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3 years ago

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Mathematics

1 answer:

Fiesta28 [93]3 years ago

8 0

For a set of trials $P_{desired}=\frac{desired}{total}$ . In this case our desired out come is all multiples of 3: 3,6,9 and 12. The total number of outcomes can be found as 100 by adding up all the frequencies. Thus our experimental probability is $P_{x=3n}=\frac{6+14+12+1}{100}=\frac{33}{100}=33\%$ .

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3 years ago

The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years

inn [45]

Answer:

b. The null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0$

c. Student's t distribution, as it is a two-sample test for the difference between means.

d. Test statistic t = 5.42.

P-value = 0

e. They appear to be significantly different, as the sample data gives strong evidence that the difference is greater than 0.

Step-by-step explanation:

This is a hypothesis test for the difference between populations means.

The claim is that the mean life spans in the county is significantly different for whites and nonwhites.

Then, the null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0$

The significance level is 0.05.

The sample 1 (white), of size n1=124 has a mean of 45.3 and a standard deviation of 12.7.

The sample 2 (non-white), of size n2=82 has a mean of 34.1 and a standard deviation of 15.6.

The difference between sample means is Md=11.2.

$M_d=M_1-M_2=45.3-34.1=11.2$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{12.7^2}{124}+\dfrac{15.6^2}{82}}\\\\\\s_{M_d}=\sqrt{1.301+2.968}=\sqrt{4.269}=2.066$

Then, we can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{11.2-0}{2.066}=\dfrac{11.2}{2.066}=5.42$

The degrees of freedom for this test are:

df=n_1+n_2-2=124+82-2=204

This test is a two-tailed test, with 204 degrees of freedom and t=5.42, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>5.42)=0.0000002$

As the P-value (0.0000002) 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 the mean life spans in the county is significantly different for whites and nonwhites.

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3 years ago