I know the answer is 60 but how would I put it in words

Leviafan [203]

Answer:

177 hope this helps

5 0

2 years ago

A manufacturer of a new medication on the market for Alzheimer's disease makes a claim that the medication is effective in 65% o

NeX [460]

Answer:

$z=\frac{0.639 -0.65}{\sqrt{\frac{0.65(1-0.65)}{180}}}=-0.309$

$p_v =P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults with the medication was effective is not significantly less than 0.65

Step-by-step explanation:

Data given and notation

n=180 represent the random sample taken

X=115 represent the adults with the medication was effective

$\hat p=\frac{115}{180}=0.639$ estimated proportion of adults with the medication was effective

$p_o=0.65$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

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 proportion is less than 0.65.:

Null hypothesis: $p \geq 0.65$

Alternative hypothesis: $p < 0.65$

When we conduct a proportion test we need to use the z statisitc, 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:

$z=\frac{0.639 -0.65}{\sqrt{\frac{0.65(1-0.65)}{180}}}=-0.309$

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 $\alpha=0.05$ . The next step would be calculate the p value for this test.

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

$p_v =P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults with the medication was effective is not significantly less than 0.65

3 0

2 years ago

- a - 5 + 12a - 9 + 7

11Alexandr11 [23.1K]

The answer is 11a -7 don’t go on brainly for math problems go on photo math. On photo math all you have to do is take a pic of the equation and it will answer it to you right away.

5 0

3 years ago

The Jones family and the Smith family both went out for a night at the movies. The Jones family bought 5 drinks and 2 bags of po

kolbaska11 [484]

It's simple. You have to subtract FIRST 18.75-5=13.75 for the Jones Family. Then you divide 13.75 by 2 bags of popcorn which equals $6.87 for the Jones Family.

For Smith Family, subtract 31.50-4=27.50. Then you divide 27.50 by 6 bags of popcorn which equals $4.58

4 0

2 years ago

A ship sails from harbour h on a bearing of 084 for 340 km until it reaches point P. It the sails on a bearing of 210 for 160km

sattari [20]

Answer:

QH = 227.8 km ≅ 228 km

Step-by-step explanation:

∵ The bearing from H to P is 084°

∵ The bearing from P to Q is 210°

∵ The distance from H to P = 340 km

∵ The distance from P to Q = 160 km

∴ The angle between 340 and 160 = 360 - 210 - (180 - 84) = 54°

( 180 - 84) ⇒ interior supplementary

By using cos Rule:

(QH)² = (PH)² + (PQ)² - 2(PH)(PQ)cos∠HPQ

(QH)² = 340² + 160² - 2(340)(160)cos(54) = 51904.965

∴ QH = 227.8 km ≅ 228 km

7 0

2 years ago

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