BRAINIEST PLZZZ

Clarinex is a drug used to treat asthma. In clinical tests of this drug, 1655 patients were treated with 5- mg doses of Clarinex

bagirrra123 [75]

Answer:

$z=\frac{0.021 -0.012}{\sqrt{\frac{0.012(1-0.012)}{1655}}}=3.363$

$p_v =P(z>3.363)=0.00039$

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interest is significantly higher than 0.012 (1.2%)

Step-by-step explanation:

Data given and notation

n=1655 represent the random sample taken

$\hat p=0.021$ estimated proportion of interest

$p_o=0.012$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

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

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that true proportions is higher than 0.012.:

Null hypothesis: $p \leq 0.012$

Alternative hypothesis: $p > 0.012$

When we conduct a proportion test we need to use the z statisitic, and the is given by:

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

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.021 -0.012}{\sqrt{\frac{0.012(1-0.012)}{1655}}}=3.363$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.01$ . The next step would be calculate the p value for this test.

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

$p_v =P(z>3.363)=0.00039$

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interest is significantly higher than 0.012 (1.2%)

3 0

3 years ago

Cómo resolver sopa de números encontrando resultados de divisiones por ejemplo 436/2

Rzqust [24]

Answer:

Lo resuelves como de costumbre :)

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

NutritionCalc Plus uses the equations from the National Academy of Medicine to calculate the Estimated Energy Requirement (EER)

olga nikolaevna [1]

Answer:

The correct option is (A).

Step-by-step explanation:

The equation for the Estimated Energy Requirement (EER) of 19 or more years older men is:

$EER = 662 - (9.53 \times AGE) + PA \times (15.91 \times WT + 539 \times HT)$

Here,

AGE = age counted in years

PA = appropriate physical activity factor

WT = weight in kilograms

HT = height in meters

The physical activity factor for men and women is:

Activity Level PA (men) PA (Women)

Sedentary 1.00 1.00

Low 1.11 1.12

Active 1.25 1.27

Very Active 1.48 1.45

Given:

AGE = 22 year-old male college student

PA = low = 1.11

WT = 180 pounds = 81.6466 kg

HT = 5'10" = 1.778 m

Compute the value of EER as follows:

$EER = 662 - (9.53 \times AGE) + PA \times (15.91 \times WT + 539 \times HT)$

$= 662 - (9.53 \times 22) + 1.11 \times (15.91 \times 81.6466 + 539 \times 1.778)\\\\=662-209.66+1.11\times (1298.997406+958.342)\\\\=452.34+2505.64674066\\\\=2957.98674066\\\\\approx 2962$