Question

In: Statistics and Probability

A random sample of 28 shoppers spent an average of $21.53 per visit at Barnes &...

A random sample of 28 shoppers spent an average of $21.53 per visit at Barnes & Noble with a standard deviation of $2.31. Find the 95% confidence interval of the true mean cost per visit.

Solutions

Expert Solution

Solution :

Given that,

= 21.53

s =2.31

n = Degrees of freedom = df = n - 1 = 28- 1 = 27

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,27 = 2.052 ( using student t table)

Margin of error = E = t/2,df * (s /n)

= 2.052* (2.31 / 28)

= 0.896

The 95% confidence interval mean is,

- E < < + E

21.53 -0.896 < < 21.53+ 0.896

20.634 < < 22.426

orchestra answered 2 years ago
Next > < Previous

Related Solutions

A random sample of 89 tourist in the Chattanooga showed that they spent an average of...
A random sample of 89 tourist in the Chattanooga showed that they spent an average of $2,860 (in a week) with a standard deviation of $126; and a sample of 64 tourists in Orlando showed that they spent an average of $2,935 (in a week) with a standard deviation of $138. We are interested in determining if there is any significant difference between the average expenditures of those who visited the two cities? a) Determine the degrees of freedom for...
1. At store A, the sample average amount spent by a random sample of 32 customers...
1. At store A, the sample average amount spent by a random sample of 32 customers was $31.22 with a sample standard deviation of $6.48. At store B, the sample average amount spent by a random sample of 36 customers was $34.68 with a sample standard deviation of $5.15. Test, using the 4-step procedure and a 5% level of significance, whether or not there is a difference between the population mean amount spent per customer at store A and the...
A random sample of 25 CCRI students found that the average amount spent on textbooks for...
A random sample of 25 CCRI students found that the average amount spent on textbooks for a course was $180 with a standard deviation of $30. Find a 95% 2-tailed confidence interval for the population mean (μ) of the amount a CCRI student spends on textbooks.
A random sample of 5 months in an urban area showed an average of 28 mail...
A random sample of 5 months in an urban area showed an average of 28 mail carriers were bitten by dogs. If the population is normal and the standard deviation of the sample is 3 find the 95% confidence interval for the population mean.
A random sample of 200 customers who visited a local grocery spent an average of $97....
A random sample of 200 customers who visited a local grocery spent an average of $97. You find a reputable source stating that the current population standard deviation of all such grocery bills in similar grocery stores is equal to $19. Using this information, find the 90% confidence interval for the population mean: $95.28 to $98.72 $94.79 to $99.21 $94.37 to $99.63 $93.87 to $100.13
A random sample of 200 customers who visited a local grocery spent an average of $97....
A random sample of 200 customers who visited a local grocery spent an average of $97. You find a reputable source stating that the current population standard deviation of all such grocery bills in similar grocery stores is equal to $19. Using this information, find the 90% confidence interval for the population mean: $95.28 to $98.72 $94.79 to $99.21 $94.37 to $99.63 $93.87 to $100.13
According to​ research, holiday shoppers spent an average of ​$366 over a holiday weekend in 2008....
According to​ research, holiday shoppers spent an average of ​$366 over a holiday weekend in 2008. The accompanying data show the amount spent by a random sample of holiday shoppers during the same weekend in 2009. Complete parts a and b below. $421 ​$278 ​$376 ​$232 ​$221 ​$219 ​$358 ​$433 ​$388 ​$391 ​$328 ​$65 ​$414 ​$434 ​$376 ​$270 critical values? Test statistic? What is the conclusion? P-value?
According to​ research, holiday shoppers spent an average of ​$376 over a holiday weekend in 2008....
According to​ research, holiday shoppers spent an average of ​$376 over a holiday weekend in 2008. The accompanying data show the amount spent by a random sample of holiday shoppers during the same weekend in 2009. Complete parts a and b below. LOADING... Click the icon to view the data table. ​$430 ​$280 ​$370 ​$237 ​$223 ​$215 ​$360 ​$435 ​$392 ​$389 ​$319 ​$67 ​$416 ​$432 ​$383 ​$281 a. Retailers were concerned that the recent economic downturn was going to affect...
A random sample of 28 policies form insurance company A shows an average net profit of...
A random sample of 28 policies form insurance company A shows an average net profit of $39 per policy, with a standard deviation of $3.70 per policy. A random sample of 34 policies from insurance company B shows an average net profit of $47 per policy with a standard deviation of $3.90 per policy. Assume the policy profits are normally distributed and have unequal variances.   Use a 0.01 level of significance to test whether the profits per policy are significantly...
The ages​ (in years) of a random sample of shoppers at a gaming store are shown....
The ages​ (in years) of a random sample of shoppers at a gaming store are shown. Determine the​ range, mean,​ variance, and standard deviation of the sample data set. 12​,17​,23​,15​,13​,18​,22​,18​,13​,17 The range is _ ​(Simplify your​ answer.) The mean is _ (Simplify your answer. Round to the nearest tenth as​ needed.) The variance is _ ​(Simplify your answer. Round to the nearest hundredth as​ needed.) The standard deviation is _ ​(Simplify your answer. Round to the nearest tenth as​ needed.)
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT