A recent national survey found that high school students watched an average (mean) of 6.8 videos per month. A random sample of 36 college students revealed that the mean number of videos watched last month was 6.2, with a standard deviation of 0.5. At the .05 significance level, can we conclude (test) that college students watch fewer videos a month than high school students?
HS-High School Students
It is a one tail test because the key word is that college students watch fewer videos a month than high school students.
Step 1: State the hypothesis Ho: CS≥6.8 (High School Students)
Step 2: Significance Level α=.05
Step 3: Test Statistic Z-Test because n>30
Step 4: Since one tailed test .5-.05=.45 & the closest number on the Z table is .4505
Which yields a Z= 1.65. Will also be on the negative side of the number line.
Step 5: Calculate Z test result
z= Xbar-μ = 6.2-6.8 = - .6/.0833= -7.202
Reject Ho | Accept Ho
z ---------- (-7.202) ---------- (-1.65) ---------0--------- (1.65) ------ We Reject Ho
Problem # 2
A recent national survey found that high school students watched an average (mean) of 5 videos per month. A random sample of 48 college students revealed that the mean number of videos watched last month was 5.7, with a standard deviation of 0.4. At the .05 significance level, can we conclude (test) that college students watch fewer videos a month than high school students?
Use the template I followed in the first problem and show all the problem information and all 5 steps of the hypothesis testing process. Show a number line along with your final decision, please.
Problem A. At the time she was hired as a server at the Grumney Family Restaurant, Beth Brigden was told,
"You can average more than $20 a day in tips." Over the first 35 days she was employed at the restaurant, the mean daily amount of her tips was $24.85, with a standard deviation of $3.24. At the .05 significance level, can Ms. Brigden conclude that she is earning an average of more than $20 in tips? Show all 5 steps
Problem B. According to the local union president, the mean gross income of plumbers in the Salt Lake City area is normally distributed, with a mean of $30,000 and a standard deviation of $3,000. A recent investigative reporter for KYAK TV found, for a sample of 18 plumbers, the mean gross income was $30,500. At the .10 significance level, is it reasonable to conclude that the mean income is not equal to $30,000? Show all 5 steps.
Problem C. Tina Dennis is the comptroller for Meek Industries. She believes that the current cash-flow problem at Meek is due to the slow collection of accounts receivable. She believes that more than 60 percent of the accounts are in arrears more than three months. A random sample of 200 accounts showed that 140 were more than three months old. At the .05 significance level, can she conclude that more than 60 percent of the accounts are in arrears for more than three months?
Problem 1. What is the cutoff/critical F value for an ANOVA problem where the degrees of freedom in the numerator are equal to 6 and the degrees of freedom for the denominator are equal to 15. Use the 0.01 significance level, and then use the 1 percent F table.
Problem 2. For the three treatments shown below, conduct an ANOVA Test. Use the 0.01 significance level. Use Excel and show all 5 steps of the hypothesis test. If you do not have the Excel Data Analysis tool, use F=21.9 for this problem. If you have the tool, show me the work, and of course it should match F=21.9.
treatment 1 treatment 2 treatment 3
8 3 3
6 2 4
10 4 5
9 3 4
X: 4 5 3 6 10
Y: 4 6 5 7 7
Problem 4. Determine the regression formal for this data. USE .05
X: 5 5 4 3 6 10 11
Y: 5 6 5 7 7 9 12
Hypothesis testing is investigated. The solution is detailed and well presented. The response received a rating of "5/5" from the student who originally posted the question.