STAT 200 Week 7 Homework STAT 200 Week 7 Homework

Week 7 HW

2. The formula for a regression equation is Y’ = 2X + 9.

a. What would be the predicted score for a person scoring 6 on X?

X = 2*6 + 9 = 21

b. If someone’s predicted score was 14, what was this person’s score on X?

Y’ = 2X + 9

14 = 2X + 9

X = (14-9)/2 = 2.5



a. r and determine if it is significantly different from zero.

r = 112 / √((98.833)(176)) = 0.8492

he P-Value is 0.032396. The result at .05 is significantly different than zero.

b. the slope of the regression line and test if it differs significantly from zero.

Slope(b) = (NΣXY – (ΣX)(ΣY)) / (NΣX2 – (ΣX)2)

Slope = 1.1332

The output suggests that the slope is 1.1332 and as the p-value for the t-test for slope is 0.032 which is smaller than the significance level of 0.05 thus the conclusion is that the slope differs significantly from zero.

c. the 95% confidence interval for the slope.

The 95% confidence interval for the slope is (1.070, 1.715).

5. At a school pep rally, a group of sophomore students organized a free raffle for prizes. They claim that they put the names of all of the students in the school in the basket and that they randomly drew 36 names out of this basket. Of the prize winners, 6 were freshmen, 14 were sophomores, 9 were juniors, and 7 were seniors. The results do not seem that random to you. You think it is a little fishy that sophomores organized the raffle and also won the most prizes. Your school is composed of 30% freshmen, 25% sophomores, 25% juniors, and 20% seniors.

a. What are the expected frequencies of winners from each class?

Class Winners (Observed) Proportion Expected Frequencies
Freshmen 6 .30 10.8
Sophomores 14 .25 9
Juniors 9 .25 9
Seniors 7 .2 7.2

b. Conduct a significance test to determine whether the winners of the prizes were distributed throughout the classes as would be expected based on the percentage of students in each group. Report your Chi Square and p values.

Chi squared equals 4.917 with 3 degrees of freedom.
The two-tailed P value equals 0.1780

c. What do you conclude?

The P value answers this question: A small P value is evidence that the data are not sampled from the distribution you expected.

14. A geologist collects hand-specimen sized pieces of limestone from a particular area. A qualitative assessment of both texture and color is made with the following results:

Is there evidence of association between color and texture forthese limestones? Explain your answer.

Chi-Square = 17.727, DF = 4, P-Value = 0.0014

Level of significance = 5% = 0.05

Due to the p-value of 0.0014being smaller than the significance level of 0.05, we must conclude that the null hypothesis is rejected. In conclusion that there is evidence of association between color and texture for these limestone’s.

70. TRUE OR FALSE?The standard deviation of the chi-square distribution is twice the mean.


102. Do men and women select different breakfasts? The breakfasts ordered by randomly selected men and women at a popular breakfast place is shown in Table 11.55. Conduct a test for homogeneity at a 5% level of significance.

Chi-Square = 4.013, DF = 3, P-Value = 0.260

Level of significance = 5% = 0.05

The p-value is 0.260 larger than the significance level of 0.05 concluding that men and women do not select significantly different breakfasts.

Use the following information to answer the next two exercises: Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes.

113. df= (25-1) (2-1) = 24

117. Let α = 0.05

Decision: Do not reject the null hypothesis since a

Conclusion: The data does not provide enough evidence to reject the claim that the variance is no more than 150 minutes.

66. Can a coefficient of determination be negative? Why or why not?

No, a squared variable can only be positive.


The cost of a leading liquid laundry detergent in different sizes is given in Table 12.31.

Size (Ounces) (X) Cost ($) (Y) Cost per Ounce
16 3.99 .2494
32 4.99 .1559
64 5.99 .0936
200 10.99 .0550

b. Does it appear from inspection that there is a relationship between the variables? Why or why not?

  • Using “size” as the independent variable and “cost” as the dependent variable, draw a scatter plot.

Yes, there is a positive linear relationship between size and cost.

c.Calculate the least-squares line. Put in the form of: ŷ = a + bx

Slope : 0.037 Y-Intercept: 3.598

The output suggests that the regression equation is,

ŷ= 3.598 + 0.03707x

d. Find the correlation coefficient. Is it significant?

t = .999 * (Sqrt 4-2 / Sqrt 1 – 0.9992)

t = 1.413/0.0447

t = 31.60

Critical region = 4.303

Since the p-value is greater than the critical region, 5% level of significance is rejected.

e. If the laundry detergent were sold in a 40-ounce size, find the estimated cost.

Y = a+bx

Y = 3.598 + 0.037*40

Estimated cost = $5.078

f. If the laundry detergent were sold in a 90-ounce size, find the estimated cost.

Y = 3.598 + 0.037*90

Estimated cost = $6.928

g. Does it appear that a line is the best way to fit the data? Why or why not?

Yes because of the linear relationship between Y and the variable X. The scatter plot shows a linear trend.

h. Are there any outliers in the given data?

No outliers is found in the given data.

i.Is the least-squares line valid for predicting what a 300-ounce size of the laundry detergent would you cost? Why or why not?

y = 3.598+0.037*300


j. What is the slope of the least-squares (best-fit) line? Interpret the slope.

The slope is 0.03707 implying per 1 ounce increase in the size increases the cost by $0.03707 on an average.

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