In a sample of 100 M&M's, 8 of them were brown. Test the claim of the Mars candy company that the percentage of brown M&M's is equal to $13\%$. Use the critical value method with significance level $\alpha=.06$.

Assumptions: $np=13\geq 5$ and $nq=87\geq 5$

$H_0: p=.13,\ H_1: p\not=.13$

Test statistic: $z=-1.487$

Critical values: $\pm 1.88$

Fail to reject the null hypothesis because the test statistic is not in the rejection region.

There is not enough evidence to reject the claim that the percentage of brown M&M's is equal to $13\%$.A company claims to sell a coin that lands heads $60\%$ of the time. To test this claim, you obtain one of these coins and toss it 100 times. The coin lands heads 48 times. Using the critical value method, test the company's claim at significance level $\alpha=.03$.

Assumptions: $np=60\geq 5$ and $nq=40\geq 5$

$H_0: p=.6,\ H_1: p\not=.6$

Test statistic: $z=-2.45$

Critical values: $\pm 2.17$

Reject the null hypothesis because the tests statistic is in the rejection region.

There is enough evidence to reject the claim that the coin lands heads $60\%$ of the time.The American Automobile Association claims that $54\%$ of fatal car/truck accidents are caused by driver error. A researcher studies 35 randomly selected accidents and finds that 14 were caused by driver error. Test the claim at significance level $\alpha=.08$ using the critical value method.

Assumptions: $np=18.9\geq 5$ and $nq=16.1\geq 5$

$H_0: p=.54,\ H_1: p\not=.54$

Test statistic: $z=-1.66$

Critical values: $\pm 1.75$

Fail to reject the null hypothesis because the tests statistic is not in the rejection region.

There is not enough evidence to reject the claim that $54\%$ of fatal car/truck accidents are caused by driver error.A survey showed that among 785 randomly selected subjects who completed four years of college, $18.3\%$ smoke. Test the claim that the rate of smoking among those with four years of college is different from the $27\%$ rate for the general population using the critical value method with significance level $\alpha=.04$.

Assumptions: $np=212\geq 5$ and $nq=573\geq 5$

$H_0: p=.27,\ H_1: p\not=.27$

Test statistic: $z=-5.49$

Critical values: $\pm 2.05$

Reject the null hypothesis because the tests statistic is in the rejection region.

There is enough evidence to**support**the claim that the rate of smoking among those with four years of college is different from $27\%$.A costume company claims that $30\%$ of children dress as favorite movie characters for Halloween. In a random sample of 250 children, it was found that 87 children were dressed as movie characters. Test the costume company's claim at significance level $\alpha=.08$ using the critical value method.

Assumptions: $np=75\geq 5$ and $nq=175\geq 5$

$H_0: p=.3,\ H_1: p\not=.3$

Test statistic: $z=1.66$

Critical values: $\pm 1.75$

Fail to reject the null hypothesis because the tests statistic is not in the rejection region.

There is not enough evidence to reject the claim that $30\%$ of children dress as favorite movie characters for Halloween.A right-tailed test has test statistic $z=1.3$. Find the $P$-value.

$P(z>1.3)=.0968$, so $P\text{-value}=.0968$A 2-tailed test has test statistic $z=2.1$. Find the $P$-value.

$P(z>2.1)=.0179$, so $P\text{-value}=.0358$A 2-tailed test has test statistic -1.26. Find the $P$-value.

$P(z<-1.26)=.1038$, so $P\text{-value}=.2076$A test of the hypotheses $H_0: p=.44$ and $H_1:p>.44$ has test statistic $z=1.40$. Find the $P$-value.

$P$=value=.0808.A hypothesis test with significance level .05 has a $P$-value $=.03$. Should we reject the null hypothesis? Explain.

Reject the null hypothesis because the $P$-value is less than the significance level.A hypothesis test with significance level .02 has $P$-value$=.04$. Should be reject the null hypothesis? Explain.

Fail to reject the null hypothesis because the $P$-value is greater than the significance level.In testing the claim that a majority of the population prefers vanilla ice cream, the test statistic is $z=2.10$. Find the hypotheses and the $P$-value. Using a 0.01 significance level, should we reject the null hypothesis? Explain.

