Listening to music has long been thought to enhance intelligence, especially during infancy andchildhood.To test whether this is true, a researcher records the number of hours that eighthigh-performing students listened to music per day for 1 week. The data are listed in the tableUsing SPSS, find the confidence limits at a 95% CI for this sampleSuppose the null hypothesis states that students listen to 4.5 hours of music per day.What would the decision be for a two-tailed hypothesis test at a .05 level ofsignificance?i.ii.Music Listening per Day (in hours)3.43.95.51.65.33.31.92.6

A random sample of 8 observations is given.We have to test whether students listen to 4.5 hours of…

Answered: Listening to music has long been…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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You are comparing men and women on hours spent watching TV. You pick a sample of 12 men and 14 women and calculate a confidence interval on the difference between means. How many degrees of freedom does your t value have?
A psychologist is studying the effects of lack of sleep on the performance of various perceptual-motor tasks. After a given period of sleep deprivation, a measurement of reaction time to an auditory stimulus was taken for each of 36 adult male subjects. The reaction times (in seconds) are summarized as follows:x = 1.82, S=0.22. Find a 99% confidence interval for the mean reaction time for sleep-deprived male subject
Find the 95% confidence interval for the percentage of leaky diapers in Example 4.4.4 assuming 400 diapers were tested and 20% were leaky. Note that it is 8 percentage points long.
Example 5: Find a 90% confidence interval for the mean difference in the mileage obtained for tires with the new material and tires with the old material. Following are the paired differences (new minus old) for each of the 17 riders, chosen at random, who took part in the experiment: 50, 45, 50, 50, 100, 100, 100, -10, 75, -25, 0, 25, 75, 25, 50, 40, 35
Example 5: Find a 90% confidence interval for the mean difference in the mileage obtained for tires with the new material and tires with the old material. Following are the paired differences (new minus old) for each of the 17 riders, chosen at random, who took part in the experiment: 50, 45, 50, 50, 100, 100, 100, -10, 75, -25, 0, 25, 75, 25, 50, 40, 35
Suppose a study reported that the average person watched 3.34 hours of television per day. A random sample of 15 people gave the number of hours of television watched per day shown. At the 10% significance level, do the data provide sufficient evidence to conclude that the amount of television watched per day last year by the average person differed from the value reported in the study? (Note: x = 4.400 hours and s= 1.262 hours.) Set up the hypotheses for the one-mean t-test. HoiH Ha: H 2.3 4.0 4.5 4.9 5.0 4.5 3.3 2.0 5.7 5.9 6.2 3.8 5.9 3.9 4.1
In a survey funded by the UW school of medicine, 750 of 1000 adult Seattle residents said they did not believe they could come down with a sexually transmitted infection (STI). Construct a 95% confidence interval estimage of the proportion of adult Seattle residents who don't believe they can contract an STI. (Use a z score of 1.96 for your computations.) (.728, .772) (.723, .777) (.718, .782) (.713, .878) (.665, .835)
The coach of a very popular men’s basketball team claims that the average distance the fans travel to the campus to watch a game is 35 miles. The team members feel otherwise. A sample of 16 fans who travel to games was randomly selected and yielded a mean of M= 36 miles and s= 5 miles. Test the coach’s claim at the 5% (.05) level of significance. one-tailed or two-tailed test: State the hypotheses: df= tα or t value for the critical region = sM = t (test statistic)= Decision:
A survey of 2450 golfers showed that 281 of them are left-handed. Construct a 90% confidence interval for the proportion of golfers that are left-handed. Question 10 options: (0.104, 0.125) (0.369, 0.451) (0.100, 0.130) (0.203, 0.293)
Example 5: Find a 90% confidence interval for the mean difference in the mileage obtained for tires with the new material and tires with the old material. Following are the paired differences (new minus old) for each of the 17 riders, chosen at random, who took part in the experiment: 50, 45, 50, 50, 100, 100, 100, -10, 75, -25, 0, 25, 75, 25, 50, 40, 35
Early-childhood-development studies indicate that the more often a child is read to from birth, the earlier the child begins to read. A local parents’ group wants to test this theory and samples families with young children. They find the following results. Construct a 99% confidence interval to estimate the true difference between the proportions of children who read at an early age when they are read to frequently compared to those who were read to less often, as described in the table of results. Let Population 1 be the children who were read to frequently and Population 2 be the children who were read to less often. Round the endpoints of the interval to three decimal places, if necessary. Ages when Children Begin to Read Read to at Least Three Times per Week Read to Fewer than Three Times per Week Started Reading by age 4 60 35 Started Reading after age 4 36 51 Lower end point: Upper end point:

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