## Exercises - Confidence Intervals & Hypothesis Testing (One Sample, Means and Proportions)

When 16 seventeen-year-old women were randomly selected and tested with an aerobics test, the mean of their scores was found to be 37.9 with a sample standard deviation of 7.3. Construct both a 90% confidence interval and a 99% confidence interval for the mean. Write out the meaning of your findings.

You are watching: Test the claim by constructing an appropriate confidence interval.

This is a small sample since $n = 16$. The Student-$t$ distribution is needed. The 90% confidence interval is from 34.7 to 41.1. The 99% confidence interval is from 32.5 to 43.4. Interpretation: You can be 90% (or 99%) confident that the true mean of the population is inside the interval 34.7 to 41.1 (or 32.5 to 43.4) OR there is approximately a 0.90 (or 0.99) probability that the true mean is in the given interval.

The school board is interested in the proportion of middle school students who do not live with both biological parents. How many students will need to be surveyed so that the board can be 90% confident in the results, with no more than 0.05 error?

Use $0.50$ for the proportion in the formula, since no previous information is available for the proportion. $n = 271$ approximately. In reality, one should use about $300$ names and hope for 95% return rate which will give 285 surveys.

Verbal SAT scores of a random sample of 250 high school students has a mean of 470.3 and a standard deviation of 138. Construct a 95% and a 99% confidence interval and explain the reason for the difference in these intervals.

95% interval is 453.2 to 487.4, 99% interval is 447.8 to 492.8. The more confident (higher percentage), the wider the interval of values.

The President claims that a majority of voters are in favor of his foreign aid program. A random sample of 1000 voters showed 545 who favored the foreign aid program. Is there sufficient evidence to support the President"s claim?

State the null hypothesis, Give the appropriate test statistic, Use alpha of 0.005 to determine the appropriate critical value, Find the p-value for the test, State your conclusions, and Interpret your results.This is a significant majority test. $H_0: p \le 0.50, z = 2.846$ is the test statistic. Critical value is 2.575, p-value is 0.0022 (the area in the right tail of the standard normal with $z = 2.85$). Reject the null hypothesis at $\alpha = 0.005$ (since the test statistic, 2.846, is in the rejection region that begins with the critical value of 2.575 and extends to the right). It seems as if the President does have a significant majority in favor of the foreign aid program (with possible error of less than 0.005).

Organic chemists often purify organic compounds by fractional crystallization. A laboratory technician desired to prepare and purify several samples, each of 4.85 grams aniline. He claims that his procedure theoretically will produce 3.43 grams of acetanilide. In a sample of 16, the mean was found to be 3.50 with a standard deviation of 0.55. Test the claim made by the chemist.

$H_0: \mu = 3.43, t = 0.51$ (test statistic). There is insufficient evidence to reject the null hypothesis; i.e., fail to reject. The laboratory technician seems to be correct.

Suppose that you take a random sample of 36 athletes who try out for the Mexican Olympic men"s basketball team. Your sample averages 75 inches tall with a standard deviation of 3 inches. Find a 95% confidence interval and interpret your findings.

$E = 0.98$. There is a 95% probability that the true mean of the athletes who try out for the Mexican Olympic men"s basketball team is between 74 inches and 76 inches.

Southern Bell wants to determine the proportion of cars with cellular phones. How many cars must be sampled in order to be 90% confident that the sample proportion is in error by no more than 0.02?

$n = 1691.27$, or about $1692$. In reality, use between 1780 and 1800. A 95% return rate on 1780 will yield 1691 subjects.

The final exam in Math 107 usually produces a mean score of 78 (percent). Twenty students in an above-average class take this final exam and produce a mean score of 82 and a standard deviation of 6.2 (with no outliers). The professor claims that the above-average class scored significantly higher than the previously accepted average of 78. Evaluate this situation with an appropriate hypothesis testing procedure.

$H_0: \mu \le 78$. The test statistic is $t = 2.885$ (small sample). The critical value is $t(19) = 2.540$ for $\alpha = 0.01$. Reject thenull hypothesis at $\alpha = 0.01$. The professor"s claim seems to be true, the above average class scored significantly higher than the usual average on the final exam.

In a local survey, 813 of the 1084 respondents indicated support for a ban on household aerosols. It is believed that more than 70% of the population supports the ban on household aerosols. Provide appropriate hypothesis testing to respond to the above claim.

$H_0: p \le 0.70$. The test statistic is $z = 3.59$ with $p$-value of $0.0001$. Reject the null hypothesis. The survey supports the claim that more than 70% of the population supports the ban on household aerosols.

You suspect that the coin used to begin the local football game is not a fair coin; i.e., that there are too many instances of "heads". (A fair coin would give approximately half "heads" and half "tails") You take the suspected coin and toss it 100 times. You get 65 heads. Is this result significantly higher than should be produced by a fair coin? As you make your determination:

State the null hypothesis,Find the value of the appropriate test statistic,Find the p-level associated with the test,State your conclusion and inference, and Evaluate the experiment.$H_0: p \le 0.50$ (for appearance of heads). Test statistic is $z = 3.00$, $p$-value or $p$-level of $0.0013$. Reject the null hypothesis. The coin seems to be unfair. How the coin is tossed could effect the outcome, etc. Also note that an appropriate alpha level for rejection of this hypothesis is $0.005$ that gives a critical value of $2.575$.

In a random sample of 18 high school juniors, the mean of scores on the verbal portion of the SAT scores was found to be 510 with a standard deviation for the sample of 140 points.

Construct a 98% confidence interval for the mean. Interpret your findings.The mean for these scores is believed to be 500. Test the claim that the sample mean is not significantly different from the value 500. State the null hypothesis, test statistic, critical value for alpha of 0.01, conclusion, and interpretation.Explain how the results of the above two parts are related.(a) $425 \lt \mu \lt 595$. 98% confident that the population mean of the verbal SAT is in the interval; (b) $H_0 : \mu = 500, t = 0.303$ (test statistic), $t(17) = \pm 2.898$ (critical value). Fail to reject the null hypothesis. The sample has a mean that is not significantly different from $500$; (c) The value $500$ (part b) is inside the interval $425 \lt \mu \lt 595$ (part a).

Describe the role of ethics in the use of statistics in experimentation. Give examples in your explanation.

One could discuss ethics in reporting results of experimentation in scientific journals or in reporting results of surveys done by television stations or in commercials, etc. Also there is always a question of whether subjects respond honestly.

To determine the percentage of Democrats in a certain state, a random sample of 2,200 was obtained. Of these 1,328 were found to be Democrats. Create a 90% confidence interval and a 99% confidence interval for ""p"", the true proportion of Democrats. Interpret your results.

90%: $0.5864 \lt p \lt 0.6208$ (90% confident that the true proportion of Democrats in the state is between $0.586$ and $0.621$) 99%: $0.5767 \lt p \lt 0.6305$ (99% confident that the true proportion of Democrats in the state is between $0.577$ and $0.631$). You can be extremely confident that there is a majority of Democrats in this state!

A school system wants to determine the proportion of students who come from broken homes, those households that have gone through divorce. How many students would need to be randomly selected in order to be 98% confident that the sample proportion is in error of no more than 0.05% In reality, how many should be in your sample to assure an adequate number given that around 90% of the sample will produce responses?

$n = 543$. Solve: $0.90x = 543$ to get $603$ students. Sample size in reality should be about 610. A 90% return rate presents other problems. What would these be?

The president of a large company (5,000 employees) is certain that a significant majority of his employees is in favor of the new health plan that will go into effect in January. The labor union (3,000 members) is not so certain that a significant majority of labor union members are in favor of this health plan. A stratified random sample of 300 union members and 200 non-union members (from the remaining 2,000 employees) indicated that 153 union members were in favor of the new health plan and 122 non-union members were in favor of the new health plan.

Is there a significant majority of union members in favor of the new health plan?Is there a significant majority of non-union members in favor of the new health plan?Was the president of the company correct in his assumption?(a) $H_0 : p \le 0.50$, test statistic of $z = 0.35$, $p$-value of $0.3632$, fail to reject $H_0$. There is not a significant majority of union members in favor of the new health plan; (b) $H_0 : p \le 0.50$, test statistic of $z = 3.11$, $p$-value of $0.0001$ (approximately), reject $H_0$. There is a significant majority of non-union members in favor of the new health plan; (c) If both groups are combined to form a group of all employees, there would be 275 out of 500 in favor of the health plan. Both the labor union and the non-union members had 10% of the employees sampled. $H_0 : p \le 0.50$, test statistic of $z = 2.24$, $p$-value of $0.0125$, reject $H_0$. There is a significant majority in favor of the new health plan.

The U.S. Department of the Treasury claims that the mean weight of quarters minted is 5.670 grams. A random sample of 50 quarters is gathered at a local bank. The mean weight is found to be 5.638 with a standard deviation of 0.072 grams.# Is the random sample significantly different from the weight claimed by the U.S. Department of the Treasury?

