuals with a mean BAC of .145. Based on past data, the researcher assumes a population standard deviation of .065.What's the margin of error for a 90% confidence interval in this scenario Select one: a. 0.0068 b. 0.0041 c. Not enough information to compute the margin of error d. 0.107 e. 0.008

what is the p value？

In a sample of 30 people, the average cost of a latte is $3.55. The standard deviation for the sample is $1.46. What is the margin of error for a 99% confidence interval? Group of answer choices $0.73 $0.69 $0.27 $0.66 thanks

Tom wants to estimate the percentage of students who live more than 3 miles from the school.He wants to create a 95% confidence interval which has a margin of error of at most 3% .How many students must be polled to create the confidence interval?

The average growth rate of income per person was nearly two-digit for decades in China after the economic reform in 1978. Since China has become a middle- income economy, observers claim that growth has slowed to 6% a year. Analysis of a sample of 31 mainland Chinese provinces in 2018 shows that annual economic growth is, on average, 6.07% with a standard deviation of 1.12%. a) Based on this cross-section evidence, estimate a 95% confidence interval of the average provincial economic growth. b) Interpret the confidence interval. Does it support the claim that economic growth has slowed in China? (up to 80 words)

If the level of confidence is lowered from 95% to 90 % , but the allowable error  and the standard deviation remain the same, what happens to the required sample size?

Given that X̄=20, σ=9 and n=81, construct the following confidence intervals: the 90% confidence interval for μ is:  a (18.355, 21.645)   b (18.040, 21.960) c (17.840, 36.450) d (18.750, 23.450)

The owner of a large car dealership believes that the financial crisis decreased the number of customers visiting her dealership. The dealership has historically had 800 customers per day. The owner takes a sample of 100 days and finds the average number of customers visiting the dealership per day was 750. Assume that the population standard deviation is 350. At a 5% significance level, the critical value(s) is(are) 1.645 -2.030 and 2.030 -1.645 -1.96 and 1.96

A sample of 27 elements from a normally distributed population is selected. The sample mean is 14 with a sample standard deviation of 5. The 95% confidence interval for u is._____ a. 12.022 to 15.978 b. 9:000 to 19.000 c. 13.791 to 14.209 d. 15.886 to 12.114

A sample size of 200 light bulbs was tested and found that 11 were defective. What is the 95%c confidence interval around this sample proportion? o 0.055±0.0079 O 0.055 ±0.0180 O 0.055±0.0316 O 0.055 + 0.0158

Goofy Marketing Solutions conducted a survey among 16 randomly selected visitors to the newly opened Outlet Shoppes at Atlanta to measure their satisfaction with the shopping experience. The sample satisfaction score is normally distributed around a mean of �¯=67with a sample standard deviation of s=7.509. Determine the 95% confidence interval for the population mean from the sample.   The upper limit of the 95% confidence interval is (rounded)....

sssss A consumer affairs investigator records the repair cost for 4 randomly selected TVs. A sample mean of $75.89 and standard deviation of $13.53 are subsequently computed. Determine the 80% confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal. Step 1 of 2: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Cell phones have become an essential part of our everyday lives. We use them for communication, photography, browsing, and banking, among other things. Choosing the right cell phone is important, and is based on several factors, including weight. Suppose the weight of a cell phone is distributed normally with a known population standard deviation of 18 g. A random sample of 15 cell phones was conducted to calculate a sample mean weight of 155.7 g. Construct a 95% confidence interval for the population mean weight, mu. Choose the correct answer below based on constructing this interval (it may be a good idea to write your final interval down, as the next question may relate to this one). Hint: although the sample size is small, you may use the standard normal distribution, as the population standard deviation is known. Hint: we cover this in the video "CI 2020 2" a. the interval is [146.59, 164.81); the critical value used is 1.96 O a) O b) b. the interval is (146.59, 164.81] ; the…

The sugar content in a one-cup serving of a certain breakfast cereal was measured for a sample of 140 servings. The average was 11.9 g and the standard deviation was 1.1 g. a. Find a 95% confidence interval for the mean sugar content. b. Find a 99% confidence interval for the mean sugar content. c. What is the confidence level of the interval (11.81, 11.99)? d. How large a sample is needed so that a 95% confidence interval specifies the mean to within±0.1? e. How large a sample is needed so that a 99% confidence interval specifies the mean to within±0.1? f. Find a 95% upper confidence bound for the mean sugar content. g. The claim is made that the mean sugar content is greater than 11.7g. With what level of confidence can this statement be made?

