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MakcuM [25]
3 years ago
14

A social psychologist predicted that ratings of an individual’s social desirability would be influenced by their physical attrac

tiveness. Participants received photos of attractive, average, or unattractive individuals. The researcher found that the more attractive the individual, the higher their rating of social desirability. This difference was significant at the 0.05 level.
Required:
a. What is the null hypothesis?
b. What is the research hypothesis?
c. What would be the Type I error?
d. What would be the Type II error?
e. What is the probability of a Type I error?
Mathematics
1 answer:
marshall27 [118]3 years ago
3 0

Answer:

It is stated in the query that the more attractive the individual, the higher their rank of desirability.

Step-by-step explanation:

This means that we should conclude that the alternative explanation of "higher degree of desire contributes to higher levels of social attractiveness"

In the event of a null hypothesis, we will believe that the degree of attractiveness has no impact on the score of cultural desirability.

(1) Null hypothesis:- The level of attraction does not have an impact on the rating of social desirability.

(2) Alternate hypothesis:- The level of attraction has a positive effect on the ranking of social desirability.

(3) The error of Type I is characterized as the rejection of the true null hypothesis. Therefore, in this case, the type I mistake would be that "the level of attraction has a positive effect on the rating of social desirability while the level of attraction does not have an impact on the rating of social desirability."

(4) The error of type II is characterized as the failure to reject the false null hypothesis. Thus, in this situation, the Type II error would be that "the level of attraction has no impact on the rating of social desirability while the real level of attraction has a positive effect on the rating of social desirability."

(5) We know that the likelihood of having a Form I error is equal to the degree of significance (in percent terms). The significance level, in this case, is 0.05, so the likelihood of making a form I error is 0.05 * 100=5 percent.

Therefore, there is a 5 percent risk of making a Type I mistake.

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In a quadrilateral ABCD, AO and BO are bisectors of angle A and angle B respectively
worty [1.4K]

Answer: ABCD is a quadrilateral

To prove : ∠AOB=

2

1

​

(∠C+∠D)

AO and BO is bisector of A and B

∠1=∠2∠3=∠4...(1)

∠A+∠B+∠C+∠D=360

(Angle sum property)

2

1

​

(∠A+∠B+∠C+∠D)=180...(2)

In △AOB

∠1+∠3+∠5=

2

1

​

(∠A+∠B+∠C+∠D)

∠1+∠3+∠5=∠1+∠3+

2

1

​

(∠C+∠D)

∠AOB=

2

1

​

(∠C+∠D)

​

Explanation: In a quadrilateral ABCD. AO and BO are bisectors of angle A and angle B respectively. Prove that ∠AOB=

2

1

​

{∠C+∠D}.

3 0
3 years ago
The mark in a subject for 12 students are as follows 31`, 37 ,35,38 ,42 ,23,17,18 ,35,25, 35,29
just olya [345]

Answer:

Mean (Average): 30.416666666667

Median: 33

Range: 25

Mode: 35, appeared 3 times

Step-by-step explanation:

8 0
2 years ago
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3 years ago
What is 24/16 in simplest form?
-Dominant- [34]
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6 0
3 years ago
Eric's class consists of 12 males and 16 females. If 3 students are selected at random, find the probability that they
Reptile [31]

Answer:

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

Step-by-step explanation:

Let 'M' be the event of selecting males n(M) = 12

Number of ways of choosing 3 students From all males and females

n(M) = 28C_{3} = \frac{28!}{(28-3)!3!} =\frac{28 X 27 X 26}{3 X 2 X 1 } = 3,276

Number of ways of choosing 3 students From all males

n(M) = 12C_{3} = \frac{12!}{(12-3)!3!} =\frac{12 X 11 X 10}{3 X 2 X 1 } =220

The probability that all are male of choosing '3' students

P(E) = \frac{n(M)}{n(S)} = \frac{12 C_{3} }{28 C_{3} }

P(E) =  \frac{12 C_{3} }{28 C_{3} } = \frac{220}{3276}

P(E) = 0.067 = 6.71%

<u><em>Final answer</em></u>:-

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

3 0
4 years ago
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