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Bingel [31]
3 years ago
10

350 people visited a zoo the zoo has four types of tickets the manager wants a sample of size 30, stratified by ticket type. how

man teenagers should be be in the sample
Mathematics
2 answers:
hram777 [196]3 years ago
7 0

Answer:

10

Step-by-step explanation:

NARA [144]3 years ago
5 0

Answer:

The number of teenagers in the stratified sample of equal proportion is 30 teenagers

Step-by-step explanation:

Whereby tickets are sold to only adults male and female and teenagers, boys and girls, we have the following groups

Group 1: Female adult

Group 2: Male adult

Group 3: Teenage boys

Group 4: Teenage girls

In stratified sampling, the types of people that visit the zoo (which is the target population) are identified and the appropriate proportion of each of the identified types is determined such that the sample is representative of the population

Where equal number of each group are observed to have visited the zoo, then, the appropriate sample size of the teenager is found as follows;

Number of groups identified = 4

Sample size = 30

Appropriate proportion of each group = 1/4

Number of teenage boys in the sample = 1/4×30 = 15

Number of teenage girls in the sample = 1/4×30 = 15

Total number of teenagers in the sample = 15 + 15 = 30 teenagers.

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3 years ago
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3 years ago
Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoc
Bingel [31]

Answer:

We need a sample size of at least 719

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?

This is at least n, in which n is found when M = 0.3, \sigma = 4.103. So

M = z*\frac{\sigma}{\sqrt{n}}

0.3 = 1.96*\frac{4.103}{\sqrt{n}}

0.3\sqrt{n} = 1.96*4.103

\sqrt{n} = \frac{1.96*4.103}{0.3}

(\sqrt{n})^{2} = (\frac{1.96*4.103}{0.3})^{2}

n = 718.57

Rouding up

We need a sample size of at least 719

6 0
3 years ago
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