$H_0: p=.5,\ H_1: p \gt .5$

$P$-value = $ .0179$

Fail to reject the null hypothesis because the $P$-value is greater than the significance level.For a left-tailed test of the null hypothesis $p=.4$ with .03 significance level, find the alternative hypothesis and the critical value.

$H_1: p \lt .4$, CV: $z=-1.88$We want to use the confidence interval method for a 2-tailed test at significance level 0.12. Find the confidence level and $z_{\alpha/2}$.

Confidence level: $88\%$, $z_{\alpha/2}=1.55$The American Automobile Association claims that $54\%$ of fatal car/truck accidents are caused by driver error. A researcher studies 35 randomly selected accidents and obtains the $92\%$ confidence interval $(0.255, 0.545)$ for the proportion of fatal car/truck accidents caused by driver error. Should we reject the AAA's claim?

$H_0: p=.54, H_1: p\not=.54$

Fail to reject the null hypothesis because .54 is in the confidence interval.A survey of Oxford students is conducted to determine their general satisfaction with the food choices at Lil's. Of 80 students surveyed, 52 said they were satisfied. Test the claim that a majority of Oxford students is satisfied with the food choices at Lil's. Use $\alpha = 0.01$ and use the $P$-value method.

Assumptions: $np=40\geq 5$ and $nq=40\geq 5$

$H_0: p=.5,\ H_1: p \gt .5$

Test statistic: $z=2.683$

$P$-value $=.0037$

Reject the null hypothesis because $P$-value $<\alpha$.

There is enough evidence to support the claim that a majority of Oxford students is satisfied with the food choices at Lil's.In a 2004 survey of undergraduate students throughout the United States, $89\%$ of the respondents said they owned a cell phone. Recently, in a survey of 1200 randomly selected undergraduate students across the United States, it was found that 1098 of such students own a cell phone. Is the proportion of undergraduate students who own a cell phone now significantly different from $89\%$? Use the critical value method with $\alpha=0.02$.

$n=1200 \qquad p=.89\qquad q=.11$

Assumptions: $np=1068\geq 5, \qquad nq=132\geq 5$

$H_0:p=.89$; $H_1:p\neq 0.89$

$\widehat{p}=\dfrac{1098}{1200}=0.915\qquad \widehat{q}=0.085$

Test statistic: $z=\displaystyle {0.915 - .89 \over \sqrt{(.89)(.11)/1200}}\approx 2.77$

$\alpha/2=0.01$ Critical values: $\pm 2.33$

Reject $H_0$ because the test statistic falls within the rejection region.

There is enough evidence to support the claim that the proportion of undergraduate students who own a cell phone now is significantly different from $89\%$.A friend claims to have a die that rolls a six $30\%$ of the time. You roll the die 200 times and it shows a six 50 times. Test the your friend's claim at significance level $\alpha=0.07$ using the critical value method, the $P$-value method and the confidence interval method.

**Critical Value Method**

Assumptions: $np=60\geq 5$ and $nq=140\geq 5$

$H_0: p=.3,\ H_1: p\not=.3$

Test statistic: $z=-1.54$

Critical values: $\pm 1.81$

Fail to reject the null hypothesis because the tests statistic is not in the rejection region.

**$P$-Value Method**

Assumptions: $np=60\geq 5$ and $nq=140\geq 5$

$P(z \lt -1.54)=.0618$, so $P$-value $=.1236$

Fail to reject the null hypothesis because $P$-value$>\alpha$.

**Confidence Interval Method**

Assumptions: $n\widehat{p}=50 \geq 5$ and $n\widehat{q}=150 \geq 5$

$E=.055$, Confidence interval: $(.195, .305)$

Fail to reject the null hypothesis because .3 is in the confidence interval.

There is not enough evidence to reject the claim that the die rolls a six $30\%$ of the time.An apple farmer has in the past lost an average of $3\%$ of his trees each year. He notices that he has been losing more trees lately. In a sample of 200 trees, 11 have died. Test the claim that his loss rate is greater than $3\%$, perhaps due to some new environmental factor.

Test the claim using the critical value method and the $P$-value method at significance level $\alpha=.02$.

If the sample size had been 100 instead of 200, how would that change the situation?

Assumptions: $np=6\geq 5$ and $nq=194\geq 5$

$H_0: p=.03,\ H_1: p>.03$

Test statistic: $z=2.07$

Critical value: $2.05$

Reject the null hypothesis because the tests statistic is in the rejection region.

