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viktelen [127]
3 years ago
8

The results from a random sample of 150 students' after-school activities at a school are shown. 68 students prefer soccer. 59 s

tudents prefer basketball. 23 students prefer swimming. Part A If there are a total of 500 students in the school, what is the best approximation for how many students in the school prefer basketball? 197 students 197 students 227 students 227 students 295 students 295 students 303 students 303 students
Mathematics
1 answer:
VladimirAG [237]3 years ago
7 0

Answer:

≈ 197 students

Step-by-step explanation:

Let's start by writing what we know.

150 students sampled

68 prefer soccer

59 prefer basketball

23 prefer swimming

Part A

If 59 students prefer basketball out of the 150 students sampled then we divide 59/150 in order to get the percentage of students that prefer basketball.

59/150 ≈ 39.33%       Remember this is only a sample of the students so it is approximate.

Now that we have the percentage, we can multiply it by the entire student body to get an approximate number of all students that prefer basketball.

39.33% = .3933

.3933 * 500 = 196.66  ≈ 197 students   (because we can't have a partial student we round up)

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tia_tia [17]

Answer:

Using a formula, the standard error is: 0.052

Using bootstrap, the standard error is: 0.050

Comparison:

The calculated standard error using the formula is greater than the standard error using bootstrap

Step-by-step explanation:

Given

Sample A                          Sample B

x_A = 30                              x_B = 50

n_A = 100                             n_B =250

Solving (a): Standard error using formula

First, calculate the proportion of A

p_A = \frac{x_A}{n_A}

p_A = \frac{30}{100}

p_A = 0.30

The proportion of B

p_B = \frac{x_B}{n_B}

p_B = \frac{50}{250}

p_B = 0.20

The standard error is:

SE_{p_A-p_B} = \sqrt{\frac{p_A * (1 - p_A)}{n_A} + \frac{p_A * (1 - p_B)}{n_B}}

SE_{p_A-p_B} = \sqrt{\frac{0.30 * (1 - 0.30)}{100} + \frac{0.20* (1 - 0.20)}{250}}

SE_{p_A-p_B} = \sqrt{\frac{0.30 * 0.70}{100} + \frac{0.20* 0.80}{250}}

SE_{p_A-p_B} = \sqrt{\frac{0.21}{100} + \frac{0.16}{250}}

SE_{p_A-p_B} = \sqrt{0.0021+ 0.00064}

SE_{p_A-p_B} = \sqrt{0.00274}

SE_{p_A-p_B} = 0.052

Solving (a): Standard error using bootstrapping.

Following the below steps.

  • Open Statkey
  • Under Randomization Hypothesis Tests, select Test for Difference in Proportions
  • Click on Edit data, enter the appropriate data
  • Click on ok to generate samples
  • Click on Generate 1000 samples ---- <em>see attachment for the generated data</em>

From the randomization sample, we have:

Sample A                          Sample B

x_A = 23                              x_B = 57

n_A = 100                             n_B =250

p_A = 0.230                          p_A = 0.228

So, we have:

SE_{p_A-p_B} = \sqrt{\frac{p_A * (1 - p_A)}{n_A} + \frac{p_A * (1 - p_B)}{n_B}}

SE_{p_A-p_B} = \sqrt{\frac{0.23 * (1 - 0.23)}{100} + \frac{0.228* (1 - 0.228)}{250}}

SE_{p_A-p_B} = \sqrt{\frac{0.1771}{100} + \frac{0.176016}{250}}

SE_{p_A-p_B} = \sqrt{0.001771 + 0.000704064}

SE_{p_A-p_B} = \sqrt{0.002475064}

SE_{p_A-p_B} = 0.050

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Answer:No

Step-by-step explanation:

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