Question

In: Statistics and Probability

A marketing researcher wants to estimate the mean amount spent​ ($) on a certain retail website...

A marketing researcher wants to estimate the mean amount spent​ ($) on a certain retail website by members of the​ website's premium program. A random sample of94members of the​ website's premium program who recently made a purchase on the website yielded a mean of$1700and a standard deviation of​$250

a. Construct a 99%confidence interval estimate for the mean spending for all shoppers who are members of the​ website's premium program.

Solutions

Expert Solution

Solution :

Given that,

Point estimate = sample mean = = $1700

Population standard deviation =    =$250
Sample size = n =94

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576   ( Using z table )

Margin of error = E = Z/2* ( /n)

= 2.576* ( 250/ 94)

= 66.4235

At 99% confidence interval estimate of the population mean is,

- E < < + E

1700- 66.4235< < 1700+66.4235

1633.5765< < 1766.4235

(1633.5765, 1766.4235 )


Related Solutions

A marketing researcher wants to estimate the mean amount spent​ ($) on a certain retail website...
A marketing researcher wants to estimate the mean amount spent​ ($) on a certain retail website by members of the​ website's premium program. A random sample of 92 members of the​ website's premium program who recently made a purchase on the website yielded a mean of ​$1400 and a standard deviation of ​$200. Complete parts​ (a) and​ (b) below. A. Construct a 99​% confidence interval estimate for the mean spending for all shoppers who are members of the​ website's premium...
A marketing researcher wants to estimate the mean amount spent​ ($) on a certain retail website...
A marketing researcher wants to estimate the mean amount spent​ ($) on a certain retail website by members of the​ website's premium program. A random sample of 99 members of the​ website's premium program who recently made a purchase on the website yielded a mean of ​$1900 and a standard deviation of ​$200. Complete parts​ (a) and​ (b) below. a. Construct a 95​% confidence interval estimate for the mean spending for all shoppers who are members of the​ website's premium...
A marketing researcher wants to estimate the mean amount spent​ ($) on a certain retail website...
A marketing researcher wants to estimate the mean amount spent​ ($) on a certain retail website by members of the​ website's premium program. A random sample of 96 members of the​ website's premium program who recently made a purchase on the website yielded a mean of ​$1400 and a standard deviation of ​$200. Complete parts​ (a) Construct a 95​% confidence interval estimate for the mean spending for all shoppers who are members of the​ website's premium program. (Round to two...
A marketing researcher wants to estimate the mean amount spent​ ($) on a certain retail website...
A marketing researcher wants to estimate the mean amount spent​ ($) on a certain retail website by members of the​ website's premium program. A random sample of94members of the​ website's premium program who recently made a purchase on the website yielded a mean of$1700and a standard deviation of​$250 a. Construct a 99%confidence interval estimate for the mean spending for all shoppers who are members of the​ website's premium program.
A marketing researcher wants to estimate the mean amount spent​ ($) on a certain retail website...
A marketing researcher wants to estimate the mean amount spent​ ($) on a certain retail website by members of the​ website's premium program. A random sample of 97 members of the​ website's premium program who recently made a purchase on the website yielded a mean of $2000 and a standard deviation of $150. Complete parts​ (a) and​ (b) below. a. Construct a 95​% confidence interval estimate for the mean spending for all shoppers who are members of the​ website's premium...
A marketing researcher wants to estimate the mean amount spent​ ($) on a certain retail website...
A marketing researcher wants to estimate the mean amount spent​ ($) on a certain retail website by members of the​ website's premium program. A random sample of 99 members of the​ website's premium program who recently made a purchase on the website yielded a mean of ​$1900 and a standard deviation of 300. Complete parts​ (a) and​ (b) below. a. Construct a 90% confidence interval estimate for the mean spending for all shoppers who are members of the​ website's premium...
A marketing researcher wants to estimate the mean amount spent​ ($) on a certain retail website...
A marketing researcher wants to estimate the mean amount spent​ ($) on a certain retail website by members of the​ website's premium program. A random sample of 99 members of the​ website's premium program who recently made a purchase on the website yielded a mean of ​$1700 and a standard deviation of ​$150. Complete parts​ (a) and​ (b) below. a. Construct a 90​%confidence interval estimate for the mean spending for all shoppers who are members of the​ website's premium program....
A marketing researcher wants to estimate the mean amount spent per year​ ($) on a web...
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 $54. a. Is there evidence that the population mean amount spent per year on the web site by membership member shoppers is different from $47? (Use .10 level of...
A marketing researcher wants to estimate the mean amount spent per year​ ($) on a web...
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 $57 and a standard deviation of ​$ 54 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...
A marketing researcher wants to estimate the mean amount spent per year​ ($)on a web site...
A marketing researcher wants to estimate the mean amount spent per year​ ($)on a web site by membership member shoppers. Suppose a random sample of 100membership member shoppers who recently made a purchase on the web site yielded a mean amount spent of $56and 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 $53?(Usea 0.01 level of​significance.) State the null...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT