If the United States is approximately 3.8 x 10 square miles and France is approximately 2.1x10' square

uppose that we have collected information on how much a sample of households spend on clothing per year. If we are 90% confident

Verizon [17]

Answer:

the chance that the population mean will either be less than $1,500 or above $2,100 is 10%

Step-by-step explanation:

We are 90% confident that the true population mean willl be between $1,500 and $2,100

This means that there is a 90% probability that the true population mean will be between $1,500 and $2,100.

So 100-90 = 10% probability that it is outside of this interval, that is, either be less than $1,500 or above $2,100.

So the correct answer is:

the chance that the population mean will either be less than $1,500 or above $2,100 is 10%

5 0

3 years ago

Which inequality is shown in the accompanying diagram?

Olegator [25]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

In order to either condense or expand logarithms using their properties, what must be?

jasenka [17]

Answer:

Answer C

Step-by-step explanation:

The rules of logarithmic functions require the two log functions to have the same base. If they do not have the same base, you cannot apply any of the rules.

5 0

2 years ago

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A sample of 12 radon detectors of a certain type was selected, and each was exposed to 100 pCI/L of radon. The resulting reading

valkas [14]

Answer:

Null hypothesis: $\mu = 100$

Alternative hypothesis: $\mu \neq 100$

$t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921$

$p_v =2*P(t_{11}$

Step-by-step explanation:

1) Data given and notation

Data: 105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6, 107.7, 103.3, 92.4

We can calculate the sample mean and deviation for this data with the following formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}$

The results obtained are:

$\bar X=98.375$ represent the sample mean

$s=0.6.109$ represent the sample standard deviation

$n=12$ sample size

$\mu_o =100$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is equal to 100pCL/L, the system of hypothesis are :

Null hypothesis: $\mu = 100$

Alternative hypothesis: $\mu \neq 100$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

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

$t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921$

4) P-value

First we need to find the degrees of freedom for the statistic given by:

$df=n-1=12-1=11$

Since is a two sided test the p value would given by:

$p_v =2*P(t_{11}$

5) Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significant different from 100 at 5% of significance.