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Anit [1.1K]
3 years ago
15

The amount that two groups of students spent on snacks in one day is shown in the dot plots below. Which statements about the me

asures of center are true? Select three choices. The mean for Group A is less than the mean for Group B. The median for Group A is less than the median for Group B. The mode for Group A is less than the mode for Group B. The median for Group A is 2. The median for Group B is 3.

Mathematics
1 answer:
RUDIKE [14]3 years ago
8 0

Answer:

  • The mean for Group A is less than the mean for Group B.
  • The median for Group A is less than the median for Group B.
  • The mode for Group A is less than the mode for Group B.

Step-by-step explanation:

First, we can find the measures of center for each group.

<u>Group A</u>

Mode: 1

Median: (1 + 2) / 2 = 3 / 2 = 1.5

Mean: (1 * 5 + 2 * 4 + 3) / 10 = (5 + 8 + 3) / 10 = 16 / 10 = 1.6

<u>Group B</u>

Mode: 3

Median: 92 + 3) / 2 = 5 / 2 = 2.5

Mean: (1 * 3 + 2 * 2 + 3 * 4 + 5) / 10 = (3 + 4 + 12 + 5) / 10 = 24 / 10 = 2.4

From here, we can see that...

  • The mean for Group A is less than the mean for Group B.
  • The median for Group A is less than the median for Group B.
  • The mode for Group A is less than the mode for Group B.

Hope this helps!

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Answer: 3/5

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

This is a hypothesis test for the difference between populations means.

The claim is that the liquid diet yields a higher mean weight loss than the powder diet.

Then, the null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2< 0

The significance level is 0.05.

The sample 1 (powder diet group), of size n1=49 has a mean of 42 and a standard deviation of 12.

The sample 2 (liquid diet group), of size n2=36 has a mean of 45 and a standard deviation of 14.

The difference between sample means is Md=-3.

M_d=M_1-M_2=42-45=-3

The estimated standard error of the difference between means is computed using the formula:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{12^2}{49}+\dfrac{14^2}{36}}\\\\\\s_{M_d}=\sqrt{2.939+5.444}=\sqrt{8.383}=2.895

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