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Dominik [7]
3 years ago
15

Assume that a hypothesis test of the given claim will be conducted. Identify the type I or type II error for the test. The princ

ipal of a school claims that the percentage of students at his school that come from single- parent homes is 11 %. Identify the type II error for the test.
Mathematics
1 answer:
ololo11 [35]3 years ago
5 0

Answer:

Type I error: Concluded that p ≠ 11% when it is 11%.

Type II error: Concluded that p = 11% when it is not 11%.

Step-by-step explanation:

A <em>type I error</em> is the rejection of the null hypothesis (<em>H₀</em>) when indeed the null hypothesis is true. It is symbolized by <em>α</em>.

A <em>type II error</em> is failing to discard the null hypothesis when indeed the null hypothesis is false. It is symbolized by <em>β</em>.

In this case a  principal of a school claims that the percentage of students at his school that come from single-parent homes is 11%.

The hypothesis to test this claim is:

<em>H₀</em>: The proportion of students at the school that come from single-parent homes is 11%, i.e. <em>p</em> = 0.11.

<em>Hₐ</em>: The proportion of students at the school that come from single-parent homes is not 11%, i.e. <em>p</em> ≠ 0.11.

  • The type I error will be committed when it is concluded that the proportion of students at the school that come from single-parent homes is not 11% when in fact it is 11%.
  • The type II error will be committed when it is concluded that the proportion of students at the school that come from single-parent homes is 11% when it is not.
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igomit [66]

Answer:

The smallest positive integer solution to the given system of congruences is 30.

Step-by-step explanation:

The given system of congruences is

x=0(mod5)

x=8(mod11)

where, m and n are positive integers.

It means, if the number divided by 5, then remainder is 0 and if the same number is divided by 11, then the remainder is 8. It can be defined as

x=5m

x=11n+8

5m\cong 11n+8

Now, we can say that m>n because m and n are positive integers.

For n=1,

5m=11(1)+8=19

5m=19

19 is not divisible by 5 so m is not an integer for n=1.

For n=2,

5m=11(2)+8

5m=30

m=6

The value of m is 6 and the value of n is 2. So the smallest positive integer solution to the given system of congruences is

x=5(6)=30

Therefore the smallest positive integer solution to the given system of congruences is 30.

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3 years ago
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