A study was designed to compare the attitudes of two groups of nursing students towards computers. Group 1 had previously taken a statistical methods course that involved significant computer interaction. Group 2 had taken a statistic methods course that did not use computers. The students' attitudes were measured by administering the Computer Anxiety Rating Scale (CARS). A random sample of 1010 nursing students from Group 1 resulted in a mean score of 55.755.7 with a standard deviation of 7.97.9. A random sample of 1515 nursing students from Group 2 resulted in a mean score of 65.665.6 with a standard deviation of 6.16.1. Can you conclude that the mean score for Group 1 is significantly lower than the mean score for Group 2? Let μ1�1 represent the mean score for Group 1 and μ2�2 represent the mean score for Group 2. Use a significance level of α=0.05�=0.05 for the test. Assume that the population variances are equal and that the two populations are normally distributed. Step 3 of 4:…

A study was designed to compare the attitudes of two groups of nursing students towards computers. Group 1 had previously taken a statistical methods course that involved significant computer interaction. Group 2 had taken a statistic methods course that did not use computers. The students' attitudes were measured by administering the Computer Anxiety Rating Scale (CARS). A random sample of 1010 nursing students from Group 1 resulted in a mean score of 55.755.7 with a standard deviation of 7.97.9. A random sample of 1515 nursing students from Group 2 resulted in a mean score of 65.665.6 with a standard deviation of 6.16.1. Can you conclude that the mean score for Group 1 is significantly lower than the mean score for Group 2? Let μ1�1 represent the mean score for Group 1 and μ2�2 represent the mean score for Group 2. Use a significance level of α=0.05�=0.05 for the test. Assume that the population variances are equal and that the two populations are normally distributed. Step 2 of 4:…

You are the manager of a restaurant for a fast-food franchise. Last month, the mean waiting time at the drive-through window for branches in your geographical region, as measured from the time a customer places an order until the time the customer receives the order, was 3.9 minutes. You select a random sample of 64 orders. The sample mean waiting time is 3.63 minutes, with a sample standard deviation of 0.8 minute. Complete parts (a) and (b) below. ..... a. At the 0.05 level of significance, is there evidence that the population mean waiting time is different from 3.9 minutes? State the null and alternative hypotheses. Ho: H (Type integers or decimals.)

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The body temperatures in degrees Fahrenheit of a sample of adults in one small town are: 97.8 97 97.5 99.1 98.4 97.9 97.7 Assume body temperatures of adults are normally distributed. Based on this data, find the 80% confidence interval of the mean body temperature of adults in the town. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places. Assume the data is from a normally distributed population.80% C.I. =

A national random sample of 20 ACT scores from 2010 is as follows: (29, 26, 13, 23, 23, 25, 17, 22, 17, 19, 12, 26, 30, 30, 18, 14, 12, 26, 17, 18). Calculate the sample mean and standard deviation. Assuming that the ACT is normally distributed, what is the 95% confidence interval for the population ACT score. Group of answer choices 18.25 to 23.45 15.32 to 23.05 19.52 to 22.19 14.32 to 23.15

(a) A sample of 150 students at University of Sussex has a mean income of £62 per week with a standard deviation of £7. Calculate the 95% confidence interval for the true population mean. Show your method in full and explain your answer.

In 2017, Americans spent a record-high $9.1 billion on Halloween- related purchases (the balance website). Sample data showing the amount, in dollars, that 16 adults spent on a Halloween costume are as follows. 13 69 23 65 32 36 31 43 52 17 13 95 46 32 66 24 a. What is the estimate of the population mean amount adults spend on a Halloween costume (to 2 decimals)? $ 41.06 b. What is the sample standard deviation (to 2 decimals)? $ c. Provide a 95% confidence interval estimate of the population standard deviation for the amount adults spend on a Halloween costume (to 2 decimals). Use Table 3 of Appendix B. $ 17.14 ≤0≤ $

A carpet company advertises that it will deliver your carpet within 15 days of purchase. A sample of (n) 49 past customers is taken. The average delivery time in the sample was () 16.2 days. Assume the population standard deviation is known to be (o) 5.6 days. Note that the significance level is x=0.05. What are the z-score and the p-value for this sample?

According to R.L. Polk & Co., Americans held onto their new vehicles for an average of 71.4 months. A random sample of 60 adults who recently sold their new vehicle was selected. Assuming the standard deviation for this population is 15.8 months, the standard error for this sample is

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