A marketing researcher wants to estimate the mean amount spent ($) on certain retail website by members of the website's premium program. A random sample of 93 members of the website's premium program who recently made a purchase on the website yielded a mean of $1200 and a standard deviation of $150. Complete parts (a) and (b) below. C a. Construct a 95% confidence interval estimate for the mean spending for all shoppers who are members of the website's premium program. Sus (Round to two decimal places as needed.)
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- The manager of the local Walmart Supercenter is studying the number of items purchased by customers in the evening hours. Listed below is the number of items for a sample of 30 customers. 9 12 5 8 4 6 6 7 11 9 8 14 9 12 5 4 10 6 18 10 6 10 11 5 What is the mean number of items? What is the median number of items? What is the range of number of items? What is the standard deviation? About 95% of the number of items are between what two values? Lower limit 10 9 13 12 13 5 Organize the number of items into a frequency distribution. What is the lower limit of the second class? What is the mean of the data organized into a frequency distribution? What is the standard deviation of the data organized into a frequency distribution? Compare these values with those computed in parts (A) and (D). Why are they different? Upper limitInsurance 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 7 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $16. For 12 randomly selected customers of Company B, you find that they pay a mean of $160 per month with a standard deviation of $14. 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 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.The question is attached in an image.
- A marketing consultant is hired by a major restaurant chain wishing to investigate the preferences and spending patterns of lunch customers. The CEO of the chain hypothesized that the average customer spends at least $13.50 on lunch. A survey of 25 customers sampled at one of the restaurants found the average lunch bill per customer to be ?¯=$14.50 . Based on previous surveys, the restaurant informs the marketing manager that the standard deviation is ?=$3.50 . To address the CEO’s conjecture, the marketing manager carried out a hypothesis test of ?0:?=13.50 vs. ??:?>13.50 and obtained a ?‑value = 0.77. The marketing chooses a significance level of ?=0.10. If he uses this significance level throughout his work, how often will he reject a true null hypothesis? Group of answer choices a.He will reject 10% of all true null hypotheses. b. He will reject 1% of all true null hypotheses. c. He will reject 5% of all true null hypotheses. d. He will not reject 10% of all true null…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 13 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $19. For 9 randomly selected customers of Company B, you find that they pay a mean of $157 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A's claim at the 0.05 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. Ho: M₁ = μ₂ Ha:M₁ •H₂ryan has collected data on individual income (reported in dollars) and has not found any outliers. ryan wants to calculate the average income in his data. which measure of central tendency should he use? -mean -variance -mode -median
- 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…1) According to Consumer Expenditures, the average consumer spent $1729 on apparel and services in 2017. In 2018, a random sample of 36 consumers had a mean of $1667.11 and a standard deviation of $351.69. Can it be said that at the 5% level of significance that the mean level of consumption differed in 2018 with respect to 2017?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 1313 people who buy insurance from Company A, the mean cost is $151$151 per month with a standard deviation of $16$16. For 99 randomly selected customers of Company B, you find that they pay a mean of $158$158 per month with a standard deviation of $19$19. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.050.05 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.
- In a recent year, Washington State public school students taking a mathematics assessment test had a mean score of 276.1 and a standard deviation of 34.4. A random sample of 64 students is drawn from this population. Identify the mean and standard deviation of the sample mean scores. 2. Find the probability that the sample mean score is at least 285.A college in the northeastern United States used SAT scores to help decide which applicants to admit. To determine whether the SAT was useful in predicting success, the college examined the relationship between the SAT scores and first-year GPAs of admitted students. The average math SAT score of a sample of 200 randomly selected students was 649.5 with a standard deviation of 66.2. The average GPA of the same students was 2.63 with a standard deviation of 0.58. The correlation between GPA and math SAT score was 0.19. (a) What does the correlation of 0.19 tell us about SAT scores and GPA scores? (You need to provide two answers.)(b) Find the equation of the least squares regression line.(c) Using the equation, predict the GPA score of a person with a math SAT score of 650.(d) Using the equation, predict the GPA score for a person with a math score of 800.The salaries of professional baseball players are heavily skewed right with a mean of $3.2 million and a standard deviation of $2 million. The salaries of professional football players are also heavily skewed right with a mean of $1.9 million and a standard deviation of $1.5 million. A random sample of 40 baseball players’ salaries and 35 football players’ salaries is selected. The mean salary is determined for both samples. Let represent the difference in the mean salaries for baseball and football players. Which of the following represents the shape of the sampling distribution for ? skewed right since the populations are both right skewed skewed right since the differences in salaries cannot be negative approximately Normal since both sample sizes are greater than 30 approximately Normal since the sum of the sample sizes is greater than 30