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3 years ago

Round 275,608 to the nearest thousand

Mathematics

2 answers:

sertanlavr [38]3 years ago

5 0

275,608 rounded to the nearest thousand is 276,000

Lilit [14]3 years ago

4 0

The answer is 276,000.

Hope that helps :]

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Read 2 more answers

Refer to the following scenario:You want to see if there is a difference between the exercise habits of Science majors and Math

bekas [8.4K]

Answer:

1. H0: P1 = P2

2. Ha: P1 ≠ P2

3. pooled proportion p = 0.542

4. P-value = 0.0171

5. The null hypothesis failed to be rejected.

At a signficance level of 0.01, there is not enough evidence to support the claim that there is significant difference between the exercise habits of Science majors and Math majors .

6. The 99% confidence interval for the difference between proportions is (-0.012, 0.335).

Step-by-step explanation:

We should perform a hypothesis test on the difference of proportions.

As we want to test if there is significant difference, the hypothesis are:

Null hypothesis: there is no significant difference between the proportions (p1-p2 = 0).

Alternative hypothesis: there is significant difference between the proportions (p1-p2 ≠ 0).

The sample 1 (science), of size n1=135 has a proportion of p1=0.607.

$p_1=X_1/n_1=82/135=0.607$

The sample 2 (math), of size n2=92 has a proportion of p2=0.446.

$p_2=X_2/n_2=41/92=0.446$

The difference between proportions is (p1-p2)=0.162.

$p_d=p_1-p_2=0.607-0.446=0.162$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{82+41}{135+92}=\dfrac{123}{227}=0.542$

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

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.542*0.458}{135}+\dfrac{0.542*0.458}{92}}\\\\\\s_{p1-p2}=\sqrt{0.001839+0.002698}=\sqrt{0.004537}=0.067$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.162-0}{0.067}=\dfrac{0.162}{0.067}=2.4014$

This test is a two-tailed test, so the P-value for this test is calculated as (using a z-table):

$\text{P-value}=2\cdot P(z>2.4014)=0.0171$

As the P-value (0.0171) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

At a signficance level of 0.01, there is not enough evidence to support the claim that there is significant difference between the exercise habits of Science majors and Math majors .

We want to calculate the bounds of a 99% confidence interval of the difference between proportions.

For a 99% CI, the critical value for z is z=2.576.

The margin of error is:

$MOE=z \cdot s_{p1-p2}=2.576\cdot 0.067=0.1735$

Then, the lower and upper bounds of the confidence interval are:

$LL=(p_1-p_2)-z\cdot s_{p1-p2} = 0.162-0.1735=-0.012\\\\UL=(p_1-p_2)+z\cdot s_{p1-p2}= 0.162+0.1735=0.335$

The 99% confidence interval for the difference between proportions is (-0.012, 0.335).

6 0

3 years ago