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Aleksandr-060686 [28]
3 years ago
15

HELP! Question 2: Evelyn goes to a middle school that includes grades 6, 7, and 8. She wants to estimate the percentage of stude

nts in the school who ride the bus to school. She selects a random sample of 200 students to survey who are attending an after-school activity. She determines that 38% of the students at the activity ride the bus to school. Which statement about Evelyn's sample is true? A. The sample is the percentage of students who ride the bus. B. The sample might not be representative of the population because it only includes students who are attending an after-school activity. C. The sample shows that exactly 38% of the students in the school ride the bus. 
Mathematics
2 answers:
asambeis [7]3 years ago
8 0

B would be the right answer.

Since she only randomly picks 200 out of x amount of students it is hard to say how many students actually ride the bus.


taurus [48]3 years ago
4 0
The answer is B

She cannot guarantee that the percentage taken directly corresponds to the entire student body
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Hypothesis Testing
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Answer:

<u>Problem 1</u>: We conclude that less than or equal to 50% of adult Americans without a high school diploma are worried about having enough saved for retirement.

<u>Problem 2</u>: We conclude that the volume of Google stock has changed.

Step-by-step explanation:

<u>Problem 1:</u>

We are given that in a recent survey conducted by Pew Research, it was found that 156 of 295 adult Americans without a high school diploma were worried about having enough saved for retirement.

Let p = <em>proportion of adult Americans without a high school diploma who are worried about having enough saved for retirement</em>

So, Null Hypothesis, H_0 : p \leq 50%    {means that less than or equal to 50% of adult Americans without a high school diploma are worried about having enough saved for retirement}

Alternate Hypothesis, H_A : p > 50%     {means that a majority of adult Americans without a high school diploma are worried about having enough saved for retirement}

This is a right-tailed test.

The test statistics that would be used here is <u>One-sample z-test</u> for proportions;

                       T.S.  =  \frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of adult Americans who were worried about having enough saved for retirement = \frac{156}{295} = 0.53

           n = sample of adult Americans = 295

So, <u><em>the test statistics</em></u> =  \frac{0.53-0.50}{\sqrt{\frac{0.50(1-0.50)}{295} } }

                                    =  1.03

The value of z-test statistics is 1.03.

<u>Also, the P-value of the test statistics is given by;</u>

              P-value = P(Z > 1.03) = 1 - P(Z \leq 1.03)

                           = 1 - 0.8485 = <u>0.1515</u>

Now, at a 0.05 level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since the value of our test statistics is less than the critical value of z as 1.03 < 1.645, <u><em>so we insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that less than or equal to 50% of adult Americans without a high school diploma are worried about having enough saved for retirement.

<u>Problem 2:</u>

We are given that a random sample of 35 trading days in 2014 resulted in a  sample mean of 3.28 million shares with a standard deviation of 1.68 million shares.

Let \mu = <em>mean daily volume in Google stock</em>

So, Null Hypothesis, H_0 : \mu = 5.44 million shares    {means that the volume of Google stock has not changed}

Alternate Hypothesis, H_A : \mu \neq 5.44 million shares     {means that the volume of Google stock has changed}

This is a two-tailed test.

The test statistics that would be used here is <u>One-sample t-test statistics</u> because we don't know about the population standard deviation;

                       T.S.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean volume in Google stock = 3.28 million shares

            s = sample standard deviation = 1.68 million shares

           n = sample of trading days = 35

So, <u><em>the test statistics</em></u> =  \frac{3.28-5.44}{\frac{1.68}{\sqrt{35} } }  ~ t_3_4

                                    =  -7.606

The value of t-test statistics is -7.606.

<u>Also, the P-value of the test statistics is given by;</u>

              P-value = P(t_3_4 < -7.606) = Less than 0.05%

Now, at a 0.05 level of significance, the t table gives a critical value of -2.032 and 2.032 at 34 degrees of freedom for the two-tailed test.

Since the value of our test statistics doesn't lie within the range of critical values of t, <u><em>so we sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.

Therefore, we conclude that the volume of Google stock has changed.

8 0
3 years ago
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