Construct a 99% confidence interval for the mean weight of quarters found by the bank.Explain how the results of the above two answers are consistentExplain how the U.S. Treasury is or is not correct in the claim that the weight of quarters minted has a mean of 5.670 grams.(a) $H_0 : \mu = 5.670$, test statistic of $z = -3.14$, critical value at $\alpha = 0.01$ is $\pm 2.575$, $p$-value is $0.0002$ (but really approximately $0.0001$ based on the high value of the test statistic), reject $H_0$. There is a significant difference between the weight of the quarters gathered by a local bank and the claim by the U.S. Department of the Treasury. (b) $5.638 \pm 0.026$ gives $5.612 \lt \mu \lt 5.664$; (c) The claimed value of $5.760$ grams is NOT in the confidence interval of $5.664$ to $5.612$ and the null hypothesis that the mean equals $5.670$ grams was rejected. These are consistent results; (d) The U.S. Treasury could be correct in that originally quarters weighed an average of $5.670$. From use in circulation, the coins could have been worn down and could have weighed less. The U.S. Treasury is incorrect in assuming that the quarters weigh $5.670$ g each. More information is needed.

A Ford engineer claims that a new fuel injection design increases the mean mileage on the Taurus above its current 30 miles per gallon level. Twenty of the new designs were checked and the mean recorded as 32.00 miles per gallon with a standard deviation of 3.87 miles per gallon. Evaluate this claim.

$H_0 : \mu \le 30$, test statistic of $t = 2.311, t(19) = 2.093$ for alpha of $0.025$, reject $H_0$. The new fuel injection engine gives significantly better gas mileage than the old fuel injection design.

Weights of cereal in 16-ounce boxes are normally distributed with a mean of 16 ounces and a standard deviation of 0.12 ounce. Respond to the following:

What is the probability that a cereal box selected at random will have at least 15.95 ounces?What is the probability that the mean of a sample of 16 boxes will be at least 15.95 ounces?In a production of 10,000 boxes, how many would you expect to be below 15.95 ounces?The production manager was concerned that his boxes of cereal had significantly lower weight than the expected value of 16 ounces. A sample of 50 boxes was obtained and the mean weight found to be 15.95 ounces with a standard deviation of 0.12 ounces. Evaluate this situation using appropriate hypothesis testing.(a) $P(x \gt 15.95) = P(z \gt -0.42) = 0.50 + 0.1628$; (b) Central Limit Theorem. $P(z \gt -1.67) = 0.9525$; (c) $P(x \lt 15.95) = P(z \gt -0.42) = 0.5 - 0.1628 = 0.3372$. So $(10000)(0.3372)$ boxes; (d) $H_0: \mu \ge 16; z = -2.95$, $p$-value is $0.0016$. Reject the null hypothesis. The evidence suggests the average weight is significantly lower than 16 oz.

It is believed that a significant majority of voters are dissatisfied with the local school system. A random sample of 1400 voters was obtained. Of those samples, 746 stated that they were dissatisfied with the school system.

Evaluate the belief with appropriate hypothesis testing.Find the 95% confidence interval for the sample proportion.Are your results consistent? Explain.(a) $H_0 : p \le 0.50; z = 2.46$; $p$-value is $0.0070$; Reject the null hypothesis. There seems to be a significant majority dissatisfied with the school system. (b) $0.507 \lt p \lt 0.559$ is the 95% confidence interval. (c) The interval does not contain $p = 0.50$ and the null hypothesis was rejected; therefore, the results are consistent.

When 16 seventeen-year-old women were randomly selected and tested with an aerobics test, the mean of their scores was found to be 37.9 with a sample standard deviation of 7.3.

Construct both a 90% and a 99% confidence interval for the mean. Interpret the 99% interval.Explain why one interval is larger than the otherHow many people would need to be interviewed so that one could be 95% confident that the mean is in error by no more than 1.5 units?If you could be guaranteed that a minimum of 95% of the data would be available, how many would you need in your sample?It is believed that this sample of seventeen-year-olds are not as aerobic as the national average of those participating in aerobics. The national average on this test is 40. Use appropriate hypothesis testing techniques to check the claim. Use an alpha of 0.05.See more: If You Have Beaten This, How Long To Beat Shadow Of War ? How Long Does It Take To Beat Middle

(a) 90%: $34.7 \lt \mu \lt 41.1$ and 99%: $32.5 \lt \mu \lt 43.3$. There is 99% confidence that the population mean is in the interval $32.5$ to $43.3$. (b) The more confident you want to be, the larger the interval. (c) 91 (d) 96 (e) $H_0 : \mu \ge 40; t = -1.15$ is the test statistic; $t = -1.753$ is the critical value; fail to reject the null hypothesis. Ther is insufficient evidence to conclude that this group is not as aerobic as the national population.