A marketing researcher wants to estimate the mean amount spent ($) on Amazon.com by Amazon Prime member shoppers. Suppose a random sample of 100 Amazon Prime member shop pers who recently made a purchase on Amazon.com yielded a mean of $1,500 and a standard deviation of $200. a. Construct a 95% confidence interval estimate for the mean spending for all Amazon Prime member shoppers. b. Interpret the interval constructed in (a).
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A marketing researcher wants to estimate the
a. Construct a 95% confidence
b. Interpret the interval constructed in (a).
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- A new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements are played. They found that for 11 days following no advertisements, the mean was 16.6 purchasing customers with a standard deviation of 1.1 customers. On 10 days following advertising, the mean was 17.7 purchasing customers with a standard deviation of 1.4 customers. Test the claim, at the 0.02 level, that the mean number of customers who make a purchase in the store is lower for days following no advertising compared to days following advertising. Assume that both populations are approximately normal and that the population variances are equal. Let days following no advertisements be Population 1 and let days following advertising be Population 2. Step 2 of 3 : Compute…For a study conducted by the research department of a pharmaceutical company, 235 randomly selected individuals were asked to report the amount of money they spend annually on prescription allergy relief medication. The sample mean was found to be $17.80 with a standard deviation of $4.30 . A random sample of 250 individuals was selected independently of the first sample. These individuals reported their annual spending on non-prescription allergy relief medication. The mean of the second sample was found to be $18.20 with a standard deviation of $4.00 . As the sample sizes were quite large, it was assumed that the respective population standard deviations of the spending for prescription and non-prescription allergy relief medication could be estimated as the respective sample standard deviation values given above. Construct a 90% confidence interval for the difference −μ1μ2 between the mean spending on prescription allergy relief medication ( μ1 ) and the mean…Assume the average selling price for houses in a certain county is $308,000 with a standard deviation of $57, 000. Using Chebychev's Theorem, determine the range of prices that includes at least 91% of the homes around the mean.
- Suppose that the quarterly sales levels among health care information systems companies are approximately normally distributed with a mean of 12 million dollars and a standard deviation of 1.3 million dollars. One health care information systems company considers a quarter a "failure" if its sales level that quarter is in the bottom 20% of all quarterly sales levels. Determine the sales level (in millions of dollars) that is the cutoff between quarters that are considered "failures" by that company and quarters that are not. Carry your intermediate computations to at least four decimal places. Round your answer to one decimal place. million dollars X SFrom checking Facebook while walking to class to posting photos on Instagram while eating in the Russell House - social media plays a significant role in the daily lives of college students. The average amount of time a college student spends on social media in the morning (6:00 am - 11:59 am) is normally distributed with a mean of 48 minutes per day and a standard deviation of 12 minutes per day. Suppose on a given morning a college student spends 85 minutes on social-media. Is this unusual?. OA. Not enough information OB. Yes OC. NoSuppose that a recent article stated the mean time spent in jail by a first-time convicted burglar is 2.7 years. A study was then done to see if the mean time has increased in the new century. A random sample of 27 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was 3.01 with a standard deviation of 1.78 years. Assume that the distribution of jail times is approximately normal. Test the claim at the level of significance 0.1. Select one: a) t_stat = 0.9 and t_cr = 1.31 : The mean time increased significantly in the new century b) t_stat = 0.9 and t_cr = 1.31 : The mean time didn’t increase in the new century c) t_stat = 0.99 and t_cr = 1.71 : The mean time didn’t increase in the new century d) t_stat = 0.99 and t_cr = 1.71 : The mean time increased significantly in the new century
- A banker finds that the number of times people use automated-teller machines (ATMs) in a year are normally distributed with a mean of 40.0 and a standard deviation of 11.2 (based on data from Maritz Marketing Research, Inc.). Find the proportion of customers who use ATMs between 20 and 60 times. Use four decimal places for your answerThe 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.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 ?=$3.50 . To address the CEO’s conjecture, the marketing manager carried out a hypothesis test of ?0:?=13.50 vs. ??:?>13.50 and obtained a ?‑value = 0.77. The marketing chooses a significance level of ?=0.10. If he uses this significance level throughout his work, how often will he reject a true null hypothesis? Group of answer choices a.He will reject 10% of all true null hypotheses. b. He will reject 1% of all true null hypotheses. c. He will reject 5% of all true null hypotheses. d. He will not reject 10% of all true null…
- A marketing researcher wants to estimate the mean amount spent per year ($) on a web site by membership member shoppers. Suppose a random sample of 100 membership member shoppers who recently made a purchase on the web site yielded a mean amount spent of $55 and a standard deviation of $52. Complete parts (a) and (b) below. a. Is there evidence that the population mean amount spent per year on the web site by membership member shoppers is different from $51? (Use a 0.01 level of significance.) State the null and alternative hypotheses. Ho: H (Type integers or decimals. Do not round. Do not include the $ symbol in your answer.) Identify the critical value(s). The critical value(s) is/are (Type an integer or a decimal. Round to two decimal places as needed. Use a comma to separate answers as needed.) Determine the test statistic. The test statistic, 1STAT, is (Type an integer or a decimal. Round to two decimal places as needed.) State the conclusion. Ho. There is evidence that the…Suppose that the quarterly sales levels among health care information systems companies are approximately normally distributed with a mean of 9 million dollars and a standard deviation of 1.4 million dollars. One health care information systems company considers a quarter a "failure" if its sales level that quarter is in the bottom 10% of all quarterly sales levels. Determine the sales level (in millions of dollars) that is the cutoff between quarters that are considered "failures" by that company and quarters that are not. Carry your intermediate computations to at least four decimal places. Round your answer to one decimal place.A marketing researcher wants to estimate the mean amount spent per year ($) on a web site by membership member shoppers. Suppose a random sample of 100 membership member shoppers who recently made a purchase on the web site yielded a mean amount spent of $58and a standard deviation of $55.Complete parts (a) and (b) below. a. Is there evidence that the population mean amount spent per year on the web site by membership member shoppers is different from $51? (Use a 0.01 level of significance.) State the null and alternative hypotheses.
