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user100 [1]
3 years ago
12

You wonder if one's age is independent of whether or not one's willingness to wear masks in public. You ask 100 people whether t

hey are wearing masks in public or not and whether they are under 20 years old, between 20 and 39, between 40 and 59, or 60 or above. Using your data, you calculate a test statistic of 16.5. Using a critical value of 7.82, what is your statistical conclusion
Mathematics
1 answer:
Sladkaya [172]3 years ago
7 0

Answer:

One's age is not independent of one's willingness to wear masks in public.

Step-by-step explanation:

The hypothesis for the study can be defined as follows:

<em>H</em>₀: One's age is independent of one's willingness to wear masks in public.

<em>Hₐ</em>: One's age is not independent of one's willingness to wear masks in public.

It is provided that the test statistic value is, <em>t</em> = 16.5.

And the critical value of the test is, <em>c</em> = 7.82.

t = 16.5 > c = 7.82

Since the  test statistic value is more than the critical value, the null hypothesis will be rejected.

Thus, concluding that one's age is not independent of one's willingness to wear masks in public.

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6. Suppose that a fair coin is tossed 2 times, and the result of each toss (H or T) is recorded.
nekit [7.7K]

Answer:

a) S= {HH, HT, TH, TT}

b) P(X=0) = (2C0) (0.5)^0 (1-0.5)^{2-0}= 0.25

P(X=1) = (2C1) (0.5)^1 (1-0.5)^{2-1}= 0.5

P(X=2) = (2C2) (0.5)^2 (1-0.5)^{2-2}= 0.25

And we have the following table:

X     |     0   |     1   |      2

P(X) |  0.25 |  0.5 |  0.25

Step-by-step explanation:

Let's define first some notation

H= represent a head for the coin tossed

T= represent tails for the coin tossed

We are going to toss a coin 2 times so then the size of the sample size is 2^2 = 4

a. What is the sample space for this chance experiment?

The sampling space on this case is given by:

S= {HH, HT, TH, TT}

b. For this chance experiment, give the probability distribution for the random variable of the total number of heads observed.

The possible values for the number of heads are X=0,1,2. If we assume a fair coin then the probability of obtain heads is the same probability of obtain tails and we can find the distribution like this:

P(X=0) = (2C0) (0.5)^0 (1-0.5)^{2-0}= 0.25

P(X=1) = (2C1) (0.5)^1 (1-0.5)^{2-1}= 0.5

P(X=2) = (2C2) (0.5)^2 (1-0.5)^{2-2}= 0.25

And we have the following table:

X     |     0   |     1   |      2

P(X) |  0.25 |  0.5 |  0.25

5 0
3 years ago
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V125BC [204]
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Answer:

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Step-by-step explanation:

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