Question

An instructor at a major research university occasionally teaches summer session and notices that that there are often students repeating the class. On the first day of class, she counts 105 students enrolled, of which 19 are repeating the class. The university enrolls 15,000 students.
1. An estimate of the population proportion repeating the class is given by:

a) 0.0013.

b) 0.007.

c) 0.181.

d)0.046.

2. The instructor wishes to estimate the proportion of students across campus who repeat a course during summer sessions and decides to do so on the basis of this class. Would you advise the instructor against it and why:

a) No, because 105 students is a pretty large class.

b) Yes, because this class is too small.

c) No, because it is completely arbitrary who takes this class.

d) Yes, because this class is not a random sample of students.

3. The standard error for the estimated sample proportion is given by:

a) 0.025.

b) 0.0014.

c) 0.0376.

d) 0.005.

4. A 95% confidence interval is given by:

a) 0.15, 0.23.

b) 0.345, 0.453.

c) 0.107, 0.255.

d) 0.09, 0.272.

5. She hypothesizes that, in general, 10% of students repeat a course. The hypotheses to be tested are:

a) H0:p? =0.181 vs. H?:p=0.1 .

b) H0:p=0.1 vs. H?:p?0.18 .

c) H0:p=0.1 vs. H?:p? =0.181 .

d) H0:p=0.1 vs. H?:p?0.1 .

6. The test statistic for this hypothesis is given by:

a) 2.575.

b) 2.332.

c) 2.765.

d) 1.96.

7. The P?value for this test is:

a) 0.05>P>0.01 .

b) P>0.05 .

c) 0.01>P>0.005 .

d) 0.005>P .

8. Based on the p-value found:

a) we have convincing proof that the proportion of students repeating a class during summer sessions is not 10%.

b) we have strong evidence that the proportion of students repeating a class during summer sessions is not 10%.

c) we have strong evidence that the proportion of students repeating a class during summer sessions is 10%.

d) we have no doubt that the proportion of students repeating a class during summer sessions is not 10%.

Answer 1

Answer:

Check the explanations

Step-by-step explanation:

According to given information, an instructor at a major research university occasionally teaches summer session and notices that that there are often students repeating the class. On the first day of class, she counts 105 students enrolled, of which 19 are repeating the class. The university enrolls 15,000 students.

Therefore the number of observation n 105 and enrolled x=19

1. An estimate of the population proportion repeating the class is given by:

c) 0.181.

Explanation:

19 --0.18090,181 n 105

2. The instructor wishes to estimate the proportion of students across campus who repeat a course during summer sessions and decides to do so on the basis of this class. Would you advise the instructor against it and why:

d) Yes, because this class is not a random sample of students.

3. The standard error for the estimated sample proportion is given by:

c) 0.0376.

Explanation:

SE  $(\hat{p})=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

0.181 × (1-0.181) 105

0,0376 0.0376

4. A 95% confidence interval is given by:

c) 0.107, 0.255.

Explanation:

In order to determine the 95% confidence interval we follow the following step:

Where the z value is determined from the standard normal table as ~ 1.96

$\hat{p}\pm \left [z\times SE(\hat{p}) \right ]$

0.181 ± (1.96 × 0.0376)

0.181 ± 0.0737

Therefore the lower confidence interval is

LCI= 0.181- 0.0737

LCI= 0.107

Therefore the upper confidence interval is

UCI = 0.181 + 0.0737

UCI0.255

Therefore 95% confidence interval is

$\left (0.107,0.255 \right )$

5. She hypothesizes that, in general, 10% of students repeat a course. The hypotheses to be tested are:

d)$0 :P = 0.10 and H_{a}:p\neq 0.10$

Explanation:

She hypothesizes that, in general, 10% of students repeat a course. The hypotheses to be tested are defined on the basis of observation,

Null hypothesis as:

$H_{0}$:p= 0.10

and alternative hypothesis as:

$H_{a}:p\neq 0.10$

6. The test statistic for this hypothesis is given by:

c) 2.765.

Explanation:

In order to determine the z test statistics as:

Z=$\frac{\hat{p}-p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}}$

 0.181 - 0.10 0.10x (1-0.10)

 =2.765

7. The P value for this test is:

c) 0.01>P>0.005 .

Explanation:

P value is calculated as:

P(Z > 2.765) 0.002845 for one tail test and

P(Z > 2.765) 0.005692 for two tail test.

8. Based on the p-value found:

b) we have strong evidence that the proportion of students repeating a class during summer sessions is not 10%.

Explanation:

As the z observed is more than the tabulated z value at 95% as:

$Z_{observed}=2.765> Z_{tabulated}=1.96$

and also P value is less than the $\alpha =1-0.95=0.05$

$P(Z\geq 2.765)=0.005692< \alpha =0.05$

Therefore we accept the alternative hypothesis and we may conclude that we have strong evidence that the proportion of students repeating a class during summer sessions is not 10%.