All categories

3 years ago

The population of a town increased from 3500 in 2005 to 5600 in 2009. Find the absolute and relative (percent) increase.

Mathematics

1 answer:

Ket [755]3 years ago

6 0

Answer:

Absolute increase in population will be 2100

And percentage increase will be 60 %

Step-by-step explanation:

We have given population in 2005 is 3500

And population in 2009 is 5600

So absolute increase in population = 5600-3500 = 2100

We have to find the percentage increase in population

So relative percentage increase in population will be $\frac{2100}{3500}\times 100=60$ %

So population will increase by 60 %

You might be interested in

A two-factor study with two levels of factor A and three levels of factor B uses a separate group of n = 5 participants in each

Mashcka [7]

Answer: There are 30 participants are needed for the entire study.

Step-by-step explanation:

Since we have given that

Number of levels of factor A = 2

Number of levels of factor B = 3

Number of participants in each treatment condition = 5

So, the number of participants are needed for the entire study is given by

$2\times 3\times 5\\\\=6\times 5\\\\=30$

Hence, there are 30 participants are needed for the entire study.

7 0

3 years ago

A random sample of 49 measurements from one population had a sample mean of 15, with sample standard deviation 5. An independent

kotykmax [81]

Answer:

Comparing the p value with the significance level $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly different.

Step-by-step explanation:

$\bar X_{1}=15$ represent the mean for sample 1

$\bar X_{2}=18$ represent the mean for sample 2

$s_{1}=5$ represent the sample standard deviation for 1

$s_{f}=6$ represent the sample standard deviation for 2

$n_{1}=49$ sample size for the group 2

$n_{2}=64$ sample size for the group 2

$\alpha=0.01$ Significance level provided

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the difference in the population means, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}=0$

Alternative hypothesis: $\mu_{1} - \mu_{2}\neq 0$

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=49+64-2=111$

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:

$t=\frac{(15-18)-0}{\sqrt{\frac{5^2}{49}+\frac{6^2}{64}}}}=-2.897$

P value

Since is a bilateral test the p value would be:

$p_v =2*P(t_{111}$

7 0

3 years ago

Claim: The standard deviation of pulse rates of adult males is less than 11 bpm. For a random sample of 132 adult males, the pu

seraphim [82]

Answer:

$\chi^2 =\frac{132-1}{121} 114.49 =123.952$

Step-by-step explanation:

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

$n=132$ represent the sample size

$\alpha$ represent the confidence level

$s^2 =10.7^2 =114.49$ represent the sample variance obtained

$\sigma^2_0 =11^2=121$ represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance specification is violated, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \geq 121$

Alternative hypothesis: $\sigma^2$

Calculate the statistic

For this test we can use the following statistic:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

$\chi^2 =\frac{132-1}{121} 114.49 =123.952$

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 131. And since is a right tailed test the p value would be given by:

$p_v =P(\chi^2$