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s2008m [1.1K]
3 years ago
15

In a study of parents' perceptions of their children's size, researchers asked parents to estimate their youngest child's height

. The researchers hypothesize that parents tend to underestimate their youngest child's size because the youngest child is the baby of the family and all other family members appear larger compared to the baby. The sample of 39 parents who were surveyed underestimated their youngest child's height by 7.5 cm on average; the standard deviation for the difference in actual heights and estimated heights was 7.2 cm, and the data were approximately normal.
Determine the TS and p-value for this study.
Mathematics
1 answer:
tatyana61 [14]3 years ago
5 0

Answer:

Test statistic t=6.505

P-value=0

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that parents tend to underestimate their youngest child's size.

Then, the null and alternative hypothesis are:

H_0: \mu=0\\\\H_a:\mu> 0

The significance level is assumed to be 0.05.

The sample has a size n=39.

The sample mean is M=7.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=7.2.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{7.2}{\sqrt{39}}=1.153

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{7.5-0}{1.153}=\dfrac{7.5}{1.153}=6.505

The degrees of freedom for this sample size are:

df=n-1=39-1=38

This test is a right-tailed test, with 38 degrees of freedom and t=6.505, so the P-value for this test is calculated as (using a t-table):

P-value=P(t>6.505)=0

As the P-value (0) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is  enough evidence to support the claim that parents tend to underestimate their youngest child's size.

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Answer:

Gọi AH là đường cao kẻ từ A đến BC

Xét tam giác AHC vuông tại H

Suy ra AH = HC. cotA= 9.cot35° = 12,9

Xét tam giác AHB vuông tại H

Suy ra BH = AH.tanA = 12,9.tan61°=23,37

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Answer:

0.8973

Step-by-step explanation:

Relevant data provided in the question as per the question below:

Free throw shooting percentage = 0.906

Free throws = 6

At least = 5

Based on the above information, the probability is

Let us assume the X signifies the number of free throws

So,  Then X ≈ Bin (n = 6, p = 0.906)

P = (X = x) =  $\sum\limits_{x}^6 (0.906)^x (1 - 0.906)^{6-x}, x = 0,1,2,3,.., 6

Now

The Required probability = P(X ≥ 5) = P(X = 5) + P(X = 6)

=  $\sum\limits_{5}^6 (0.906)^5 (1 - 0.906)^{6-5} + $\sum\limits_{6}^6 (0.906)^6 (1 - 0.906)^{6-6}

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3 years ago
Consider the following probability distribution function of the random variable X which represents the number of people in a gro
Troyanec [42]

Answer:

(a) 3.55

(b) 3.45 and 1.86

(c) 0.25

(d) 0.016

Step-by-step explanation:

The random variable <em>X</em> denotes the number of people in a group (party) at a restaurant.

(a)

The formula to compute the mean is:

\mu=\sum {x\cdot P(X=x)}

Consider the Excel sheet attached.

The mean is, 3.55.

(b)

The formula to compute the variance is:

\sigma^{2}=[\sum {x^{2}\cdot P(X=x)}]-(\mu)^{2}

Consider the Excel sheet attached.

Compute the variance as follows:

\sigma^{2}=[\sum {x^{2}\cdot P(X=x)}]-(\mu)^{2}\\\\=16.05-(3.55)^{2}\\\\=3.4475\\\\\approx 3.45

The variance is, 3.45.

Compute the standard deviation as follows:

\sigma=\sqrt{\sigma^{2}}\\\\=\sqrt{3.45}\\\\=1.85742\\\\\approx 1.86

The standard deviation is, 1.86.

(c)

Compute the probability that the next party will be over 4 people as follows:

P(X>4)=P(X=5)+P(X=6)+P(X=7)+P(X=8)\\\\=0.10+0.05+0.05+0.05\\\\=0.25

Thus, the probability that the next party will be over 4 people is 0.25.

(d)

Compute the probability that the next three parties will each be over 4 people as follows:

It is provided that the three parties are independent.

P (Next 3 parties will be each over 4) = [P (X > 4)]³

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Thus, the probability that the next three parties will each be over 4 people is 0.016.

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3 years ago
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