Exam 9: Inferences Based on a Two Samples: Confidence Intervals and Tests of Hypotheses

In a controlled laboratory environment, a random sample of 10 adults and a random sample of 10 children were tested by a psychologist to determine the room temperature that each person finds most comfortable. The data are summarized below: Sample Mean Sample Variance Adults (1) 77. 4.5 Children (2) 74. 2.5 Suppose that the psychologist decides to construct a 99% confidence interval for the difference in mean comfortable room temperatures instead of proceeding with a test of hypothesis. The 99% confidence interval turns out to be (-2.9, 3.1). Select the correct statement.

University administrators are trying to decide where to build a new parking garage on campus. The state legislature has budgeted just enough money for one parking structure on campus. The administrators have determined that the parking garage will be built either by the college of engineering or by the college of business. To help make the final decision, the university has randomly and independently asked students from each of the two colleges to estimate how long they usually take to find a parking spot on campus (in minutes). Based on their sample, the following 95% confidence interval (for μe - μb) was created - (4.20, 10.20). What conclusion can the university make about the population mean parking times based on this confidence interval?

A paired difference experiment yielded the following results. $n _ { d } = 50 , \quad \sum { d } = 967 , \quad \sum d ^ { 2 } = 19,201$ Test $H _ { 0 } : \mu _ { d } = 20$ against $H _ { \mathrm { a } } : \mu _ { d } \neq 20$ , where $\mu _ { d } = \mu _ { 1 } - \mu _ { 2 }$ , using $\alpha = .05$ .

Given $\mathrm { v } _ { 1 } = 15 \text { and } \mathrm { v } _ { 2 } = 20 , \text { find } \mathrm { P } ( \mathrm { F } > 1.84 )$

The data for a random sample of six paired observations are shown below. Pair Observation 1 Observation 2 1 1 3 2 2 4 3 3 5 4 4 6 5 5 7 6 6 8 a. Calculate the difference between each pair of observations by subtracting observation 2 from observation 1 . Use the differences to calculate $s ^ { 2 }$ . b. Calculate the standard deviations $s _ { 1 } { } ^ { 2 }$ and $s _ { 2 } { } ^ { 2 }$ of each column of observations. Then find pooled estimate of the variance $s p ^ { 2 }$ . c. Comparing $s _ { d } { } ^ { 2 }$ and $s p ^ { 2 }$ , explain the benefit of a paired difference experiment.

A new weight-reducing technique, consisting of a liquid protein diet, is currently undergoing tests by the Food and Drug Administration (FDA) before its introduction into the market. The weights of a random sample of five people are recorded before they are introduced to the liquid protein diet. The five individuals are then instructed to follow the liquid protein diet for 3 weeks. At the end of this period, their weights (in pounds) are again recorded. The results are listed in the table. Let μ1 be the true mean weight of individuals before starting the diet and let μ2 be the true mean weight of individuals after 3 weeks on the diet. Person Weight Before Diet Weight After Diet 1 165 158 2 210 205 3 203 200 4 212 206 5 219 215 Summary information is as follows: $\bar { d } = 5 , s _ { d } = 1.58$ . Test to determine if the diet is effective at reducing weight. Use $\alpha = .10$ .

Data was collected from CEOs of companies within both the low-tech industry and the consumer products industry. The following printout compares the mean return-to-pay ratios between CEOs in the low-tech industry with CEOs in the consumer products industry. HYPOTHESIS: MEAN X = MEAN Y SAMPLES SELECTED FROM RETURN industry 1 (low tech) (NUMBER =15 ) industry 3 (consumer products) (NUMBER =15 ) X= industry1 Y= industry 3 SAMPLE MEAN OF X =157.286 SAMPLE VARIANCE OF X=1563.45 SAMPLE SIZE OF X =14 SAMPLE MEAN OF Y=217.583 SAMPLE VARIANCE OF Y=1601.54 SAMPLE SIZE OF Y =12 MEAN X - MEAN Y=-60.2976 t=-4.23468 -=0.000290753 P-VALUE /2=0.000145377 SD. =14.239 Using the printout, which of the following assumptions is not necessary for the test to be valid?

A paired difference experiment yielded $\mathrm { n } \mathrm { d }$ pairs of observations. For the given case, what is the rejection region for testing $\mathrm { H } _ { 0 } : \mu _ { \mathrm { d } } = 15$ against $\mathrm { Ha } : \mu _ { \mathrm { d } } < 15$ ? $\mathrm { n } _ { \mathrm { d } } = 11 , \alpha = 0.05$ A) $\mathrm { t } < - 1.812$ B) $\mathrm { t } < - 1.796$ C) $\mathrm { t } < 1.812$ D) $\mathrm { t } < 2.228$

Specify the appropriate rejection region for testing $\mathrm { H } _ { 0 : } : \sigma _ { 1 } ^ { 2 } = \sigma _ { 2 } ^ { 2 }$ in the given situation. - $\mathrm { H } _ { \mathrm { a } } : \sigma _ { 1 } ^ { 2 } > \sigma _ { 2 } ^ { 2 } ; \alpha = 0.01 , \mathrm { n } _ { 1 } = 10 , \mathrm { n } _ { 2 } = 21$ A) $\mathrm { F } > 3.46$ B) $\mathrm { F } > 4.81$ C) $\mathrm { F } > 2.84$ D) $| \mathrm { F } | > 3.46$

A cola manufacturer invited consumers to take a blind taste test. Consumers were asked to decide which of two sodas they preferred. The manufacturer was also interested in what factors played a role in taste preferences. Below is a printout comparing the taste preferences of men and women. HYPOTHESIS: PROP. $\mathrm { X } =$ PROP. $Y$ SAMPLES SELECTED FROM soda(brand1,brand2) males ( sex =0, males ) (NUMBER=115) females ( sex =1, females ) (=56) X= males Y= females SAMPLE PROPORTION OF X =0.422018 SAMPLE SIZE OF X=109 SAMPLE PROPORTION OF Y =0.25 SAMPLE SIZE OF Y =52 PROPORTION X - PROPORTIOI Y=0.172018 Z=2.11825 Suppose the manufacturer wanted to test to determine if the males preferred its brand more than the females. Using the test statistic given, compute the appropriate p-value for the test.

