How to do 6-12÷3+8-9

• A researcher claims that less than 40% of U.S. cell phone owners use their phone for most of their online browsing. In a rando

antiseptic1488 [7]

Answer:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p > α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

Step-by-step explanation:

Set up hypotheses:

Null hypotheses = H₀: p = 0.40

Alternate hypotheses = H₁: p < 0.40

Determine the level of significance and Z-score:

Given level of significance = 1% = 0.01

Since it is a lower tailed test,

Z-score = -2.33 (lower tailed)

Determine type of test:

Since the alternate hypothesis states that less than 40% of U.S. cell phone owners use their phone for most of their online browsing, therefore we will use a lower tailed test.

Select the test statistic:

Since the sample size is quite large (n > 30) therefore, we will use Z-distribution.

Set up decision rule:

Since it is a lower tailed test, using a Z statistic at a significance level of 1%

We Reject H₀ if Z < -1.645

We Reject H₀ if p ≤ α

Compute the test statistic:

$Z = \frac{\hat{p} - p}{ \sqrt{\frac{p(1-p)}{n} }}$

$Z = \frac{0.31 - 0.40}{ \sqrt{\frac{0.40(1-0.40)}{100} }}$

$Z = \frac{- 0.09}{ 0.048989 }$

$Z = - 1.84$

From the z-table, the p-value corresponding to the test statistic -1.84 is

p = 0.03288

Conclusion:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p > α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

8 0

4 years ago

5ab^2 - 12a^2b + 3ab + ab^2 + 4a^2b

Natali [406]

By combining like terms, the answer is a

8 0

3 years ago

Solve for r<br> 2r = 7/10

deff fn [24]

Answer:

7/20

Step-by-step explanation:

given,

2r = 7/10

r= 7/10*1/2

r= 7/20

6 0

3 years ago

A wheel has 10 equally sized slices numbered from 1 to 10.

Ksenya-84 [330]

The random probability of event X, wheel stopping on a white slice is 0.9 while the probability of not X, wheel stopping on Grey slice is 0.1

Total numbers on wheel = total possible outcomes = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Grey colored portion = {3}

White colored portion = {1, 2, 4, 5, 6, 7, 8, 9, 10}

X = Event that wheel stops on a white slice

P(X) = number of white slices ÷ total number of slices

P(X) = 9 / 10 = 0.9

P(not X) = 1 - P(X)

P(not X) = 1 - 0.9 = 0.1

Therefore, the probability that wheel stops on a white slice is 0.9 while, the probability that wheel does not stop on a white slice is 0.1

Learn more :brainly.com/question/18153040

4 0

3 years ago

Suppose we are interested in analyzing the weights of NFL players. We know that on average, NFL players weigh 247 pounds with a

tigry1 [53]

Answer:

$SE= \frac{\sigma}{\sqrt{n}} = \frac{47}{\sqrt{30}}= 8.58$

$ME= 1.64 *\frac{\sigma}{\sqrt{n}} = 1.64*\frac{47}{\sqrt{30}}= 14.073$

$\bar X - ME = 237- 14.073 = 222.927$

$\bar X + ME = 237+ 14.073 = 251.073$

$n=(\frac{1.640(47)}{5})^2 =237.65 \approx 238$

So the answer for this case would be n=238 rounded up to the nearest integer

Null hypothesis: $\mu \geq 247$

Alternative hypothesis: $\mu$

$z=\frac{237-247}{\frac{47}{\sqrt{30}}}=-1.165$

$p_v =P(z$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is lower than 247 at 10% of significance.

Step-by-step explanation:

For this case we have the following data given:

$\bar X =237$ represent the sample mean

$\sigma = 47$ represent the population deviation

$n =30$ represent the sample size selected

$\mu_0 = 247$ represent the value that we want to test.

The standard error for this case is given by:

$SE= \frac{\sigma}{\sqrt{n}} = \frac{47}{\sqrt{30}}= 8.58$

For the 90% confidence the value of the significance is given by $\alpha=1-0.9 = 0.1$ and $\alpha/2 = 0.05$ so we can find in the normal standard distribution a quantile that accumulates 0.05 of the area on each tail and we got:

$z_{\alpha/2}= 1.64$

And the margin of error would be:

$ME= 1.64 *\frac{\sigma}{\sqrt{n}} = 1.64*\frac{47}{\sqrt{30}}= 14.073$

The confidence interval for this case would be given by:

$\bar X - ME = 237- 14.073 = 222.927$

$\bar X + ME = 237+ 14.073 = 251.073$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

Replacing into formula (b) we got:

$n=(\frac{1.640(47)}{5})^2 =237.65 \approx 238$

So the answer for this case would be n=238 rounded up to the nearest integer

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is lower than 247 pounds, the system of hypothesis would be:

Null hypothesis: $\mu \geq 247$

Alternative hypothesis: $\mu$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-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".

Calculate the statistic

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

$z=\frac{237-247}{\frac{47}{\sqrt{30}}}=-1.165$

P-value

Since is a left tailed test the p value would be:

$p_v =P(z$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is lower than 247 at 10% of significance.

5 0

3 years ago