SOLUTION: 1) A major department store chain is interested in estimating the average amount its credit card customers spent on their first visit to the chain's new store in the mall. Fifteen
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-> SOLUTION: 1) A major department store chain is interested in estimating the average amount its credit card customers spent on their first visit to the chain's new store in the mall. Fifteen
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Question 1207859: 1) A major department store chain is interested in estimating the average amount its credit card customers spent on their first visit to the chain's new store in the mall. Fifteen credit card accounts were randomly sampled and analyzed with the following results: x = $50.50 and s2 = 400. A 95% confidence interval for the average amount the credit card customers spent on their first visit to the chain's new store in the mall is:
a. $50.50 ± $10.12
b. $50.50 ± $11.79
c. $50.50 ± $9.09
d. $50.50 ± $11.08
e. $50.50 ± $11.00
2) Which of the following is not true about the Student's t distribution?
a. It has more area in the tails and less in the center than does the normal distribution.
b. It is used to construct confidence intervals for the population mean when the population standard deviation is known.
c. It is bell shaped and symmetrical.
d. As the number of degrees of freedom increases, the t distribution approaches the normal distribution.
e. both (a) and (b)
3) Researchers determined that 60 Kleenex tissues is the average number of tissues used during a cold. Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues used during a cold: x = 52 and s = 22. Suppose the alternative we wanted to test was H1: 𝜇 < 60. The correct rejection region for 𝛼 = 0.05 is
a. reject H0 if t < -1.6604
b. reject H0 if t > 1.9842 or Z < -1.9842
c. reject H0 if t > -1.9842
d. reject H0 if t > 1.6604
e. reject H0 if t < -1.9842
n = 15 = sample size
xbar = 50.50 = sample mean
s^2 = 400 which leads to s = 20 = sample standard deviation
sigma = population standard deviation = unknown
We don't know the value of sigma and n > 30 is not the case, so we must use the T distribution.
df = degrees of freedom
df = n-1
df = 15-1
df = 14
We'll need to find the t critical value.
To do so, you could use a stats calculator like a TI83.
However, I'll use a T table such as this https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
Such a table can be found at the back of your stats textbook.
Highlight the df = 14 row and the column that mentions "confidence level = 95%" (mentioned at the bottom of the table)
The intersection of this row and column yields the approximate t critical value t = 2.145
What does this tell us?
It tells us that P(-2.145 < t < 2.145) = 0.95 approximately when df = 14.
The 0.95 is the area of the main body while 1-0.95 = 0.05 is the combined area of the two tails.
The 0.95 refers to the confidence level 95%.
E = margin of error
E = tCritical*s/sqrt(n)
E = 2.145*20/sqrt(15)
E = 11.08 approximately
The confidence interval is centered at xbar.
The radius of the interval is the margin of error.
We'll add and subtract (i.e. plus minus) the value of E to xbar so we can determine the boundaries.
L = lower bound = xbar-E
U = upper bound = xbar+E
confidence interval = xbar ± E = 50.50 ± 11.08
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