Suppose you want to estimate the difference between two population proportions correct to within 0.03 with probability 0.90. If prior information suggests that p1 ≈ 0.4 and p2 ≈ 0.8, and you want to select independent random samples of equal size from the populations, how large should the sample sizes be? A) $\mathrm { n } _ { 1 } = \mathrm { n } _ { 2 } = 1708$ B) $\mathrm { n } _ { 1 } = \mathrm { n } _ { 2 } = 1925$ C) $\mathrm { n } _ { 1 } = \mathrm { n } _ { 2 } = 963$ D) $\mathrm { n } _ { 1 } = \mathrm { n } _ { 2 } = 1203$

A paired difference experiment has 75 pairs of observations. What is the rejection region for testing Ha: μd > 0? Use α = .01.

Determine whether the sample sizes are large enough to conclude that the sampling distributions are approximately normal. $n 1 = 48 , n 2 = 55 , \hat { p } _ { 1 } = .4 , \hat { p } _ { 2 } = .7$

An inventor has developed a new spray coating that is designed to improve the wear of bicycle tires. To test the new coating, the inventor randomly selects one of the two tires on each of 50 bicycles to be coated with the new spray. The bicycle is then driven for 100 miles and the amount of the depth of the tread left on the two bicycle tires is measured (in millimeters). It is desired to determine whether the new spray coating improves the wear of the bicycle tires. The data and summary information is shown below: 1 1.452 0.785 Bicycle Coated Tire (C) Non-Coated Tire (N) 1 1.452 0.785 2 1.634 0.844 \downarrow \downarrow \downarrow 50 1.211 0.954 Coated Non-Coated Difference Mean 1.38 0.85 0.53 Std. Dev. 0.12 0.11 0.06 Sample Size 50 50 50 Use the summary data to calculate the test statistic to determine if the new spray coating improves the mean wear of the bicycle tires. A) $z = 62.46$ B) $z = 23.02$ C) $z = 55.26$ D) $z = 34.25$

Consider the following set of salary data: Men (1) Women (2) Sample Size 100 80 Mean \ 12,850 \ 13,000 Standard Deviation \ 345 \ 500 To determine if women have a higher mean salary than men, we would test: A) $H _ { 0 } : \mu _ { 1 } - \mu _ { 2 } = 0$ vs. $H _ { \mathrm { a } } : \mu _ { 1 } - \mu _ { 2 } < 0$ B) $H _ { 0 } : \mu _ { 1 } - \mu _ { 2 } = 0$ vs. $H _ { a } : \mu _ { 1 } - \mu _ { 2 } \neq 0$ C) $H _ { 0 } : \mu _ { 1 } - \mu _ { 2 } = 0$ vs. $H _ { a } : \mu _ { 1 } - \mu _ { 2 } > 0$ D) $H _ { 0 } : \mu _ { 1 } - \mu _ { 2 } = 0$ vs. $H _ { a } : \mu _ { 1 } - \mu _ { 2 } = 0$

Suppose you want to estimate the difference between two population means correct to within 2.5 with probability 0.95. If prior information suggests that the population variances are both equal to the value 20, and you want to select independent random samples of equal size from the populations, how large should the sample sizes be? A) $\mathrm { n } _ { 1 } = \mathrm { n } _ { 2 } = 62$ B) $\mathrm { n } _ { 1 } = \mathrm { n } _ { 2 } = 25$ C) $\mathrm { n } _ { 1 } = \mathrm { n } _ { 2 } = 44$ D) $\mathrm { n } _ { 1 } = \mathrm { n } _ { 2 } = 18$

A new type of band has been developed for children who have to wear braces. The new bands are designed to be more comfortable, look better, and provide more rapid progress in realigning teeth. An experiment was conducted to compare the mean wearing time necessary to correct a specific type of misalignment between the old braces and the new bands. One hundred children were randomly assigned, 50 to each group. A summary of the data is shown in the table. Old Braces New Bands 410 days 380 days s 43 days 65 days How many patients would need to be sampled to estimate the difference in means to within 28 days with probability 99%?

Using paired differences removes sources of variation that tend to inflate $\sigma ^ { 2 }$ .

Which supermarket has the lowest prices in town? All claim to be cheaper, but an independent agency recently was asked to investigate this question. The agency randomly selected 100 items common to each of two supermarkets (labeled A and B) and recorded the prices charged by each supermarket. The summary results are provided below: =2.09 =1.99 =.10 =0.22 =0.19 =.03 Assuming the data represent a matched pairs design, calculate the confidence interval for comparing mean prices using a $95 \%$ confidence level. A) . $10 \pm .00588$ B) $10 \pm .056975$ C). $10 \pm .1255$ D) $.10 \pm .004935$

In a controlled laboratory environment, a random sample of 10 adults and a random sample of 10 children were tested by a psychologist to determine the room temperature that each person finds most comfortable. The data are summarized below: Sample Mean Sample Variance Adults (1) 77. 4.5 Children (2) 74. 2.5 Find the standard error of the estimate for the difference in mean comfortable room temperatures between adults and children.