48,972 50,100 50,020 51.560 49,900 49,800 52,193 11. Television Viewing by Teens

Question

48,972 50,100 50,020 51.560 49,900 49,800 52,193 11. Television Viewing by Teens Teens are reported to watch the fewest total hours of television per week of all the demographic groups. The average television viewing for teens on Sunday from 1:00 to 7:00 P.M. is 58 minutes. A random sample of local teens disclosed the following times for Sunday afternoon television viewing. At = 0.01, can it be concluded that the average is greater than the national viewing time? (Note: Change all times to minutes.) 2:30 1:00 1:30 2:00 2:15 2:30 1:30 1:50 3:20 2:10 Source: World Almanac. 12. Chocolate Chip Cookie Calories The average 1-ounce chocolate chip cookie contains 110 calories. A random sample of 15 different brands of 1-ounce chocolate chip cookies resulted in the following calorie amounts. At the = 0.01 level, is there sufficient evidence that the 100 125 150 160 185 125 155 145 160 100 150 140 135 120 110 Source: The Doctor's Pocket Calorie, Fat, and Carbohydrate Counter. 13. Sleep Time A person read that the average number of hours an adult sleeps on Friday night to Saturday morn- ing was 7.2 hours. The researcher feels that college stu- dents do not sleep 7.2 hours on average. The researcher randomly selected 15 students and found that on average they slept 8.3 hours. The standard deviation of the sample nouch evidence to say

Explanation / Answer

Solution11:

televison viewing by teens:

Null hypothesis:

H0:=58 minutes

Alternative Hypothesis

Ha:>58 minutes

alpha=0.01

perfom one tail t test in R to get t and p value.

Given sample times in hours are

150,120,90,200,60,135,110,150,90,150

Enter the below code in R:

viewingtime <- c( 150,120,90,200,60,135,110,150,90,150)  
t.test(viewingtime,mu=58,conf.level=0.99,alternative = "greater")

output:

One Sample t-test

data: viewingtime

t = 5.3138, df = 9, p-value = 0.0002425

alternative hypothesis: true mean is greater than 58

99 percent confidence interval:

89.65984 Inf

sample estimates:

mean of x

125.5

Intrepretation:

t=5.3138

p=0.0002425

p<alpha

p<0.01

Reject Null hypothesis.

there is suffiicient stattistical evidence at 1% level of signiifcance to conclude that average is greater than the

national viewing time.

Solution12:

chocolate chip cookie calories:

Null hypothesis:

H0:=110

Alternative Hypothesis

Ha:>110

alpha=0.01

Carry out one tail t test in R using below command:

t.test(chocolatechipcalories,mu=110,conf.level=0.99,alternative = "greater")

output is:

One Sample t-test

data: chocolatechipcalories

t = 4.3894, df = 14, p-value = 0.0003087

alternative hypothesis: true mean is greater than 110

99 percent confidence interval:

120.9901 Inf

sample estimates:

mean of x

137.3333

From following output:

t=4.3894

p=0.0003087

p<alpha

p<0.01

Reject Null hypothesis.

there is sufficient evidence at 1% level of significance to conclude that the average calorie content is greater than

110 calories.

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