Easy question. 10 points

You are conducting a study to see if the proportion of voters who prefer Candidate A is significantly more than 47%. With Ha : p

rodikova [14]

Answer:

The p-value is 0.0229

Step-by-step explanation:

With $H_{a}: p > 0.47$ we have an upper-tail alternative. Because the p-value is defined as the probability of getting a value at least as extreme as the value observed. The observed value is given by the test statistic z = 1.997 which comes from a standard normal distribution. Therefore, we compute the p-value in the following way P(Z > 1.997) = 0.0229, i.e., the p-value is 0.0229

3 0

3 years ago

A simple random sample of size nequals40 is drawn from a population. The sample mean is found to be 103.9, and the sample stand

yan [13]

Answer:

There is enough evidence to support the claim that the population mean is greater than 100

Step-by-step explanation:

Step 1: We state the hypothesis and identify the claim

$H_0:\mu=100$ and $H_1:\mu \:>\:100$ (claim)

Step 2: Calculate the test value.

$t=\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$

$\implies t=\frac{103.9-100}{\frac{21.9}{\sqrt{40} } }=1.1263$

Step 3: Find the P-value. The p-value obtained from a calculator is using d.f=39 and test-value 1.126 is 0.134

Step 4: We fail to reject the null hypothesis since P-value is greater that the alpha level. (0.134>0.05).

Step 5: There is enough evidence to support the claim that the population mean is greater than 100.

Alternatively: We could also calculate the critical value to obtain +1.685 for $\alpha=0.05$ and d.f=39 and compare to the test-value:

The critical value (1.685>1.126) falls in the non-rejection region. See attachment.

NB: The t- distribution must be used because the population standard deviation is not known.

5 0

4 years ago

In a survey of adults aged 57 through 85 years, it was found that 86.6% of them used at least one prescription medication. Com

OleMash [197]

Answer:

a) 272 used at least one prescription medication.

b) The 90% confidence interval estimate of the percentage of adults aged 57 through 85 years who use at least one prescription medication is (85.6%, 87.6%).

Step-by-step explanation:

Question a:

86.6% out of 3149, so:

0.866*3149 = 2727.

272 used at least one prescription medication.

Question b:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 3149, \pi = 0.866$

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.866 - 1.645\sqrt{\frac{0.866*0.134}{3149}} = 0.856$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.866 + 1.645\sqrt{\frac{0.866*0.134}{3149}} = 0.876$

For the percentage:

0.856*100% = 85.6%

0.876*100% = 87.6%.

The 90% confidence interval estimate of the percentage of adults aged 57 through 85 years who use at least one prescription medication is (85.6%, 87.6%).

7 0

3 years ago

In a certain chemical, the ratio of zinc to copper is 4 to 13. A jar of the chemical contains 468 of copper. How many grams of

Gennadij [26K]

Answer:

The answer is 144

Step-by-step explanation:

7 0

3 years ago

Can someone give me answers?

Alenkasestr [34]

Answer:

Maybe if you give a better quality photo that isn't sideways! :D

Step-by-step explanation:

7 0

3 years ago

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