Question

It is known that roughly 2/3 of all human beings have a dominant right foot or eye. Is there also right-sided dominance in kissing behavior? An article reported that in a random sample of 130 kissing couples, both people in 84 of the couples tended to lean more to the right than to the left. (Use α = 0.05.) If 2/3 of all kissing couples exhibit this right-leaning behavior, what is the probability that the number in a sample of 130 who do so differs from the expected value by at least as much as what was actually observed? (Round your answer to three decimal places.)

Answer 1

Answer:

1) The calculated statistic is in the acceptance region so the decision is to not reject the null hypothesis. This means that at a 5% level the population proportion of couples that lean to right while kissing is 2/3.

2) P(84 ≤ X ≤ 87) = 0.21226

Step-by-step explanation:

Hello!

1)

The study variable is:

X: Number of couples that lean to the right while kissing out of 130 kissing couples.

This variable is discrete and has a binomial distribution, the success of the binomial experiment is "lean to the right" and the study parameter is the population proportion of couples that lean to the right while kissing (p)

The hypothesis is that 2/3 of people leans to the right while kissing, symbolically: p= 2/3 (or p= 0.67)

So the statistical hypothesis is:

H₀: p= 2/3

H₁: p≠2/3

α: 0.05

This test has a two-tailed critical region and the distribution to use is the standard normal.

$Z_{\alpha /2} = Z_{0.025}= -1.96$$Z_{1-\alpha /2} = Z_{0.975}= 1.96$

If Z≤ -1.96 or if Z ≥ 1.96 you will reject the null hypothesis.

Data:

Sample n= 135

Sample proportion 'p= 84/135= 0.62 (28/45)

$Z= \frac{('p - p)}{\sqrt{\frac{'p(1-'p)}{n} } }$≈ N(0;1)

$Z= \frac{(0.62 - 2/3)}{\sqrt{\frac{0.62*0.38}{130} } }$

Z= -1.10

The calculated statistic is in the acceptance region so the decision is to not reject the null hypothesis. This means that at a 5% level the population proportion of couples that lean to right while kissing is 2/3.

2)  

If 2/3 of all kissing couples exhibit this right-leaning behavior, what is the probability that the number in a sample of 130 who do so differs from the expected value by at least as much as what was actually observed? (Round your answer to three decimal places.)

The expected number of right kissers is calculated as:

'x= n*p= 130*2/3= 86.67≅ 87

You need to calculate the probability of the kissers being between the expected (86) and observed (84) coumples:

Remember using CLT we approximated the distribution of the sample proportion to normal: 'p ≈ N(np, np(1-p))

Mean= np= 86.67≅ 87

Standard deviation: √np(1-p)= √130*2/3(1-2/3)= 5.37

P(84 ≤ X ≤ 87) = P(X ≤ 87) - P(X ≤ 84)

P(Z ≤ (87-87)/5.37) - P(Z ≤ (84-87)/5.37) =  P(Z ≤ 0) - P(Z ≤ -0.56) = 0.5 - 0.28774= 0.21226

Answer 2

The probability that the number in a sample of 130 who do so differs from the expected value is 0.6198.

How to calculate probability?

From the information, the critical value is given as 84. Therefore, the mean will be:

= np = 130 × 2/3 = 86.67

The standard deviation is also 5.3748. The corresponding z score will be:

= (84 - 86.67)/5.3748

= -0.50.

Therefore, the left tailed area will be:

P(z < -0.50) = 0.3099

Since, it's two tailed, we'll multiply by 2. This will be:

= 2 × 0.3099

= 0.6198

In conclusion, the probability is 0.6198.

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