Mathematics, 19.12.2019 07:31 dakotakeating4513

Location is known to affect the number, of a particular item, sold by heb pantry. two different locations, a and b, are selected on an experimental basis. location a was observed for 18 days and location b was observed for 13 days. the number of the particular items sold per day was recorded for each location. on average, location a sold 39 of these items with a sample standard deviation of 8 and location b sold 49 of these items with a sample standard deviation of 4. select a 99% confidence interval for the difference in the true means of items sold at location a and b. a) o [-1242,-7582]

b) o132.76, 45.24]

c)。8 1.76, 94.24]

d) 0-1 6.03,-3.97]

e)。[42.76, 55.24]

f. none of the above

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Answer from: Dianar8407

d) [-16.03,-3.97]

$-16.03 \leq \mu_A -\mu_B \leq -3.97$ .

Step-by-step explanation:

Notation and previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

n_A=18 represent the sample of A

n_B =13 represent the sample of B

$\bar x_A =39$ represent the mean sample for A

$\bar x_B =49$ represent the mean sample for B

s_A =8 represent the sample deviation for A

s_B =4 represent the sample deviation for B

$\alpha=0.01$ represent the significance level

Confidence =99% or 0.99

The confidence interval for the difference of means is given by the following formula:

$(\bar X_A -\bar X_B) \pm t_{\alpha/2}\sqrt{(\frac{s^2_A}{n_A}+\frac{s^2_B}{n_B})}$ (1)

The point of estimate for $\mu_A -\mu_B$ is just given by:

$\bar X_A -\bar X_B =39-49=-10$

The appropiate degrees of freedom are df=n_1+ n_2 -2=18+13-2=29

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,29)".And we see that $t_{\alpha/2}=2.756$

The standard error is given by the following formula:

$SE=\sqrt{(\frac{s^2_A}{n_A}+\frac{s^2_B}{n_B})}$

And replacing we have:

$SE=\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=2.188$

Confidence interval

Now we have everything in order to replace into formula (1):

$-10-2.756\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=-16.03$

$-10+2.756\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=-3.97$

So on this case the 99% confidence interval for the differences of means would be given by $-16.03 \leq \mu_A -\mu_B \leq -3.97$ .

d) [-16.03,-3.97]

Answer from: Quest

i dont know sorry

step-by-step explanation:

Answer from: Quest

from the figure we get,

the x axis shows the months and,

the y-axis shows the number of satisfied customer.

each line or each cell in upward direction (y-axis) shows 500.

the appropriate scale for the y - axis is: -

a = 2000 customers

b = 4000 customers

c = 6000 customers

d = 8000 customers

e = 10000 customers

f = 12000 customers

g = 14000 customers

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