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xxTIMURxx [149]

2 years ago

14

Claim: The mean pulse rate (in beats per minute) of adult males is equal to 69 bpm. For a random sample of 171 adult males,

the mean pulse rate is 67.6 bpm and the standard deviation is 11.1 bpm. Find the value of the test statistic.

1 answer:

spayn [35]2 years ago

3 0

Answer:

-1.649313114

Step-by-step explanation:

Sample size, n=171

Sample mean, x=67.6

Standard deviation, s=11.1

Study claims mean pulse $\mu=69$

Null hypothesis

$H_0: \mu=69$

$H_a: \mu \neq 69$

Test statistic is given by

$t=\frac {x-\mu}{\frac {s}{\sqrt n}}$ hence

$t=\frac {67.6-69}{\frac {11.1}{\sqrt 171}}=-1.649313114$

The test static value is -1.649313114

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Shtirlitz [24]

Answer:

1.02% probability of spending 0 dollars on coffee over the course of a five day week

7.68% probability of spending 3 dollars on coffee over the course of a five day week

23.04% probability of spending 6 dollars on coffee over the course of a five day week

34.56% probability of spending 9 dollars on coffee over the course of a five day week

25.92% probability of spending 12 dollars on coffee over the course of a five day week

7.78% probability of spending 12 dollars on coffee over the course of a five day week

Step-by-step explanation:

For each day, there are only two possible outcomes. Either the subway is on time, or it is not. Each day, the probability of the train being on time is independent from other days. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

The probability that the subway is delayed is 40%. 100-40 = 60% of the train being on time, so $p = 0.6$

The week has 5 days, so $n = 5$

She spends 3 dollars on coffee each day the train is on time.

Probabability that she spends 0 dollars on coffee:

This is the probability of the train being late all 5 days, so it is P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.6)^{0}.(0.4)^{5} = 0.0102$

1.02% probability of spending 0 dollars on coffee over the course of a five day week

Probabability that she spends 3 dollars on coffee:

This is the probability of the train being late for 4 days and on time for 1, so it is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{5,1}.(0.6)^{1}.(0.4)^{4} = 0.0768$

7.68% probability of spending 3 dollars on coffee over the course of a five day week

Probabability that she spends 6 dollars on coffee:

This is the probability of the train being late for 3 days and on time for 2, so it is P(X = 2).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{5,2}.(0.6)^{2}.(0.4)^{3} = 0.2304$

23.04% probability of spending 6 dollars on coffee over the course of a five day week

Probabability that she spends 9 dollars on coffee:

This is the probability of the train being late for 2 days and on time for 3, so it is P(X = 3).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{5,3}.(0.6)^{3}.(0.4)^{2} = 0.3456$

34.56% probability of spending 9 dollars on coffee over the course of a five day week

Probabability that she spends 12 dollars on coffee:

This is the probability of the train being late for 1 day and on time for 4, so it is P(X = 4).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 4) = C_{5,4}.(0.6)^{4}.(0.4)^{1} = 0.2592$

25.92% probability of spending 12 dollars on coffee over the course of a five day week

Probabability that she spends 15 dollars on coffee:

Probability that the subway is on time all days of the week, so $P(X = 5)$ .

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{5,5}.(0.6)^{5}.(0.4)^{0} = 0.0778$

7.78% probability of spending 12 dollars on coffee over the course of a five day week

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3 years ago