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jok3333 [9.3K]
3 years ago
14

There are 400 animals that live at a zoo. You find that 22 of 65 randomly chosen animals are

Mathematics
1 answer:
snow_tiger [21]3 years ago
8 0

Answer:

About 135.

Step-by-step explanation:

As the sample is random the number of monkeys likely to be in the zoo

=  (22/65) * 400

= 135.38

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3 0
3 years ago
An article presents a study of the failure pressures of roof panels. A sample of 15 panels constructed with 8-inch nail spacing
wariber [46]

Answer:

There is sufficient evidence to conclude that six inches spacing has a higher mean failure pressure than 8 inches

Reject H_0\ at\ 5% level of significance

Step-by-step explanation:

From the question we are told that:

Sample size 1 n_1=15

Spacing 1 s_1=8

Mean 1 \=x_1=8.48

Standard deviation 1 \sigma_1=0.96 kPa

Sample size 2 n_2=16

Spacing 2 s_2=6

Mean 1 \=x_1=9.93

Standard deviation 1\sigma_1=1.02 kPa

Generally the hypothesis is mathematically given as

Null H_0:\mu_1-\mu_2

Alternative H_a:\mu_1\mu_2

Generally the equation for pooled estimate is mathematically given by

  S=\sqrt{\frac{(n_1-1)\sigma_1^2+(n_2-1)\sigma_2^2}{n_1+n_2-2} }

Therefore

  S=\sqrt{\frac{(15-1)0.96^2+(16-1)1.02^2}{15+15-2} }

 S=0.9905

Generally the equation for test statistics is mathematically given by

 t=\frac{\=x_1+\=x_2}{S\sqrt{\frac{1}{n _1}+\frac{1}{n_2}}}

 t=\frac{\=8.48-9.93}{0.9905\sqrt{\frac{1}{15}+\frac{1}{15}}}

 t=-4.00091

Therefore From table

 P value=P(t_{15+15-2}=4.00091)

 P value=0.00041918

 P

Therefore

There is sufficient evidence to conclude that six inches spacing has a higher mean failure pressure than 8 inches

Reject H_0\ at\ 5% level of significance

3 0
3 years ago
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