$P(z \gt 2.07)=.0192$, so $P$-value $=.0192$

Reject the null hypothesis because $P$-value $<\alpha$.

There is enough evidence to support the claim that the farmer's loss rate is greater than $3\%$.$np=3<5$, so the assumptions are not met for the hypothesis test.

A study claims that $42\%$ of children choose to eat a healthy snack after school. A school principal decides to test this claim by conducting a survey at her school. Out of 150 students surveyed, 49 of them chose a healthy snack after school. Test the claim at significance level $\alpha=.03$ using the critical value method, the $P$-value method and the confidence interval method.

Assumptions: $np=63\geq 5$ and $nq=87\geq 5$

$H_0: p=.42,\ H_1: p\not=.42$

Test statistic: $z=-2.316$

Critical values: $\pm 2.17$

Reject the null hypothesis because the tests statistic is in the rejection region.

$P(z \lt -2.310)=.0102$, so $P$-value $=.0204$

Reject the null hypothesis because $P$-value $ \lt \alpha$.

$E=.0831$, Confidence interval: $(.244, .410)$

Reject the null hypothesis because .42 is not in the confidence interval.

There is enough evidence to reject the claim that $42\%$ of children choose to eat a healthy snack after school.A survey of 1280 student loan borrowers found that $(.130,.169)$ is the $95\%$ confidence interval for those who answered "yes" to the question: "Do you owe more than $\$30,000$ for your undergraduate education?" Based on this information answer the following questions.

Out of the 1280 surveyed students, how many answered "yes" to the question posed? Justify your answer.

Use hypothesis testing to examine the claim that $20\%$ of all student loan borrowers owe more than $\$30,000$.

Recall that the confidence interval is given by $(\widehat{p}-E,\widehat{p} +E)$, where $\widehat{p}$ is the proportion of students who answered "yes" to the question. Therefore, $E=\dfrac{.169-.130}{2}=.0195$ and $\widehat{p}=.130+.0195=.169-.0195=0.1495$. Therefore, the number of students who answered "yes" to the question posed is $$1280 \cdot 0.1495=191.36\approx 191$$

Assumptions: $np=256\geq 5$ and $nq=1024\geq 5$

$H_0:p=.2\qquad H_1:p\neq .2$

Confidence interval: $(.130,.169)$

Reject $H_0$. $p=0.2$ is not in the confidence interval.

There is enough evidence to reject the claim that $20\%$ of all student loan borrowers owe more than \$30,000.

A college administration claims that $83\%$ of their students support recycling on campus. Two statistics students, Mike and Frank, conduct a survey to test this claim. They find that the sample proportion of students who support recycling is $81\%$. They each construct a confidence interval based on this sample proportion. Mike comes up with $(.74, .88)$ while Frank gets $(.76, .90)$.

- Indicate which interval is incorrect and explain why.
- Based on the correct interval, should the administration's claim be rejected? Why or why not?
- Based on the correct interval, should the administration's claim be supported? Why or why not?

$(.76,.90)$ is incorrect since $\widehat{p}=.81$ is not at the center of the interval.

The claim $p=.83$ should not be rejected since 0.83 in in the interval $(.74,.88)$.

The claim $p=.83$ should not be supported, because the claim is the null hypothesis and we should never support the null hypothesis.

Suppose that we reject a null hypothesis using a $90\%$ confidence interval. Find the probability we made a type I error.

$10\%$Suppose you are testing a hypothesis at $\alpha=.05$. If the null hypothesis is actually true, find the probability of a Type I error and of a Type II error.

$P($type I error$)=.05$ and $P($type II error$)=0$A coin is flipped 100 times to test whether it is fair and we fail to reject the null hypothesis. In each case below, determine whether we have made a type I error, type II error, or correct decision.

- The coin was unfair.
- The coin was fair.

- Type II error
- Correct decision

A study is conducted and the $P$-value for the sample is found to be 0.085. Which of the following statements can be correctly inferred?

- If $H_0$ is true, then the probability that the test statistic is in the rejection region is 0.085.
- If $H_0$ is true, then the probability of getting this sample or a sample more extreme is 0.085.
- The probability that $H_0$ is true is 0.085.
- The probability of a type II error is 0.085.

Only (b).