Question

**1.**

The average final exam score for the statistics course is 78%. A professor wants to see if the average final exam score for students who are given colored pens on the first day of class is different. The final exam scores for the 11 randomly selected students who were given the colored pens are shown below. Assume that the distribution of the population is normal.

77, 58, 53, 90, 83, 77, 51, 72, 53, 68, 81

What can be concluded at the the αα = 0.01 level of significance level of significance?

- For this study, we should use Select an answer z-test for a population proportion t-test for a population mean
- The null and alternative hypotheses would be:

H0:H0: p or μ ; > or = or < or ≠ ______

H1:H1: μ or p ; < or > or ≠ or = _______

- The test statistic? z or t = ______ (please show your answer to 3 decimal places.)
- The p-value = ______ (Please show your answer to 4 decimal places.)
- The p-value is? ≤ or > α
- Based on this, we should Select an answer; accept or reject or fail to reject, the null hypothesis.
- Thus, the final conclusion is that ...
- The data suggest the population mean is not
**significantly**different from 78 at αα = 0.01, so there is statistically insignificant evidence to conclude that the population mean final exam score for students who are given colored pens at the beginning of class is equal to 78. - The data suggest the populaton mean is
**significantly**different from 78 at αα = 0.01, so there is statistically significant evidence to conclude that the population mean final exam score for students who are given colored pens at the beginning of class is different from 78. - The data suggest that the population mean final exam score for
students who are given colored pens at the beginning of class is
not
**significantly**different from 78 at αα = 0.01, so there is statistically insignificant evidence to conclude that the population mean final exam score for students who are given colored pens at the beginning of class is different from 78.

- The data suggest the population mean is not

**2.**

The recidivism rate for convicted sex offenders is 15%. A warden suspects that this percent is lower if the sex offender is also a drug addict. Of the 318 convicted sex offenders who were also drug addicts, 29 of them became repeat offenders. What can be concluded at the αα = 0.01 level of significance?

- For this study, we should use Select an answer t-test for a population mean z-test for a population proportion
- The null and alternative hypotheses would be:

Ho: μ or p ; < or = or ≠ or > _____ (please enter a decimal)

H_{1}: μ or p ; < or = or ≠ or > _____ (Please enter a decimal)

- The test statistic? z or t = _____ (please show your answer to 3 decimal places.)
- The p-value = _____ (Please show your answer to 4 decimal places.)
- The p-value is? ≤ or > α
- Based on this, we should Select an answer; fail to reject or reject or accept, the null hypothesis.
- Thus, the final conclusion is that ...
- The data suggest the population proportion is not
**significantly**lower than 15% at αα = 0.01, so there is statistically insignificant evidence to conclude that the population proportion of convicted sex offender drug addicts who become repeat offenders is lower than 15%. - The data suggest the population proportion is not
**significantly**lower than 15% at αα = 0.01, so there is statistically significant evidence to conclude that the population proportion of convicted sex offender drug addicts who become repeat offenders is equal to 15%. - The data suggest the populaton proportion is
**significantly**lower than 15% at αα = 0.01, so there is statistically significant evidence to conclude that the population proportion of convicted sex offender drug addicts who become repeat offenders is lower than 15%.

- The data suggest the population proportion is not
- Interpret the p-value in the context of the study.
- If the population proportion of convicted sex offender drug addicts who become repeat offenders is 15% and if another 318 inner city residents are surveyed then there would be a 0.17% chance that fewer than 9% of the 318 convicted sex offender drug addicts in the study become repeat offenders.
- There is a 0.17% chance that fewer than 15% of all convicted sex offender drug addicts become repeat offenders.
- There is a 15% chance of a Type I error.
- If the sample proportion of convicted sex offender drug addicts who become repeat offenders is 9% and if another 318 convicted sex offender drug addicts are surveyed then there would be a 0.17% chance of concluding that fewer than 15% of convicted sex offender drug addicts become repeat offenders.

- Interpret the level of significance in the context of the
study.
- There is a 1% chance that Lizard People aka "Reptilians" are running the world.
- If the population proportion of convicted sex offender drug addicts who become repeat offenders is 15% and if another 318 convicted sex offender drug addicts are observed, then there would be a 1% chance that we would end up falsely concluding that the proportion of all convicted sex offender drug addicts who become repeat offenders is lower than 15%.
- If the population proportion of convicted sex offender drug addicts who become repeat offenders is lower than 15% and if another 318 convicted sex offender drug addicts are observed then there would be a 1% chance that we would end up falsely concluding that the proportion of all convicted sex offender drug addicts who become repeat offenders is equal to 15%.
- There is a 1% chance that the proportion of all convicted sex offender drug addicts who become repeat offenders is lower than 15%.

Answer #1

Q1:

Sample mean using excel function AVERAGE(), x̅ = 69.3636

Sample standard deviation using excel function STDEV.S, s = 13.6914

Sample size, n = 11

a) For this study, we should use Select : **t-test for a
population mean**

b) The null and alternative hypotheses would be:

Ho : µ = 78 ; H1 : µ ≠ 78

c) Test statistic:

t = (x̅- µ)/(s/√n) = (69.3636 - 78)/(13.6914/√11) =
**-2.092**

df = n-1 = 10

d) p-value :

Two tailed p-value = T.DIST.2T(ABS(-2.092), 10) = 0.0629

e) The p-value is **>** α.

f) **Fail to reject**, the null hypothesis.

g) Conclusion :

The data suggest that the population mean final exam score for students who are given colored pens at the beginning of class is not significantly different from 78 at α = 0.01, so there is statistically insignificant evidence to conclude that the population mean final exam score for students who are given colored pens at the beginning of class is different from 78.

**-------------------------------------**

Q2: n = 318, x = 29

p̄ = x/n = 0.0912

α = 0.01

a) For this study, we should use Select : **z-test for a
population proportion**

b) The null and alternative hypotheses would be:

Ho : p = 0.15 ; H1 : p < 0.15

c) Test statistic:

z = (p̄ -p)/√(p*(1-p)/n) = (0.0912 - 0.15)/√(0.15 * 0.85/318)
**= -2.937**

d) p-value = NORM.S.DIST(-2.9368, 1) =
**0.0017**

e) The p-value is **≤** α.

f) **Reject** the null hypothesis.

g) Conclusion :

The data suggest the population proportion is significantly lower than 15% at αα = 0.01, so there is statistically significant evidence to conclude that the population proportion of convicted sex offender drug addicts who become repeat offenders is lower than 15%.

f) Interpret the p-value :

If the population proportion of convicted sex offender drug addicts who become repeat offenders is 15% and if another 318 inner city residents are surveyed then there would be a 0.17% chance that fewer than 9% of the 318 convicted sex offender drug addicts in the study become repeat offenders.

i) Interpret the level of significance :

There is a 1% chance that the proportion of all convicted sex offender drug addicts who become repeat offenders is lower than 15%.

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