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4 years ago

Solve the inequality -16<8-4w

Mathematics

2 answers:

miskamm [114]4 years ago

5 0

Answer:

w>3

Step-by-step explanation:

Ket [755]4 years ago

4 0

To solve an inequality, you have to isolate the variable

in order to do that, we have to put the constant on the other side by using the opposite function

so: -16<8-4w ( subtract 8 from both sides) = -24 < -4w

now, divide each side by -4. because we are dividing by a negative number, the sign flips

6 > w

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3 years ago

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Step-by-step explanation:

Data given and notation

$\bar X_{m}=39.80$ represent the mean for the sample male

$\bar X_{f}=50.30$ represent the mean for the sample female

$s_{m}=10.040$ represent the sample standard deviation for the males

$s_{f}=13.215$ represent the sample standard deviation for the females

$n_{m}=10$ sample size for the group male

$n_{f}=10$ sample size for the group female

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 mean for females is higher than the mean for males, the system of hypothesis would be:

Null hypothesis: $\mu_{f}-\mu_{m}\leq 0$

Alternative hypothesis: $\mu_{f} - \mu_{m}> 0$

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

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

And the degrees of freedom are given by $df=n_m +n_f -2=10+10-2=18$

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.

What is the test statistic?

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

$t=\frac{(50.3-39.80)-0}{\sqrt{\frac{13.215^2}{10}+\frac{10.040^2}{16}}}}=2.001$

Critical value

For this case since we have a right tailed test we need to look into the t distribution with 18 degrees of freedom a value that accumulates 0.05 of the area on the right, and on this case:

$t_{crit}=1.734$

And the rejection zone of the null hypothesis would be: A. t ≥ 1.734

For our case our calculated value is higher than the critical value so we have enough evidence to reject the null hypothesis at the 5% of significance.

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