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UkoKoshka [18]
2 years ago
12

A small school with 60 total students records how many of their students attend on each of the 180 days in a school year. the ma

in number of students in attendance daily is 55 students and the standard deviation is 4 students. suppose we take a random sample of five School days and calculate the mean number of students in attendance on those days in each sample

Mathematics
1 answer:
horrorfan [7]2 years ago
5 0

Answer:

For the sampling distribution,

a) Mean = μₓ = 55.0 students.

b) Standard Deviation = 1.8 students.

Step-by-step explanation:

The complete Question is attached to this solution.

The Central limit theorem explains that for the sampling distribution, the mean is approximately equal to the population mean and the standard deviation of the sampling distribution is related to the population standard deviation through

σₓ = (σ/√n)

where σ = population standard deviation = 4

n = sample size = 5

Mean = population mean

μₓ = μ = 55 students.

Standard deviation

σₓ = (σ/√n) = (4/√5) = 1.789 students = 1.8 students to 1 d.p

Hope this Helps!!!

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J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.
melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

                      P.Q.  =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }  ~ t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean for mail-order sales = $74.50

\bar X_2 = sample mean for internet sales = $84.40

s_1 = sample standard deviation for mail-order purchases = $17.25

s_2 = sample standard deviation for internet purchases = $21.25

n_1 = sample of sales receipts for mail-order purchases = 16

n_2 = sample of sales receipts for internet purchases = 9

Also,  s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }  =  \sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} } = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by (\mu_1-\mu_2).

Now, 99% confidence interval for (\mu_1-\mu_2) is given by;

             = (\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_)  \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

          = (74.50-84.40) \pm (2.807  \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})

          = [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

5 0
2 years ago
A stadium has two sponsorship deals. Deal A had revenue of $100,000 and expense of $10,000. Deal B has revenue of $50,000 and ex
horsena [70]

Answer:

Deal A) 90% Profit ; Deal B) 60% Profit

Step-by-step explanation:

Multiple ways of solving this problem here is one way:

100k-10k = 90k(Revenue) : Then you make this as a percent by 100k then multiply it by 100.

100*((100,000-10,000)/100,000) = 90%

The same is true for the other equation:

100*((50,000-20,000)/50,000) = 60%


4 0
1 year ago
Read 2 more answers
A survey of 2645 consumers by DDB Needham Worldwide of Chicago for public relations agency Porter/Novelli showed that how a comp
Andre45 [30]

Answer:look it up

Step-by-step explanation:

5 0
2 years ago
The following are the hypotheses for a test of the claim that college graduation statue and cola preference are independent. H0:
Shkiper50 [21]

Answer:

\chi^2 = 0.579

And the critical value for the significance level used is:

\chi^2_{crit}= 5.991

Since the calculated value is less than the critical value we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the College graduation status and cola preference are independent

Step-by-step explanation:

For this case we want to test the following hypothesis:

Null hypothesis: College graduation status and cola preference are independent

Alternative hypothesis: College graduation status and cola preference are dependent

For this case we got a calculated statistic of:

\chi^2 = 0.579

And the critical value for the significance level used is:

\chi^2_{crit}= 5.991

Since the calculated value is less than the critical value we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the College graduation status and cola preference are independent

7 0
2 years ago
Jason is designing the packaging for a new brand of snack crackers. To fit on shelves, the package must not be more than 8 1/2 i
drek231 [11]

Answer:

Step-by-step explanation:

8 1/2 inches  = \frac{17}{2} inches

To fit on shelves, the package must not be more than 8 1/2 inches tall, it means the height of the package can be at \frac{17}{2} inches

The box must be able to hold 100 cubic units of crackers. it means the volume of the box is 100 cubic units.

The base of the package to be squared

And we have the formula:

V = h*S (Area of the base)

=> S = V/h = 100 :  \frac{17}{2} = \frac{200}{17}  

=> the side of square: \sqrt{\frac{200}{17} }

The dimensions for Jason's design are:

The height: \frac{17}{2} inches  

The length and width: \sqrt{\frac{200}{17} }

6 0
2 years ago
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