The claim is that the proportion of adults who smoked a cigarette in the past week is less than 0.35, and the sample statistics include n = 1168 subjects with 385 saying that they smoked a cigarette in the past week. Find the value of the test statistic

Answer:

Test statistic is z = -1.46

Step-by-step explanation:

Let's first of all define the hypotheses:

Null hypothesis:

H0: p = 0.35, i.e 35% in the sample of 1,168 adults have smoked cigarettes in the previous week.

Alternative hypothesis:

Ha: p < 0.35, i.e less than 35% in the sample of 1,168 adults have smoked cigarette in the previous week.

The sample size is, n = 1,168 while the number of adults who smoked in the previous week would be; x = 385

Therefore, the sample proportion of adults who smoked in the previous week would be calculated as;

p^ = x/n = 385/1168 ≈ 0.3296

Now, from Central Limit Theorem for large samples, The sampling distribution of the sample proportion p^, will have a mean of μ = p = 0.35

Formula for standard deviation is;

σ = √[p (1 – p)/n]

σ = √(0.35 × (1 – 0.35)/1168)

σ = √0.0001947774

σ = 0.014

Formula for test statistic is;

z = (p^ - p)/σ

z = (0.3296 - 0.35)/0.014

z = - 1.46

3 0

3 years ago

vRichard has a red card (R) and a black card (B). He picks a card randomly from the pair and then rolls a six-sided die. The sam

Alik [6]

The Sample space for three cards is

$S=\left\{\left(B,1\right),\left(R,1\right),\left(G,1\right),\left(B,2\right),\left(R,2\right),\left(G,2\right),\left(B,3\right),\left(R,3\right),\left(G,3\right),\left(B,4\right),\left(R,4\right),\left(G,4\right)$ $\left(B,5\right),\left(R,5\right),\left(G,5\right),\left(B,6\right),\left(R,6\right),\left(G,6\right) \right \}$

There are two cards red(R) and black(B) and a six-sided die.

The event is one card is picked randomly and then rolls the die.

So, the Sample space is as follows.

$S=\left\{\left(B,1\right),\left(R,1\right),\left(B,2\right),\left(R,2\right),\left(B,3\right),\left(R,3\right),\left(B,4\right),\left(R,4\right),\left(B,5\right),\left(R,5\right),\left(B,6\right),\left(R,6\right) \right \}$

<h3>Find Sample space if there are three cards.</h3>

Consider there were three cards red(R), black(B) and green(G) a six-sided die.

The event is one card is picked randomly and then rolls the die.

So, the sample space will be as follows,

$S=\left\{\left(B,1\right),\left(R,1\right),\left(G,1\right),\left(B,2\right),\left(R,2\right),\left(G,2\right),\left(B,3\right),\left(R,3\right),\left(G,3\right),\left(B,4\right),\left(R,4\right),\left(G,4\right)$ $\left(B,5\right),\left(R,5\right),\left(G,5\right)\left(B,6\right),\left(R,6\right),\left(G,6\right) \right \}$ .

Learn more about sample space here :-

brainly.com/question/24273864

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7 0

2 years ago

One pipe fills a storage pool in 8 hours. A second pipe fills the same pool in 4 hours. When a third pipe is added and all three

steposvetlana [31]

The first pipe fills 1/8 of a pool in one hr.
The second pipe fills 1/4 of a pool in one hr.

(1/8+1/4)2= 3/4 done by the first two pipes in 2 hours.
Therefore the last pipe must do 1 1/4 of the filling in 2 hours, which is 5/8 per hour.

5/8x=1
x=8/5

It takes the third pipe 1 hour and 36 minutes to fill it by itself.

7 0

3 years ago