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Leno4ka [110]
3 years ago
9

Two cards are selected from a standard deck of 52 playing cards the first card is not replaced before The second card is selecte

d find the probability of selecting a queen and then selecting a two
Mathematics
2 answers:
Kamila [148]3 years ago
8 0

Answer:

There are 4-queen's and 4-two's

4/52 * 4/51 =0.006

Step-by-step explanation:

spin [16.1K]3 years ago
4 0

Answer:

1/2652

Step-by-step explanation:

1/52*1/51=1/2652


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3 years ago
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2 years ago
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HA: p not equal to 0.30

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C. The Randomization Condition is met.

D. The Success/Failure Condition is met.

3) Test statistic z = 2.089

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4) C. Reject H0. There is sufficient evidence to suggest that the percentage of bills paid by medical insurance has changed.

Step-by-step explanation:

1) This is a hypothesis test for a proportion.

The claim is that there is a significant change in the percent of bills being paid by medical​ insurance.

As we are looking for evidence of a difference, no matter if it is higher or lower than the null hypothesis proportion, the alternative hypothesis is defined by a unequal sign.

Then, the null and alternative hypothesis are:

H_0: \pi=0.3\\\\H_a:\pi\neq 0.3

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The independence assumption and the randomization condition are met as the bills were selected randomly from the population.

The 10% condition can not be checked, as we do not know the size of the population.

The success/failure condition is met as the products np and n(1-p) are bigger than 10 (the number of successes and failures are both bigger than 10).

3) The significance level is assumed to be 0.05.

The sample has a size n=9260.

The sample proportion is p=0.31.

 

The standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.3*0.7}{9260}}\\\\\\ \sigma_p=\sqrt{0.000023}=0.005

Then, we can calculate the z-statistic as:

z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.31-0.3-0.5/9260}{0.005}=\dfrac{0.01}{0.005}=2.089

This test is a two-tailed test, so the P-value for this test is calculated as:

\text{P-value}=2\cdot P(z>2.089)=0.0367

As the P-value (0.0184) 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 there is a significant change in the percent of bills being paid by medical​ insurance.

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NNADVOKAT [17]

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