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7nadin3 [17]
3 years ago
5

Imagine that you are doing an exhaustive study on the children in all of the daycares in your school district. You are particula

rly interested in how much time children spend in active play on weekends. You find that for this population of 2,431 children, the average number of minutes spent in active play on weekends is μ = 65.87, with a standard deviation of σ = 87.08. You select a random sample of 25 children of daycare age in this same school district. In this sample, you find that the average number of minutes the children spend in active play on weekends is M = 59.28, with a standard deviation of s = 95.79.
The difference between M and mu is due to the sampling error is ______.
Mathematics
1 answer:
alisha [4.7K]3 years ago
8 0

Answer: The difference between M and mu is due to the sampling error is <u>-6.59 .</u>

Step-by-step explanation:

As per given:  

in population, the average number of minutes spent in active play on weekends is μ = 65.87

In sample, the average number of minutes the children spend in active play on weekends is M = 59.28

Now, the difference between M and μ is due to the sampling error is M- μ

= 59.28  - 65.87

= -6.59

So, the difference between M and mu is due to the sampling error is <u>-6.59 .</u>

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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
3 years ago
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