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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

3 years ago

Heidi's Accessories bought 50 necklaces for $10 each on account. The invoice included a 6% sales tax and payment terms of 2/10,

garri49 [273]

Answer:

The correct answer is B. Accounts payable for US$ 53.oo

Step-by-step explanation:

1. Let's recall what is the Journal entry when a company return part of the inventory previously purchased:

Debit

The amount Heidi's accessories owed to the supplier would have been sitting as a credit on the accounts payable account because the invoice has not been payed. In the case of this return, we should debit it because we should diminish the amount due.

Credit

The goods Heidi's accessories are returning and therefore, the asset of inventory decreases. The credit to purchase returns reduces the value of the total purchases. That's why options C and D are incorrect.

2. Now, let's calculate the amount of the return that Heidi's accessories made, this way:

Return = Number of necklaces * Price of each necklace + Sales taxes

Replacing with the real values, we have:

Return = 5 * 10 + (5 * 10) * 0.06

Return = 50 + (50) * 0.06

Return = 50 + 3 = 53

The correct answer is B. Accounts payable for US$ 53.oo

7 0

3 years ago

Help me Please help me please help me please help me please

ira [324]

Answer: