Mathematics, 23.04.2020 04:25 jmanrules200

Multiple-choice questions on Advanced Placement exams have five options: A, B, C, D, and E. A random sample of the correct choice on 400 multiple-choice questions on a variety of AP exams1 shows that B was the most common correct choice, with 90 of the 400 questions having B as the answer. Does this provide evidence that B is more likely to be the correct choice than would be expected if all five options were equally likely? Show all details of the test. The data are available in APMultipleChoice.

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Testing the hypothesis, it is found that since the p-value of the test is of 0.1056 > 0.05, it does not provide evidence that B is more likely to be the correct choice than would be expected if all five options were equally likely.

At the null hypothesis, it is tested if all of them are equally as likely, that is, the proportion of B is $p = \frac{1}{5} = 0.2$ . Thus:

H_0: p = 0.2

At the alternative hypothesis, it is tested if B is more likely, that is, if the proportion of B is more than 0.2. Thus:

H_1: p 0.2

The test statistic is given by:

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

In which:

$\overline{p}$ is the sample proportion. p is the proportion tested at the null hypothesis. n is the sample size.

In this problem, the parameters are: $p = 0.2, n = 400, \overline{p} = \frac{90}{400} = 0.225$ .

The value of the test statistic is:

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

$z = \frac{0.225 - 0.2}{\sqrt{\frac{0.2(0.8)}{400}}}$

z = 1.25

The p-value of the test is the probability of finding a sample proportion above 0.225, which is 1 subtracted by the p-value of z = 1.25.

Looking at the z-table, z = 1.25 has a p-value of 0.8944.

1 - 0.8944 = 0.1056.

Since the p-value of the test is of 0.1056 > 0.05, it does not provide evidence that B is more likely to be the correct choice than would be expected if all five options were equally likely.

A similar problem is given at link

Answer from: sweetmochi13

Option B is not the most common correct choice.

Step-by-step explanation:

The multiple-choice questions on Advanced Placement exams have five options: A, B, C, D, and E.

The probability that any of these five option is the correct answer is:

$p=\frac{1}{5}=0.20$

A random sample of 400 multiple-choice questions on Advanced Placement exam are selected.

The results showed that 90 of the 400 questions having B as the answer.

To test the hypothesis that option B is more likely the correct answer for most question, the hypothesis can be defined as:

H₀: All the options are equally probable, i.e. p = 0.20.

Hₐ: Option B is more likely the correct option, i.e. p > 0.20.

Compute the sample proportion as follows:

$\hat p=\frac{90}{400}=0.225$

The test statistic is:

$z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.225-0.20}{\sqrt{\frac{0.20(1-0.20)}{400}}}= 1.25$

The test statistic is 1.25.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

Compute the p-value as follows:

$p-value=P(Z1.25)\\=1-P(Z$

*Use a z-table.

The p-value is 0.1057.

The p-value of the test is quite large. Thus, the null hypothesis was failed to rejected.

Hence, it can be concluded that option B is not the most common correct choice.

Answer from: Quest

attachment?

step-by-step explanation:

Answer from: Quest

Iam not good at math i wish i could

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